6.1 Angle Measure 6.2 Trigonometry of Right Triangles 6.3 Trigonometric Functions of Angles 6.4 The Law of Sines 6.5 The Law of Cosines

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1 6.1 ngle Mesure 6. Trigonometr of Right Tringles 6. Trigonometric Functions of ngles 6.4 The Lw of Sines 6.5 The Lw of osines hpter Overview The trigonometric functions cn be defined in two different but equivlent ws s functions of rel numbers (hpter 5) or s functions of ngles (hpter 6). The two pproches to trigonometr re independent of ech other, so either hpter 5 or hpter 6 m be studied first. We stud both pproches becuse different pplictions require tht we view these functions differentl. The pproch in this chpter lends itself to geometric problems involving finding ngles nd distnces. Suppose we wnt to find the distnce to the sun. Using tpe mesure is of course imprcticl, so we need something besides simple mesurement to tckle this problem. ngles re es to mesure for emple, we cn find the ngle formed b the sun, erth, nd moon b simpl pointing to the sun with one rm nd the moon with the other nd estimting the ngle between them. The ke ide then is to find reltionship between ngles nd distnces. So if we hd w to determine distnces from ngles, we d be ble to find the distnce to the sun without going there. The trigonometric functions provide us with just the tools we need. If is right tringle with cute ngle u s in the figure, then we define sin u to be the rtio /r. Tringle is similr to tringle, so r r lthough the distnces nd r re different from nd r, the given rtio is the sme. Thus, in n right tringle with cute ngle u, the rtio of the side opposite ngle u to the hpotenuse is the sme nd is clled sin u. The other trigonometric rtios re defined in similr fshion. ' Gregor D. Dimijin M.D. r ' In this chpter we lern how trigonometric functions cn be used to mesure distnces on the erth nd in spce. In Eercises 61 nd 6 on pge 487, we ctull de- r' ' ' ' 467

2 468 HPTER 6 Trigonometric Functions of ngles termine the distnce to the sun using trigonometr. Right tringle trigonometr hs mn other pplictions, from determining the optiml cell structure in beehive (Eercise 67, pge 497) to eplining the shpe of rinbow (Eercise 69, pge 498). In the Focus on Modeling, pges 5 5, we see how surveor uses trigonometr to mp town. 6.1 ngle Mesure n ngle O consists of two rs R 1 nd R with common verte O (see Figure 1). We often interpret n ngle s rottion of the r R 1 onto R. In this cse, R 1 is clled the initil side, nd R is clled the terminl side of the ngle. If the rottion is counterclockwise, the ngle is considered positive, nd if the rottion is clockwise, the ngle is considered negtive. terminl side O R initil side Positive ngle R R O initil side terminl side Negtive ngle R Figure 1 ngle Mesure The mesure of n ngle is the mount of rottion bout the verte required to move R 1 onto R. Intuitivel, this is how much the ngle opens. One unit of mesurement for ngles is the degree. n ngle of mesure 1 degree is formed b rotting the initil side 60 of complete revolution. In clculus nd other brnches of mthemtics, 1 more nturl method of mesuring ngles is used rdin mesure. The mount n ngle opens is mesured long the rc of circle of rdius 1 with its center t the verte of the ngle. 1 Rdin mesure of Definition of Rdin Mesure If circle of rdius 1 is drwn with the verte of n ngle t its center, then the mesure of this ngle in rdins (bbrevited rd) is the length of the rc tht subtends the ngle (see Figure ). Figure The circumference of the circle of rdius 1 is p nd so complete revolution hs mesure p rd, stright ngle hs mesure p rd, nd right ngle hs mesure

3 SETION 6.1 ngle Mesure 469 p/ rd. n ngle tht is subtended b n rc of length long the unit circle hs rdin mesure (see Figure ). 1 rd π rd π rd 1 rd 1 Figure Rdin mesure O 1 O 1 O 1 O 1 Since complete revolution mesured in degrees is 60 nd mesured in rdins is p rd, we get the following simple reltionship between these two methods of ngle mesurement. Reltionship between Degrees nd Rdins 180 p rd 1 rd 180 p b p 1. To convert degrees to rdins, multipl b p 180 rd To convert rdins to degrees, multipl b. p Mesure of =1 rd Mesure of Å57.96* Figure 4 To get some ide of the size of rdin, notice tht 1 rd nd rd n ngle u of mesure 1 rd is shown in Figure 4. Emple 1 onverting between Rdins nd Degrees p () Epress 60 in rdins. (b) Epress rd in degrees. 6 Solution The reltionship between degrees nd rdins gives p p () (b) 6 rd p 6 b180 b b rd p rd p note on terminolog: We often use phrse such s 0 ngle to men n ngle whose mesure is 0. lso, for n ngle u, we write u 0 or u p/6 to men the mesure of u is 0 or p/6 rd. When no unit is given, the ngle is ssumed to be mesured in rdins.

4 470 HPTER 6 Trigonometric Functions of ngles ngles in Stndrd Position n ngle is in stndrd position if it is drwn in the -plne with its verte t the origin nd its initil side on the positive -is. Figure 5 gives emples of ngles in stndrd position () (b) (c) (d) Figure 5 ngles in stndrd position Two ngles in stndrd position re coterminl if their sides coincide. In Figure 5 the ngles in () nd (c) re coterminl. Emple oterminl ngles () Find ngles tht re coterminl with the ngle u 0 in stndrd position. (b) Find ngles tht re coterminl with the ngle u p in stndrd position. Solution () To find positive ngles tht re coterminl with u, we dd n multiple of 60. Thus nd re coterminl with u 0. To find negtive ngles tht re coterminl with u, we subtrct n multiple of 60. Thus nd re coterminl with u. (See Figure 6.) _0* 0* * 0 Figure 6 (b) To find positive ngles tht re coterminl with u, we dd n multiple of p. Thus p p 7p nd p 1p 4p

5 SETION 6.1 ngle Mesure 471 re coterminl with u p/. To find negtive ngles tht re coterminl with u, we subtrct n multiple of p. Thus p p 5p nd re coterminl with u. (See Figure 7.) p 4p 11p Figure 7 0 π 0 7π 5π _ 0 Emple oterminl ngles Find n ngle with mesure between 0 nd 60 tht is coterminl with the ngle of mesure 190 in stndrd position. Solution We cn subtrct 60 s mn times s we wish from 190, nd the resulting ngle will be coterminl with 190. Thus, is coterminl with 190, nd so is the ngle 190 (60) 570. To find the ngle we wnt between 0 nd 60, we subtrct 60 from 190 s mn times s necessr. n efficient w to do this is to determine how mn times 60 goes into 190, tht is, divide 190 b 60, nd the reminder will be the ngle we re looking for. We see tht 60 goes into 190 three times with reminder of 10. Thus, 10 is the desired ngle (see Figure 8). 190* 0 10* 0 Figure 8 Length of irculr rc r s n ngle whose rdin mesure is u is subtended b n rc tht is the frction u/ 1p of the circumference of circle. Thus, in circle of rdius r, the length s of n rc tht subtends the ngle u (see Figure 9) is s u circumference of circle p Figure 9 s ur u 1pr ur p

6 47 HPTER 6 Trigonometric Functions of ngles Length of irculr rc In circle of rdius r, the length s of n rc tht subtends centrl ngle of u rdins is s ru Solving for u, we get the importnt formul u s r This formul llows us to define rdin mesure using circle of n rdius r: The rdin mesure of n ngle u is s/r, where s is the length of the circulr rc tht subtends u in circle of rdius r (see Figure 10). r Figure 10 The rdin mesure of u is the number of rdiuses tht cn fit in the rc tht subtends u; hence the term rdin. 1 rd r r rd r r Emple 4 rc Length nd ngle Mesure () Find the length of n rc of circle with rdius 10 m tht subtends centrl ngle of 0. (b) centrl ngle u in circle of rdius 4 m is subtended b n rc of length 6 m. Find the mesure of u in rdins. Solution () From Emple 1(b) we see tht 0 p/6 rd. So the length of the rc is The formul s ru is true onl when u is mesured in rdins. (b) the formul u s/r, we hve s ru 110 p 6 5p m u s r 6 4 rd Figure 11 1 r u r re of irculr Sector The re of circle of rdius r is pr. sector of this circle with centrl ngle u hs n re tht is the frction u/1p of the re of the entire circle (see Figure 11). So the re of this sector is u re of circle p u p 1pr 1 r u

7 SETION 6.1 ngle Mesure 47 re of irculr Sector In circle of rdius r, the re of sector with centrl ngle of u rdins is 1 r u The formul 1 r u is true onl when u is mesured in rdins. Emple 5 re of Sector Find the re of sector of circle with centrl ngle 60 if the rdius of the circle is m. Solution To use the formul for the re of circulr sector, we must find the centrl ngle of the sector in rdins: p/180 rd p/ rd. Thus, the re of the sector is 1 r u 1 1 p b p m s r Figure 1 irculr Motion Suppose point moves long circle s shown in Figure 1. There re two ws to describe the motion of the point liner speed nd ngulr speed. Liner speed is the rte t which the distnce trveled is chnging, so liner speed is the distnce trveled divided b the time elpsed. ngulr speed is the rte t which the centrl ngle u is chnging, so ngulr speed is the number of rdins this ngle chnges divided b the time elpsed. Liner Speed nd ngulr Speed Suppose point moves long circle of rdius r nd the r from the center of the circle to the point trverses u rdins in time t. Let s ru be the distnce the point trvels in time t. Then the speed of the object is given b The smbol v is the Greek letter omeg. ngulr speed Liner speed v u t s t Emple 6 Finding Liner nd ngulr Speed bo rottes stone in -ft-long sling t the rte of 15 revolutions ever 10 seconds. Find the ngulr nd liner velocities of the stone. Solution In 10 s, the ngle u chnges b 15 p 0p rdins. So the ngulr speed of the stone is v u 0p rd p rd/s t 10 s

8 474 HPTER 6 Trigonometric Functions of ngles The distnce trveled b the stone in 10 s is s 15 pr 15 p 90p ft. So the liner speed of the stone is s 90p ft 9p ft/s t 10 s Notice tht ngulr speed does not depend on the rdius of the circle, but onl on the ngle u. However, if we know the ngulr speed v nd the rdius r, we cn find liner speed s follows: s/t ru/t r1u/t rv. Reltionship between Liner nd ngulr Speed If point moves long circle of rdius r with ngulr speed v, then its liner speed is given b rv Emple 7 Finding Liner Speed from ngulr Speed womn is riding biccle whose wheels re 6 inches in dimeter. If the wheels rotte t 15 revolutions per minute (rpm), find the speed t which she is trveling, in mi/h. Solution The ngulr speed of the wheels is p 15 50p rd/min. Since the wheels hve rdius 1 in. (hlf the dimeter), the liner speed is rv 1 # 50p 10,10. in./min Since there re 1 inches per foot, 580 feet per mile, nd 60 minutes per hour, her speed in miles per hour is 10,10. in./min 60 min/h 1 in./ft 580 ft/mi 61,61 in./h 6,60 in./mi 9.7 mi/h 6.1 Eercises 1 1 Find the rdin mesure of the ngle with the given degree mesure Find the degree mesure of the ngle with the given rdin mesure. 7p 11p p p p p.. p 4. 1p The mesure of n ngle in stndrd position is given. Find two positive ngles nd two negtive ngles tht re coterminl with the given ngle. p

9 SETION 6.1 ngle Mesure p p The mesures of two ngles in stndrd position re given. Determine whether the ngles re coterminl , 40. 0, 0 5p p. 4., 11p 6, 17p , , Find n ngle between 0 nd 60 tht is coterminl with the given ngle Find n ngle between 0 nd p tht is coterminl with the given ngle. 17p p p 6 17p 51p Find the length of the rc s s in the figure. 5. Find the length of n rc tht subtends centrl ngle of 45 in circle of rdius 10 m. 5. Find the length of n rc tht subtends centrl ngle of rd in circle of rdius mi. 54. centrl ngle u in circle of rdius 5 m is subtended b n rc of length 6 m. Find the mesure of u in degrees nd in rdins. 55. n rc of length 100 m subtends centrl ngle u in circle of rdius 50 m. Find the mesure of u in degrees nd in rdins. 56. circulr rc of length ft subtends centrl ngle of 5. Find the rdius of the circle. 57. Find the rdius of the circle if n rc of length 6 m on the circle subtends centrl ngle of p/6 rd. 58. Find the rdius of the circle if n rc of length 4 ft on the circle subtends centrl ngle of Find the re of the sector shown in ech figure. () (b) 0.5 rd 80* Find the rdius of ech circle if the re of the sector is 1. () (b) 140* rd 150* 50. Find the ngle u in the figure. 51. Find the rdius r of the circle in the figure rd r 61. Find the re of sector with centrl ngle 1 rd in circle of rdius 10 m. 6. sector of circle hs centrl ngle of 60. Find the re of the sector if the rdius of the circle is mi. 6. The re of sector of circle with centrl ngle of rd is 16 m. Find the rdius of the circle. 64. sector of circle of rdius 4 mi hs n re of 88 mi. Find the centrl ngle of the sector. 65. The re of circle is 7 cm. Find the re of sector of this circle tht subtends centrl ngle of p/6 rd. 66. Three circles with rdii 1,, nd ft re eternll tngent to one nother, s shown in the figure on the net pge. Find the re of the sector of the circle of rdius 1 tht is cut off

10 476 HPTER 6 Trigonometric Functions of ngles b the line segments joining the center of tht circle to the centers of the other two circles. of the erth from the following observtions. He noticed tht on certin d the sun shone directl down deep well in Sene (modern swn). t the sme time in lendri, 500 miles north (on the sme meridin), the rs of the sun shone t n ngle of 7. to the zenith. Use this informtion nd the figure to find the rdius nd circumference of the erth. 500 mi lendri 7.* Rs of sun pplictions 67. Trvel Distnce cr s wheels re 8 in. in dimeter. How fr (in miles) will the cr trvel if its wheels revolve 10,000 times without slipping? 68. Wheel Revolutions How mn revolutions will cr wheel of dimeter 0 in. mke s the cr trvels distnce of one mile? 69. Ltitudes Pittsburgh, Pennslvni, nd Mimi, Florid, lie pproimtel on the sme meridin. Pittsburgh hs ltitude of 40.5 N nd Mimi, 5.5 N. Find the distnce between these two cities. (The rdius of the erth is 960 mi.) Pittsburgh Mimi 70. Ltitudes Memphis, Tennessee, nd New Orlens, Louisin, lie pproimtel on the sme meridin. Memphis hs ltitude 5 N nd New Orlens, 0 N. Find the distnce between these two cities. (The rdius of the erth is 960 mi.) 71. Orbit of the Erth Find the distnce tht the erth trvels in one d in its pth round the sun. ssume tht er hs 65 ds nd tht the pth of the erth round the sun is circle of rdius 9 million miles. [The pth of the erth round the sun is ctull n ellipse with the sun t one focus (see Section 10.). This ellipse, however, hs ver smll eccentricit, so it is nerl circulr.] 7. Nuticl Miles Find the distnce long n rc on the surfce of the erth tht subtends centrl ngle of 1 minute 11 minute 1 60 degree. This distnce is clled nuticl mile. (The rdius of the erth is 960 mi.) 74. Irrigtion n irrigtion sstem uses stright sprinkler pipe 00 ft long tht pivots round centrl point s shown. Due to n obstcle the pipe is llowed to pivot through 80 onl. Find the re irrigted b this sstem. 00 ft 80* Sene 75. Windshield Wipers The top nd bottom ends of windshield wiper blde re 4 in. nd 14 in. from the pivot point, respectivel. While in opertion the wiper sweeps through 15. Find the re swept b the blde. sun 7. ircumference of the Erth The Greek mthemticin Ertosthenes (c ) mesured the circumference 14 in. 15* 4 in.

11 SETION 6.1 ngle Mesure The Tethered ow cow is tethered b 100-ft rope to the inside corner of n L-shped building, s shown in the figure. Find the re tht the cow cn grze. 8. Speed of urrent To mesure the speed of current, scientists plce pddle wheel in the strem nd observe the rte t which it rottes. If the pddle wheel hs rdius 0.0 m nd rottes t 100 rpm, find the speed of the current in m/s. 0 ft 50 ft 100 ft 60 ft 50 ft 77. Winch winch of rdius ft is used to lift hev lods. If the winch mkes 8 revolutions ever 15 s, find the speed t which the lod is rising. 84. iccle Wheel The sprockets nd chin of biccle re shown in the figure. The pedl sprocket hs rdius of 4 in., the wheel sprocket rdius of in., nd the wheel rdius of 1 in. The cclist pedls t 40 rpm. () Find the ngulr speed of the wheel sprocket. (b) Find the speed of the biccle. (ssume tht the wheel turns t the sme rte s the wheel sprocket.) 1 in. in. 4 in. 78. Fn ceiling fn with 16-in. bldes rottes t 45 rpm. () Find the ngulr speed of the fn in rd/min. (b) Find the liner speed of the tips of the bldes in in./min. 79. Rdil Sw rdil sw hs blde with 6-in. rdius. Suppose tht the blde spins t 1000 rpm. () Find the ngulr speed of the blde in rd/min. (b) Find the liner speed of the swteeth in ft/s. 80. Speed t Equtor The erth rottes bout its is once ever h 56 min 4 s, nd the rdius of the erth is 960 mi. Find the liner speed of point on the equtor in mi/h. 81. Speed of r The wheels of cr hve rdius 11 in. nd re rotting t 600 rpm. Find the speed of the cr in mi/h. 8. Truck Wheels truck with 48-in.-dimeter wheels is trveling t 50 mi/h. () Find the ngulr speed of the wheels in rd/min. (b) How mn revolutions per minute do the wheels mke? 85. onicl up conicl cup is mde from circulr piece of pper with rdius 6 cm b cutting out sector nd joining the edges s shown. Suppose u 5p/. () Find the circumference of the opening of the cup. (b) Find the rdius r of the opening of the cup. [Hint: Use pr.] (c) Find the height h of the cup. [Hint: Use the Pthgoren Theorem.] (d) Find the volume of the cup. 6 cm 6 cm h r 6 cm

12 478 HPTER 6 Trigonometric Functions of ngles 86. onicl up In this eercise we find the volume of the conicl cup in Eercise 85 for n ngle u. () Follow the steps in Eercise 85 to show tht the volume of the cup s function of u is V1u 9 p u 4p u, 0 u p (b) Grph the function V. (c) For wht ngle u is the volume of the cup mimum? Discover Discussion 87. Different Ws of Mesuring ngles The custom of mesuring ngles using degrees, with 60 in circle, dtes bck to the ncient blonins, who used number sstem bsed on groups of 60. nother sstem of mesuring ngles divides the circle into 400 units, clled grds. In this sstem right ngle is 100 grd, so this fits in with our bse 10 number sstem. Write short ess compring the dvntges nd disdvntges of these two sstems nd the rdin sstem of mesuring ngles. Which sstem do ou prefer? 88. locks nd ngles In one hour, the minute hnd on clock moves through complete circle, nd the hour hnd 1 moves through 1 of circle. Through how mn rdins do the minute nd the hour hnd move between 1:00 P.M. nd 6:45 P.M. (on the sme d)? Trigonometr of Right Tringles In this section we stud certin rtios of the sides of right tringles, clled trigonometric rtios, nd give severl pplictions. Trigonometric Rtios onsider right tringle with u s one of its cute ngles. The trigonometric rtios re defined s follows (see Figure 1). The Trigonometric Rtios hpotenuse opposite sin u opposite hpotenuse cos u djcent hpotenuse tn u opposite djcent djcent csc u hpotenuse opposite sec u hpotenuse djcent cot u djcent opposite Figure 1 The smbols we use for these rtios re bbrevitions for their full nmes: sine, cosine, tngent, cosecnt, secnt, cotngent. Since n two right tringles with

13 SETION 6. Trigonometr of Right Tringles 479 Hipprchus (circ 140..) is considered the founder of trigonometr. He constructed tbles for function closel relted to the modern sine function nd evluted for ngles t hlf-degree intervls. These re considered the first trigonometric tbles. He used his tbles minl to clculte the pths of the plnets through the hevens. ngle u re similr, these rtios re the sme, regrdless of the size of the tringle; the trigonometric rtios depend onl on the ngle u (see Figure ) Figure ß = 5 4 ç = 5 0 ß = = ç = = 50 5 œ 5 Figure Emple 1 Finding Trigonometric Rtios Find the si trigonometric rtios of the ngle u in Figure. Solution sin u csc u cos u 15 sec u 15 tn u 15 cot u 15 4 å Figure 4 œ 7 Emple Finding Trigonometric Rtios If cos 4, sketch right tringle with cute ngle, nd find the other five trigonometric rtios of. Solution Since cos is defined s the rtio of the djcent side to the hpotenuse, we sketch tringle with hpotenuse of length 4 nd side of length djcent to. If the opposite side is, then b the Pthgoren Theorem, 4 or 7, so 17. We then use the tringle in Figure 4 to find the rtios. sin 17 4 csc 4 17 Specil Tringles cos 4 sec 4 tn 17 cot 17 ertin right tringles hve rtios tht cn be clculted esil from the Pthgoren Theorem. Since the re used frequentl, we mention them here. The first tringle is obtined b drwing digonl in squre of side 1 (see Figure 5 on pge 480). the Pthgoren Theorem this digonl hs length 1. The

14 480 HPTER 6 Trigonometric Functions of ngles ristrchus of Smos (10 0..) ws fmous Greek scientist, musicin, stronomer, nd geometer. In his book On the Sizes nd Distnces of the Sun nd the Moon, he estimted the distnce to the sun b observing tht when the moon is ectl hlf full, the tringle formed b the sun, moon, nd the erth hs right ngle t the moon. His method ws similr to the one described in Eercise 61 in this section. ristrchus ws the first to dvnce the theor tht the erth nd plnets move round the sun, n ide tht did not gin full cceptnce until fter the time of opernicus, 1800 ers lter. For this reson he is often clled the opernicus of ntiquit. resulting tringle hs ngles 45,45, nd 90 (or p/4, p/4, nd p/). To get the second tringle, we strt with n equilterl tringle of side nd drw the perpendiculr bisector D of the bse, s in Figure 6. the Pthgoren Theorem the length of D is 1. Since D bisects ngle, we obtin tringle with ngles 0, 60, nd 90 (or p/6, p/, nd p/). œ 45* 1 Figure 5 45* 1 60* 1 Figure 6 0* œ We cn now use the specil tringles in Figures 5 nd 6 to clculte the trigonometric rtios for ngles with mesures 0,45, nd 60 (or p/6, p/4, nd p/). These re listed in Tble 1. D Tble 1 Vlues of the trigonometric rtios for specil ngles u in degrees u in rdins sin u cos u tn u csc u sec u cot u p p p For n eplntion of numericl methods, see the mrgin note on pge 46. It s useful to remember these specil trigonometric rtios becuse the occur often. Of course, the cn be reclled esil if we remember the tringles from which the re obtined. To find the vlues of the trigonometric rtios for other ngles, we use clcultor. Mthemticl methods (clled numericl methods) used in finding the trigonometric rtios re progrmmed directl into scientific clcultors. For instnce, when the SIN ke is pressed, the clcultor computes n pproimtion to the vlue of the sine of the given ngle. lcultors give the vlues of sine, cosine, nd tngent; the other rtios cn be esil clculted from these using the following reciprocl reltions: csc t 1 sin t sec t 1 cos t cot t 1 tn t You should check tht these reltions follow immeditel from the definitions of the trigonometric rtios. We follow the convention tht when we write sin t, we men the sine of the ngle whose rdin mesure is t. For instnce, sin 1 mens the sine of the ngle whose r-

15 SETION 6. Trigonometr of Right Tringles din mesure is 1. When using clcultor to find n pproimte vlue for this number, set our clcultor to rdin mode; ou will find tht If ou wnt to find the sine of the ngle whose mesure is 1, set our clcultor to degree mode; ou will find tht Emple Using lcultor to Find Trigonometric Rtios With our clcultor in degree mode, nd writing the results correct to five deciml plces, we find With our clcultor in rdin mode, nd writing the results correct to five deciml plces, we find pplictions of Trigonometr of Right Tringles tringle hs si prts: three ngles nd three sides. To solve tringle mens to determine ll of its prts from the informtion known bout the tringle, tht is, to determine the lengths of the three sides nd the mesures of the three ngles. Emple 4 cos Solving Right Tringle Solve tringle, shown in Figure 7. sin sin sin sec cos 88 cot tn 1.54 Solution It s cler tht 60. To find, we look for n eqution tht reltes to the lengths nd ngles we lred know. In this cse, we hve sin 0 /1, so 0* b Similrl, cos 0 b/1, so 1 sin Figure 7 b 1 cos b 61 It s ver useful to know tht, using the informtion given in Figure 8, the lengths of the legs of right tringle re Figure 8 r sin u b r cos u r b r sin u nd b r cos u The bilit to solve right tringles using the trigonometric rtios is fundmentl to mn problems in nvigtion, surveing, stronom, nd the mesurement of distnces. The pplictions we consider in this section lws involve right tringles but, s we will see in the net three sections, trigonometr is lso useful in solving tringles tht re not right tringles. To discuss the net emples, we need some terminolog. If n observer is looking t n object, then the line from the ee of the observer to the object is clled

16 48 HPTER 6 Trigonometric Functions of ngles Thles of Miletus (circ ) is the legendr founder of Greek geometr. It is sid tht he clculted the height of Greek column b compring the length of the shdow of his stff with tht of the column. Using properties of similr tringles, he rgued tht the rtio of the height h of the column to the height h of his stff ws equl to the rtio of the length s of the column s shdow to the length s of the stff s shdow: h h s s Since three of these quntities re known, Thles ws ble to clculte the height of the column. ccording to legend, Thles used similr method to find the height of the Gret Prmid in Egpt, fet tht impressed Egpt s king. Plutrch wrote tht lthough he [the king of Egpt] dmired ou [Thles] for other things, et he prticulrl liked the mnner b which ou mesured the height of the prmid without n trouble or instrument. The principle Thles used, the fct tht rtios of corresponding sides of similr tringles re equl, is the foundtion of the subject of trigonometr. the line of sight (Figure 9). If the object being observed is bove the horizontl, then the ngle between the line of sight nd the horizontl is clled the ngle of elevtion. If the object is below the horizontl, then the ngle between the line of sight nd the horizontl is clled the ngle of depression. In mn of the emples nd eercises in this chpter, ngles of elevtion nd depression will be given for hpotheticl observer t ground level. If the line of sight follows phsicl object, such s n inclined plne or hillside, we use the term ngle of inclintion. Figure 9 The net emple gives n importnt ppliction of trigonometr to the problem of mesurement: We mesure the height of tll tree without hving to climb it! lthough the emple is simple, the result is fundmentl to understnding how the trigonometric rtios re pplied to such problems. Emple 5 Finding the Height of Tree gint redwood tree csts shdow 5 ft long. Find the height of the tree if the ngle of elevtion of the sun is 5.7. Solution Line of sight ngle of elevtion Horizontl Let the height of the tree be h. From Figure 10 we see tht h tn h 5 tn Line of sight Definition of tngent Multipl b 5 Use clcultor ngle of depression Horizontl Therefore, the height of the tree is bout 56 ft. h Figure * 5 ft

17 SETION 6. Trigonometr of Right Tringles 48 Emple 6 Problem Involving Right Tringles From point on the ground 500 ft from the bse of building, n observer finds tht the ngle of elevtion to the top of the building is 4 nd tht the ngle of elevtion to the top of flgpole top the building is 7. Find the height of the building nd the length of the flgpole. 4* 7* 500 ft Figure 11 The ke lbels SIN 1 or INV SIN stnd for inverse sine. We stud the inverse trigonometric functions in Section 7.4. h k Solution Figure 11 illustrtes the sitution. The height of the building is found in the sme w tht we found the height of the tree in Emple 5. h tn 4 Definition of tngent 500 h 500 tn 4 Multipl b Use clcultor The height of the building is pproimtel ft. To find the length of the flgpole, let s first find the height from the ground to the top of the pole: k tn k 500 tn To find the length of the flgpole, we subtrct h from k. So the length of the pole is pproimtel 55 ft. In some problems we need to find n ngle in right tringle whose sides re given. To do this, we use Tble 1 (pge 480) bckwrd ; tht is, we find the ngle with the specified trigonometric rtio. For emple, if sin u 1, wht is the ngle u? From Tble 1 we cn tell tht u 0. To find n ngle whose sine is not given in the tble, we use the SIN 1 or INV SIN or RSIN kes on clcultor. For emple, if sin u 0.8, we ppl the SIN 1 ke to 0.8 to get u 5.1 or 0.97 rd. The clcultor lso gives ngles whose cosine or tngent re known, using the OS 1 or TN 1 ke. 40 ft 6 ft Figure 1 Emple 7 Solving for n ngle in Right Tringle 40-ft ldder lens ginst building. If the bse of the ldder is 6 ft from the bse of the building, wht is the ngle formed b the ldder nd the building? Solution First we sketch digrm s in Figure 1. If u is the ngle between the ldder nd the building, then sin u So u is the ngle whose sine is To find the ngle u, we use the clcultor. With our clcultor in degree mode, we get u 8.6 SIN 1 ke on

18 484 HPTER 6 Trigonometric Functions of ngles 6. Eercises 1 6 Find the ect vlues of the si trigonometric rtios of the ngle u in the tringle * 5 6* Epress nd in terms of trigonometric rtios of u Sketch tringle tht hs cute ngle u, nd find the other five trigonometric rtios of u. 17. sin u 18. cos u cot u 1 0. tn u Find () sin nd cos b, (b) tn nd cot b, nd (c) sec nd csc b å 5 å sec u 7. 8 Evlute the epression without using clcultor.. sin p 6 cos p 6 4. sin 0 csc 0 5. sin 0 cos 60 sin 60 cos 0 csc u Find the side lbeled. In Eercises 1 nd 14 stte our nswer correct to five deciml plces * 1 45* sin 60 1cos 60 1cos 0 1sin 0 sin p cos p 4 sin p 4 cos p b Solve the right tringle * 0* 45* 16 75* 100

19 SETION 6. Trigonometr of Right Tringles * π π π π * 65* 4. Epress the length in terms of the trigonometric rtios of u. 10 0* 5 7. Use ruler to crefull mesure the sides of the tringle, nd then use our mesurements to estimte the si trigonometric rtios of u. 44. Epress the length, b, c, nd d in the figure in terms of the trigonometric rtios of u. 8. Using protrctor, sketch right tringle tht hs the cute ngle 40. Mesure the sides crefull, nd use our results to estimte the si trigonometric rtios of Find correct to one deciml plce. 9. d c b * 0* 40. pplictions 85 60* 0* 45. Height of uilding The ngle of elevtion to the top of the Empire Stte uilding in New York is found to be 11 from the ground t distnce of 1 mi from the bse of the building. Using this informtion, find the height of the Empire Stte uilding.

20 486 HPTER 6 Trigonometric Functions of ngles 46. Gtew rch plne is fling within sight of the Gtew rch in St. Louis, Missouri, t n elevtion of 5,000 ft. The pilot would like to estimte her distnce from the Gtew rch. She finds tht the ngle of depression to point on the ground below the rch is. () Wht is the distnce between the plne nd the rch? (b) Wht is the distnce between point on the ground directl below the plne nd the rch? 47. Devition of Lser em lser bem is to be directed towrd the center of the moon, but the bem strs 0.5 from its intended pth. () How fr hs the bem diverged from its ssigned trget when it reches the moon? (The distnce from the erth to the moon is 40,000 mi.) (b) The rdius of the moon is bout 1000 mi. Will the bem strike the moon? 48. Distnce t Se From the top of 00-ft lighthouse, the ngle of depression to ship in the ocen is. How fr is the ship from the bse of the lighthouse? 49. Lening Ldder 0-ft ldder lens ginst building so tht the ngle between the ground nd the ldder is 7. How high does the ldder rech on the building? 50. Lening Ldder 0-ft ldder is lening ginst building. If the bse of the ldder is 6 ft from the bse of the building, wht is the ngle of elevtion of the ldder? How high does the ldder rech on the building? 51. ngle of the Sun 96-ft tree csts shdow tht is 10 ft long. Wht is the ngle of elevtion of the sun? 5. Height of Tower 600-ft gu wire is ttched to the top of communictions tower. If the wire mkes n ngle of 65 with the ground, how tll is the communictions tower? 5. Elevtion of Kite mn is ling on the bech, fling kite. He holds the end of the kite string t ground level, nd estimtes the ngle of elevtion of the kite to be 50. If the string is 450 ft long, how high is the kite bove the ground? 54. Determining Distnce womn stnding on hill sees flgpole tht she knows is 60 ft tll. The ngle of depression to the bottom of the pole is 14, nd the ngle of elevtion to the top of the pole is 18. Find her distnce from the pole. 55. Height of Tower wter tower is locted 5 ft from building (see the figure). From window in the building, n observer notes tht the ngle of elevtion to the top of the tower is 9 nd tht the ngle of depression to the bottom of the tower is 5. How tll is the tower? How high is the window? 9* 56. Determining Distnce n irplne is fling t n elevtion of 5150 ft, directl bove stright highw. Two motorists re driving crs on the highw on opposite sides of the plne, nd the ngle of depression to one cr is 5 nd to the other is 5. How fr prt re the crs? 57. Determining Distnce If both crs in Eercise 56 re on one side of the plne nd if the ngle of depression to one cr is 8 nd to the other cr is 5, how fr prt re the crs? 58. Height of lloon hot-ir blloon is floting bove stright rod. To estimte their height bove the ground, the blloonists simultneousl mesure the ngle of depression to two consecutive mileposts on the rod on the sme side of the blloon. The ngles of depression re found to be 0 nd. How high is the blloon? 59. Height of Mountin To estimte the height of mountin bove level plin, the ngle of elevtion to the top of the mountin is mesured to be. One thousnd feet closer to the mountin long the plin, it is found tht the ngle of elevtion is 5. Estimte the height of the mountin. 60. Height of loud over To mesure the height of the cloud cover t n irport, worker shines spotlight upwrd t n ngle 75 from the horizontl. n observer 600 m w mesures the ngle of elevtion to the spot of light to be 45. Find the height h of the cloud cover. 5* 5 ft h 45* 75* 600 m

21 SETION 6. Trigonometr of Right Tringles Distnce to the Sun When the moon is ectl hlf full, the erth, moon, nd sun form right ngle (see the figure). t tht time the ngle formed b the sun, erth, nd moon is mesured to be If the distnce from the erth to the moon is 40,000 mi, estimte the distnce from the erth to the sun. sun moon erth 64. Prll To find the distnce to nerb strs, the method of prll is used. The ide is to find tringle with the str t one verte nd with bse s lrge s possible. To do this, the str is observed t two different times ectl 6 months prt, nd its pprent chnge in position is recorded. From these two observtions, E 1 SE cn be clculted. (The times re chosen so tht E 1 SE is s lrge s possible, which gurntees tht E 1 OS is 90.) The ngle E 1 SO is clled the prll of the str. lph enturi, the str nerest the erth, hs prll of Estimte the distnce to this str. (Tke the distnce from the erth to the sun to be mi.) E 6. Distnce to the Moon To find the distnce to the sun s in Eercise 61, we needed to know the distnce to the moon. Here is w to estimte tht distnce: When the moon is seen t its zenith t point on the erth, it is observed to be t the horizon from point (see the figure). Points nd re 6155 mi prt, nd the rdius of the erth is 960 mi. () Find the ngle u in degrees. (b) Estimte the distnce from point to the moon. erth moon O E Distnce from Venus to the Sun The elongtion of plnet is the ngle formed b the plnet, erth, nd sun (see the figure). When Venus chieves its mimum elongtion of 46., the erth, Venus, nd the sun form tringle with right ngle t Venus. Find the distnce between Venus nd the sun in stronomicl Units (U). ( definition, the distnce between the erth nd the sun is 1 U.) S 6155 mi 6. Rdius of the Erth In Eercise 7 of Section 6.1 method ws given for finding the rdius of the erth. Here is more modern method: From stellite 600 mi bove the erth, it is observed tht the ngle formed b the verticl nd the line of sight to the horizon is Use this informtion to find the rdius of the erth. Venus sun 1 U å erth Discover Discussion 66. Similr Tringles If two tringles re similr, wht properties do the shre? Eplin how these properties mke it possible to define the trigonometric rtios without regrd to the size of the tringle.

22 488 HPTER 6 Trigonometric Functions of ngles 6. Trigonometric Functions of ngles In the preceding section we defined the trigonometric rtios for cute ngles. Here we etend the trigonometric rtios to ll ngles b defining the trigonometric functions of ngles. With these functions we cn solve prcticl problems tht involve ngles which re not necessril cute. Trigonometric Functions of ngles Let POQ be right tringle with cute ngle u s shown in Figure 1(). Plce u in stndrd position s shown in Figure 1(b). P P(, ) hpotenuse opposite r Figure 1 O djcent () Q O (b) Q Then P P1, is point on the terminl side of u. In tringle POQ, the opposite side hs length nd the djcent side hs length. Using the Pthgoren Theorem, we see tht the hpotenuse hs length r. So sin u r cos u r tn u The other trigonometric rtios cn be found in the sme w. These observtions llow us to etend the trigonometric rtios to n ngle. We define the trigonometric functions of ngles s follows (see Figure ). P(, ) r 0 Definition of the Trigonometric Functions Let u be n ngle in stndrd position nd let P1, be point on the terminl side. If r is the distnce from the origin to the point P1,, then Figure sin u r cos u r tn u 1 0 csc u r 1 0 sec u r 1 0 cot u 1 0

23 SETION 6. Trigonometric Functions of ngles 489 Reltionship to the Trigonometric Functions of Rel Numbers You m hve lred studied the trigonometric functions defined using the unit circle (hpter 5). To see how the relte to the trigonometric functions of n ngle, let s strt with the unit circle in the coordinte pln. P(, ) 0 1 P(, ) is the terminl point determined b t. Let P 1, be the terminl point determined b n rc of length t on the unit circle. Then t subtends n ngle u t the center of the circle. If we drop perpendiculr from P onto the point Q on the -is, then tringle OPQ is right tringle with legs of length nd, s shown in the figure. t Now, b the definition of the trigonometric functions of the rel number t, we hve the definition of the trigonometric functions of the ngle u, we hve If u is mesured in rdins, then u t. (See the figure below.) ompring the two ws of defining the trigonometric functions, we see tht the re identicl. In other words, s functions, the ssign identicl vlues to given rel number (the rel number is the rdin mesure of u in one cse or the length t of n rc in the other). sin t cos t sin u opp hp 1 cos u dj hp 1 P(, ) t 0 1 P(, ) r 0 1 Tringle OPQ is right tringle. The rdin mesure of ngle is t. Wh then do we stud trigonometr in two different ws? ecuse different pplictions require tht we view the trigonometric functions differentl. (See Focus on Modeling, pges 459, 5, nd 575, nd Sections 6., 6.4, nd 6.5.)

24 490 HPTER 6 Trigonometric Functions of ngles O P'(', ') P(, ) Q Q' Since division b 0 is n undefined opertion, certin trigonometric functions re not defined for certin ngles. For emple, tn 90 / is undefined becuse 0. The ngles for which the trigonometric functions m be undefined re the ngles for which either the - or -coordinte of point on the terminl side of the ngle is 0. These re qudrntl ngles ngles tht re coterminl with the coordinte es. It is crucil fct tht the vlues of the trigonometric functions do not depend on the choice of the point P1,. This is becuse if P 1, is n other point on the terminl side, s in Figure, then tringles POQ nd P OQ re similr. Figure The following mnemonic device cn be used to remember which trigonometric functions re positive in ech qudrnt: ll of them, Sine, Tngent, or osine. Sine Tngent ll osine You cn remember this s ll Students Tke lculus. _ (_, ) Figure 4 90* 0 r 15* 0 0* r 45* (, ) (, ) Evluting Trigonometric Functions t n ngle From the definition we see tht the vlues of the trigonometric functions re ll positive if the ngle u hs its terminl side in qudrnt I. This is becuse nd re positive in this qudrnt. [Of course, r is lws positive, since it is simpl the distnce from the origin to the point P1,.] If the terminl side of u is in qudrnt II, however, then is negtive nd is positive. Thus, in qudrnt II the functions sin u nd csc u re positive, nd ll the other trigonometric functions hve negtive vlues. You cn check the other entries in the following tble. Signs of the Trigonometric Functions Qudrnt Positive functions Negtive functions We now turn our ttention to finding the vlues of the trigonometric functions for ngles tht re not cute. Emple 1 I ll none II sin, csc cos, sec, tn, cot III tn, cot sin, csc, cos, sec IV cos, sec sin, csc, tn, cot Find () cos 15 nd (b) tn 90. Finding Trigonometric Functions of ngles Solution () From Figure 4 we see tht cos 15 /r. ut cos 45 /r, nd since cos 45 1/, we hve cos 15 1 (b) The ngles 90 nd 0 re coterminl. From Figure 5 it s cler tht tn 90 tn 0 nd, since tn 0 1/, we hve Figure 5 tn 90 1

25 SETION 6. Trigonometric Functions of ngles 491 From Emple 1 we see tht the trigonometric functions for ngles tht ren t cute hve the sme vlue, ecept possibl for sign, s the corresponding trigonometric functions of n cute ngle. Tht cute ngle will be clled the reference ngle. Reference ngle Let u be n ngle in stndrd position. The reference ngle u ssocited with u is the cute ngle formed b the terminl side of u nd the -is. Figure 6 shows tht to find reference ngle it s useful to know the qudrnt in which the terminl side of the ngle lies. Figure 6 The reference ngle u for n ngle u 0 = Emple Finding Reference ngles Find the reference ngle for () u 5p nd (b) u π 0 Solution () The reference ngle is the cute ngle formed b the terminl side of the ngle 5p/ nd the -is (see Figure 7). Since the terminl side of this ngle is in qudrnt IV, the reference ngle is Figure 7 u p 5p p 0 870* (b) The ngles 870 nd 150 re coterminl [becuse ]. Thus, the terminl side of this ngle is in qudrnt II (see Figure 8). So the reference ngle is u Evluting Trigonometric Functions for n ngle Figure 8 To find the vlues of the trigonometric functions for n ngle u, we crr out the following steps. 1. Find the reference ngle u ssocited with the ngle u.. Determine the sign of the trigonometric function of u b noting the qudrnt in which u lies.. The vlue of the trigonometric function of u is the sme, ecept possibl for sign, s the vlue of the trigonometric function of u.

26 49 HPTER 6 Trigonometric Functions of ngles 40* 0 Figure 9 S sin 40 is negtive. T Emple Find () sin 40 nd (b) cot 495. Using the Reference ngle to Evlute Trigonometric Functions Solution () This ngle hs its terminl side in qudrnt III, s shown in Figure 9. The reference ngle is therefore , nd the vlue of sin 40 is negtive. Thus sin 40 sin 60 1 Sign Reference ngle 0 495* (b) The ngle 495 is coterminl with the ngle 15, nd the terminl side of this ngle is in qudrnt II, s shown in Figure 10. So the reference ngle is , nd the vlue of cot 495 is negtive. We hve cot 495 cot 15 cot 45 1 Figure 10 S tn 495 is negtive, T so cot 495 is negtive. 4π Figure 11 S T sin 16p 0 is negtive. Emple 4 Using the Reference ngle to Evlute Trigonometric Functions 16p Find () sin nd (b) sec p. 4 b Solution () The ngle 16p/ is coterminl with 4p/, nd these ngles re in qudrnt III (see Figure 11). Thus, the reference ngle is 14p/ p p/. Since the vlue of sine is negtive in qudrnt III, we hve oterminl ngles Sign Reference ngle sin 16p sin 4p sin p 1 oterminl ngles Sign Reference ngle 0 π _ 4 Figure 1 S cos1 p 4 is positive, T so sec1 p 4 is positive. (b) The ngle p/4 is in qudrnt IV, nd its reference ngle is p/4 (see Figure 1). Since secnt is positive in this qudrnt, we get sec p 4 b sec p 4 Sign 1 Reference ngle Trigonometric Identities The trigonometric functions of ngles re relted to ech other through severl importnt equtions clled trigonometric identities. We ve lred encountered the

27 SETION 6. Trigonometric Functions of ngles 49 reciprocl identities. These identities continue to hold for n ngle u, provided both sides of the eqution re defined. The Pthgoren identities re consequence of the Pthgoren Theorem.* Fundmentl Identities Reciprocl Identities csc u 1 sin u sec u 1 cos u cot u 1 tn u tn u sin u cos u cot u cos u sin u Pthgoren Identities sin u cos u 1 tn u 1 sec u 1 cot u csc u r 0 (, ) Proof Let s prove the first Pthgoren identit. Using r (the Pthgoren Theorem) in Figure 1, we hve sin u cos u r b r b r r r 1 Thus, sin u cos u 1. (lthough the figure indictes n cute ngle, ou should check tht the proof holds for ll ngles u.) See Eercises 59 nd 60 for the proofs of the other two Pthgoren identities. Figure 1 Emple 5 Epressing One Trigonometric Function in Terms of nother () Epress sin u in terms of cos u. (b) Epress tn u in terms of sin u, where u is in qudrnt II. Solution () From the first Pthgoren identit we get sin u 1 cos u where the sign depends on the qudrnt. If u is in qudrnt I or II, then sin u is positive, nd hence sin u 1 cos u wheres if u is in qudrnt III or IV, sin u is negtive nd so sin u 1 cos u * We follow the usul convention of writing sin u for 1sin u. In generl, we write sin n u for 1sin u n for ll integers n ecept n 1. The eponent n 1 will be ssigned nother mening in Section 7.4. Of course, the sme convention pplies to the other five trigonometric functions.

28 494 HPTER 6 Trigonometric Functions of ngles (b) Since tn u sin u/cos u, we need to write cos u in terms of sin u. prt () cos u 1 sin u nd since cos u is negtive in qudrnt II, the negtive sign pplies here. Thus tn u sin u cos u sin u 1 sin u If ou wish to rtionlize the denomintor, ou cn epress cos u s # Emple 6 Evluting Trigonometric Function If tn u nd u is in qudrnt III, find cos u. Solution 1 We need to write cos u in terms of tn u. From the identit tn u 1 sec u, we get sec u tn u 1. In qudrnt III, sec u is negtive, so Thus sec u tn u 1 cos u 1 sec u 1 tn u œ 1 Figure 14 œ 1 Figure 15 Solution This problem cn be solved more esil using the method of Emple of Section 6.. Recll tht, ecept for sign, the vlues of the trigonometric functions of n ngle re the sme s those of n cute ngle (the reference ngle). So, ignoring the sign for the moment, let s sketch right tringle with n cute ngle u stisfing tn u (see Figure 14). the Pthgoren Theorem the hpotenuse of this tringle hs length 11. From the tringle in Figure 14 we immeditel see tht cos u / 11. Since u is in qudrnt III, cos u is negtive nd so Emple 7 Evluting Trigonometric Functions If sec u nd u is in qudrnt IV, find the other five trigonometric functions of u. Solution We sketch tringle s in Figure 15 so tht sec u. Tking into ccount the fct tht u is in qudrnt IV, we get sin u 1 csc u 1 res of Tringles cos u 11 cos u 1 tn u 1 sec u cot u 1 1 We conclude this section with n ppliction of the trigonometric functions tht involves ngles tht re not necessril cute. More etensive pplictions pper in the net two sections.

29 SETION 6. Trigonometric Functions of ngles 495 h () b The re of tringle is 1 bse height. If we know two sides nd the included ngle of tringle, then we cn find the height using the trigonometric functions, nd from this we cn find the re. If u is n cute ngle, then the height of the tringle in Figure 16() is given b h b sin u. Thus, the re is 1 bse height 1 b sin u Figure 16 (b) b h =180* _ If the ngle u is not cute, then from Figure 16(b) we see tht the height of the tringle is h b sin1180 u b sin u This is so becuse the reference ngle of u is the ngle 180 u. Thus, in this cse lso, the re of the tringle is 1 bse height 1 b sin u re of Tringle The re of tringle with sides of lengths nd b nd with included ngle u is 1 b sin u Emple 8 Finding the re of Tringle Find the re of tringle shown in Figure 17. Figure cm 10* cm Solution The tringle hs sides of length 10 cm nd cm, with included ngle 10. Therefore 1 b sin u sin sin 60 Reference ngle cm 6. Eercises 1 8 Find the reference ngle for the given ngle. 1. () 150 (b) 0 (c) 0. () 10 (b) 10 (c) 780. () 5 (b) 810 (c) () 99 (b) 199 (c) () 11p (b) 11p (c) 11p 4 6 4p p 6. () (b) (c) 4 p 6 5p 7. () (b) 1.4p (c) ().p (b). (c) 10p

30 496 HPTER 6 Trigonometric Functions of ngles 9 Find the ect vlue of the trigonometric function. 9. sin sin cos cos tn sec csc cot cos sec tn cos sin p. sin 5p. sin p 4. cos 7p 5. cos 7p b 6. tn 5p 6 7. sec 17p 8. csc 5p 4 9. cot p 4 b 0. cos 7p 1. tn 5p. 4 sin 11p 6 6 Find the qudrnt in which u lies from the informtion given.. sin u 0 nd cos u 0 4. tn u 0 nd sin u 0 5. sec u 0 nd tn u 0 6. csc u 0 nd cos u Write the first trigonometric function in terms of the second for u in the given qudrnt. 7. tn u, cos u; u in qudrnt III 8. cot u, sin u; u in qudrnt II 9. cos u, sin u; u in qudrnt IV 40. sec u, sin u; u in qudrnt I 41. sec u, tn u; u in qudrnt II 4. csc u, cot u; u in qudrnt III 4 50 Find the vlues of the trigonometric functions of u from the informtion given. 4. sin u 5, u in qudrnt II 44. cos u 7 1, u in qudrnt III 45. tn u 4, cos u sec u 5, sin u csc u, u in qudrnt I 48. cot u 1 4, sin u cos u 7, tn u tn u 4, sin u If u p/, find the vlue of ech epression. () sin u, sin u (b) sin 1 u, 1 sin u (c) sin u, sin1u 5. Find the re of tringle with sides of length 7 nd 9 nd included ngle Find the re of tringle with sides of length 10 nd nd included ngle Find the re of n equilterl tringle with side of length tringle hs n re of 16 in, nd two of the sides of the tringle hve lengths 5 in. nd 7 in. Find the ngle included b these two sides. 56. n isosceles tringle hs n re of 4 cm, nd the ngle between the two equl sides is 5p/6. Wht is the length of the two equl sides? Find the re of the shded region in the figure Use the first Pthgoren identit to prove the second. [Hint: Divide b cos u.] 60. Use the first Pthgoren identit to prove the third. pplictions 10* 61. Height of Rocket rocket fired stright up is trcked b n observer on the ground mile w. () Show tht when the ngle of elevtion is u, the height of the rocket in feet is h 580 tn u. (b) omplete the tble to find the height of the rocket t the given ngles of elevtion. u h 1 mi h π 1

31 SETION 6. Trigonometric Functions of ngles Rin Gutter rin gutter is to be constructed from metl sheet of width 0 cm b bending up one-third of the sheet on ech side through n ngle u. () Show tht the cross-sectionl re of the gutter is modeled b the function 1u 100 sin u 100 sin u cos u (b) Grph the function for 0 u p/. (c) For wht ngle u is the lrgest cross-sectionl re chieved? Find the rnge nd height of shot put thrown under the given conditions. () On the erth with 0 1 ft/s nd u p/6 (b) On the moon with 0 1 ft/s nd u p/6 H R 10 cm 10 cm 6. Wooden em rectngulr bem is to be cut from clindricl log of dimeter 0 cm. The figures show different ws this cn be done. () Epress the cross-sectionl re of the bem s function of the ngle u in the figures. (b) Grph the function ou found in prt (). (c) Find the dimensions of the bem with lrgest crosssectionl re. 10 cm 66. Sledding The time in seconds tht it tkes for sled to slide down hillside inclined t n ngle u is d t 16 sin u where d is the length of the slope in feet. Find the time it tkes to slide down 000-ft slope inclined t 0. d width depth 0 cm 0 cm 67. eehives In beehive ech cell is regulr hegonl prism, s shown in the figure. The mount of w W in the cell depends on the pe ngle u nd is given b W cot u 0.65 csc u 64. Strength of em The strength of bem is proportionl to the width nd the squre of the depth. bem is cut from log s in Eercise 6. Epress the strength of the bem s function of the ngle u in the figures. 65. Throwing Shot Put The rnge R nd height H of shot put thrown with n initil velocit of 0 ft/s t n ngle u re given b R 0 sin1u g H 0 sin u g On the erth g ft/s nd on the moon g 5. ft/s. ees instinctivel choose u so s to use the lest mount of w possible. () Use grphing device to grph W s function of u for 0 u p. (b) For wht vlue of u does W hve its minimum vlue? [Note: iologists hve discovered tht bees rrel devite from this vlue b more thn degree or two.]

32 498 HPTER 6 Trigonometric Functions of ngles 68. Turning orner steel pipe is being crried down hllw 9 ft wide. t the end of the hll there is rightngled turn into nrrower hllw 6 ft wide. () Show tht the length of the pipe in the figure is modeled b the function L1u 9 csc u 6 sec u (b) Grph the function L for 0 u p/. (c) Find the minimum vlue of the function L. (d) Eplin wh the vlue of L ou found in prt (c) is the length of the longest pipe tht cn be crried round the corner. 9 ft 69. Rinbows Rinbows re creted when sunlight of different wvelengths (colors) is refrcted nd reflected in rindrops. The ngle of elevtion u of rinbow is lws the sme. It cn be shown tht u 4b where sin k sin b 6 ft nd 59.4 nd k 1. is the inde of refrction of wter. Use the given informtion to find the ngle of elevtion u of rinbow. (For mthemticl eplntion of rinbows see lculus, 5th Edition, b Jmes Stewrt, pges ) Discover Discussion 70. Using lcultor To solve certin problem, ou need to find the sine of 4 rd. Your stud prtner uses his clcultor nd tells ou tht On our clcultor ou get sin sin Wht is wrong? Wht mistke did our prtner mke? 71. Viète s Trigonometric Digrm In the 16th centur, the French mthemticin Frnçois Viète (see pge 49) published the following remrkble digrm. Ech of the si trigonometric functions of u is equl to the length of line segment in the figure. For instnce, sin u 0 PR 0, since from OPR we see tht sin u opp hp For ech of the five other trigonometric functions, find line segment in the figure whose length equls the vlue of the function t u. (Note: The rdius of the circle is 1, the center is O, segment QS is tngent to the circle t R, nd SOQ is right ngle.) S 0 PR 0 0 OR 0 0 PR PR 0 1 R O P Q

33 SETION 6. Trigonometric Functions of ngles 499 DISOVERY PROJET c c b Thles used similr tringles to find the height of tll column. (See pge 48.) b Similrit In geometr ou lerned tht two tringles re similr if the hve the sme ngles. In this cse, the rtios of corresponding sides re equl. Tringles nd in the mrgin re similr, so b b c c Similrit is the crucil ide underling trigonometr. We cn define sin u s the rtio of the opposite side to the hpotenuse in n right tringle with n ngle u, becuse ll such right tringles re similr. So the rtio represented b sin u does not depend on the size of the right tringle but onl on the ngle u. This is powerful ide becuse ngles re often esier to mesure thn distnces. For emple, the ngle formed b the sun, erth, nd moon cn be mesured from the erth. The secret to finding the distnce to the sun is tht the trigonometric rtios re the sme for the huge tringle formed b the sun, erth, nd moon s for n other similr tringle (see Eercise 61 in Section 6.). In generl, two objects re similr if the hve the sme shpe even though the m not be the sme size.* For emple, we recognize the following s representtions of the letter becuse the re ll similr. d d If two figures re similr, then the distnces between corresponding points in the figures re proportionl. The blue nd red s bove re similr the rtio of distnces between corresponding points is. We s tht the similrit rtio is s. To obtin the distnce d between n two points in the blue, we multipl the corresponding distnce d in the red b. So d sd or d d Likewise, the similrit rtio between the first nd lst letters is s 5, so Write short prgrph eplining how the concept of similrit is used to define the trigonometric rtios.. How is similrit used in mp mking? How re distnces on cit rod mp relted to ctul distnces?. How is our erbook photogrph similr to ou? ompre distnces between different points on our fce (such s distnce between ers, length of * If the hve the sme shpe nd size, the re congruent, which is specil cse of similrit.

34 500 HPTER 6 Trigonometric Functions of ngles nose, distnce between ees, nd so on) to the corresponding distnces in photogrph. Wht is the similrit rtio? 4. The figure illustrtes method for drwing n pple twice the size of given pple. Use the method to drw tie times the size (similrit rtio ) of the blue tie. 5. Give conditions under which two rectngles re similr to ech other. Do the sme for two isosceles tringles. 6. Suppose tht two similr tringles hve similrit rtio s. () How re the perimeters of the tringles relted? (b) How re the res of the tringles relted? c b h s sc sb sh 1 If the side of squre is doubled, its re is multiplied b. 7. () If two squres hve similrit rtio s, show tht their res 1 nd hve the propert tht s 1. (b) If the side of squre is tripled, its re is multiplied b wht fctor? (c) plne figure cn be pproimted b squres (s shown). Eplin how we cn conclude tht for n two plne figures with similrit rtio s, their res stisf s 1. (Use prt ().)

35 SETION 6.4 The Lw of Sines If the side of cube is doubled, its volume is multiplied b. 8. () If two cubes hve similrit rtio s, show tht their volumes V 1 nd V hve the propert tht V s V 1. (b) If the side of cube is multiplied b 10, b wht fctor is the volume multiplied? (c) How cn we use the fct tht solid object cn be filled b little cubes to show tht for n two solids with similrit rtio s, the volumes stisf V s V 1? 9. King Kong is 10 times s tll s Joe, norml-sized 00-lb gorill. ssuming tht King Kong nd Joe re similr, use the results from Problems 7 nd 8 to nswer the following questions. () How much does King Kong weigh? (b) If Joe s hnd is 1 in. long, how long is King Kong s hnd? (c) If it tkes squre rds of mteril to mke shirt for Joe, how much mteril would shirt for King Kong require? 6.4 The Lw of Sines Figure 1 b c In Section 6. we used the trigonometric rtios to solve right tringles. The trigonometric functions cn lso be used to solve oblique tringles, tht is, tringles with no right ngles. To do this, we first stud the Lw of Sines here nd then the Lw of osines in the net section. To stte these lws (or formuls) more esil, we follow the convention of lbeling the ngles of tringle s,,, nd the lengths of the corresponding opposite sides s, b, c, s in Figure 1. To solve tringle, we need to know certin informtion bout its sides nd ngles. To decide whether we hve enough informtion, it s often helpful to mke sketch. For instnce, if we re given two ngles nd the included side, then it s cler tht one nd onl one tringle cn be formed (see Figure ()). Similrl, if two sides nd the included ngle re known, then unique tringle is determined (Figure (c)). ut if we know ll three ngles nd no sides, we cnnot uniquel determine the tringle becuse mn tringles cn hve the sme three ngles. (ll these tringles would be similr, of course.) So we won t consider this lst cse. () S or S (b) SS (c) SS (d) SSS Figure In generl, tringle is determined b three of its si prts (ngles nd sides) s long s t lest one of these three prts is side. So, the possibilities, illustrted in Figure, re s follows.

36 50 HPTER 6 Trigonometric Functions of ngles se 1 se se se 4 One side nd two ngles (S or S) Two sides nd the ngle opposite one of those sides (SS) Two sides nd the included ngle (SS) Three sides (SSS) ses 1 nd re solved using the Lw of Sines; ses nd 4 require the Lw of osines. The Lw of Sines The Lw of Sines ss tht in n tringle the lengths of the sides re proportionl to the sines of the corresponding opposite ngles. The Lw of Sines In tringle we hve sin sin b sin c c b h=b ß Proof To see wh the Lw of Sines is true, refer to Figure. the formul 1 in Section 6. the re of tringle is b sin. the sme formul the re 1 1 of this tringle is lso c sin nd bc sin. Thus 1 bc sin 1 c sin 1 b sin Multipling b / 1bc gives the Lw of Sines. Figure b Los ngeles Figure 4 75* 60* c=40 mi Phoeni Emple 1 Trcking Stellite (S) stellite orbiting the erth psses directl overhed t observtion sttions in Phoeni nd Los ngeles, 40 mi prt. t n instnt when the stellite is between these two sttions, its ngle of elevtion is simultneousl observed to be 60 t Phoeni nd 75 t Los ngeles. How fr is the stellite from Los ngeles? In other words, find the distnce in Figure 4. Solution Whenever two ngles in tringle re known, the third ngle cn be determined immeditel becuse the sum of the ngles of tringle is 180. In this cse, (see Figure 4), so we hve sin b sin 60 b sin c b sin sin 60 sin Lw of Sines Substitute Solve for b The distnce of the stellite from Los ngeles is pproimtel 416 mi.

37 SETION 6.4 The Lw of Sines 50 5* Emple Solve the tringle in Figure 5. Solving Tringle (S) c=80.4 0* Figure 5 b Solution First, Since side c is known, to find side we use the reltion sin Similrl, to find b we use sin b sin c c sin sin sin c b c sin sin 80.4 sin 0 sin sin 15 sin Lw of Sines Solve for Lw of Sines Solve for b The mbiguous se In Emples 1 nd unique tringle ws determined b the informtion given. This is lws true of se 1 (S or S). ut in se (SS) there m be two tringles, one tringle, or no tringle with the given properties. For this reson, se is sometimes clled the mbiguous cse. To see wh this is so, we show in Figure 6 the possibilities when ngle nd sides nd b re given. In prt () no solution is possible, since side is too short to complete the tringle. In prt (b) the solution is right tringle. In prt (c) two solutions re possible, nd in prt (d) there is unique tringle with the given properties. We illustrte the possibilities of se in the following emples. b b b b Figure 6 The mbiguous cse () (b) (c) (d) 7 7 œ 45* Figure 7 Emple SS, the One-Solution se Solve tringle, where 45, 71, nd b 7. Solution We first sketch the tringle with the informtion we hve (see Figure 7). Our sketch is necessril tenttive, since we don t et know the other ngles. Nevertheless, we cn now see the possibilities. We first find. sin sin b sin b sin 7 71 sin b1 b 1 Lw of Sines Solve for sin

38 504 HPTER 6 Trigonometric Functions of ngles We consider onl ngles smller thn 180, since no tringle cn contin n ngle of 180 or lrger. The supplement of n ngle u (where 0 u 180 ) is the ngle 180 u. Which ngles hve sin 1? From the preceding section we know tht there re two such ngles smller thn 180 (the re 0 nd 150 ). Which of these ngles is comptible with wht we know bout tringle? Since 45, we cnnot hve 150, becuse So 0, nd the remining ngle is Now we cn find side c. sin b sin c c b sin sin 7 sin 105 sin 0 7 sin Lw of Sines Solve for c In Emple there were two possibilities for ngle, nd one of these ws not comptible with the rest of the informtion. In generl, if sin 1, we must check the ngle nd its supplement s possibilities, becuse n ngle smller thn 180 cn be in the tringle. To decide whether either possibilit works, we check to see whether the resulting sum of the ngles eceeds 180. It cn hppen, s in Figure 6(c), tht both possibilities re comptible with the given informtion. In tht cse, two different tringles re solutions to the problem. ln Oddie/PhotoEdit Surveing is method of lnd mesurement used for mpmking. Surveors use process clled tringultion in which network of thousnds of interlocking tringles is creted on the re to be mpped. The process is strted b mesuring the length of bseline between two surveing sttions. Then, using n instrument clled theodolite, the ngles between these two sttions nd third sttion re mesured. The Lw of Sines is then used to clculte the two other sides of the tringle formed b the three sttions. The clculted sides re used s bselines, nd the process is repeted over nd over to crete network of tringles. In this method, the onl distnce mesured is the initil bseline; ll (continued) Emple 4 SS, the Two-Solution se Solve tringle if 4.1, 186., nd b Solution From the given informtion we sketch the tringle shown in Figure 8. Note tht side m be drwn in two possible positions to complete the tringle. From the Lw of Sines Figure 8 sin b sin b= sin =186. = * There re two possible ngles between 0 nd 180 such tht sin Using the SIN 1 ke on clcultor (or INV SIN or RSIN ), we find tht one of these ngles is pproimtel The other is pproimtel We denote these two ngles b 1 nd so tht nd

39 SETION 6.4 The Lw of Sines 505 other distnces re clculted from the Lw of Sines. This method is prcticl becuse it is much esier to mesure ngles thn distnces. heck bse Thus, two tringles stisf the given conditions: tringle nd tringle. Solve tringle : Thus c 1 1 sin sin sin 1 sin 4.1 Find 1 Lw of Sines Solve tringle : seline One of the most mbitious mpmking efforts of ll time ws the Gret Trigonometric Surve of Indi (seeproblem 8, pge 55) which required severl epeditions nd took over centur to complete. The fmous epedition of 18, led b Sir George Everest, lsted 0 ers. Rnging over trecherous terrin nd encountering the dreded mlri-crring mosquitoes, this epedition reched the foothills of the Himls. lter epedition, using tringultion, clculted the height of the highest pek of the Himls to be 9,00 ft. The pek ws nmed in honor of Sir George Everest. Tod, using stellites, the height of Mt. Everest is estimted to be 9,08 ft. The ver close greement of these two estimtes shows the gret ccurc of the trigonometric method. Thus Tringles nd re shown in Figure 9. Figure c sin 186. sin sin sin 4.1 b= * = * 65.8* c Å57.8 b=48.6 Find Lw of Sines.7* =186. The net emple presents sitution for which no tringle is comptible with the given dt. 114.* 4.1* c Å105. Emple 5 SS, the No-Solution se Solve tringle, where 4, 70, nd b 1. Solution To orgnize the given informtion, we sketch the digrm in Figure 10. Let s tr to find. We hve 1 4* Figure sin sin b sin b sin 1 sin Lw of Sines Solve for sin Since the sine of n ngle is never greter thn 1, we conclude tht no tringle stisfies the conditions given in this problem.

40 506 HPTER 6 Trigonometric Functions of ngles 6.4 Eercises 1 6 Use the Lw of Sines to find the indicted side or ngle u Solve the tringle using the Lw of Sines * 5* 70* * 46* * 0* * * Sketch ech tringle nd then solve the tringle using the Lw of Sines , 68, c 0 1., 110, c , 65, b , 95, , 51, b , 100, c * 10* 8.1* 67* * 100* * 17 6 Use the Lw of Sines to solve for ll possible tringles tht stisf the given conditions , b 15, , c 40, , c 45, b 45, c 4, 8 1. b 5, c 0, 5. 75, b 100, 0. 50, b 100, , b 80, , c 15, 9 6. b 7, c 8, For the tringle shown, find () D nd (b) D. 8. For the tringle shown, find the length D. 9. In tringle, 40, 15, nd b 0. () Show tht there re two tringles, nd, tht stisf these conditions. (b) Show tht the res of the tringles in prt () re proportionl to the sines of the ngles nd, tht is, 0. Show tht, given the three ngles,, of tringle nd one side, s, the re of the tringle is pplictions 0 5* 1 re of ^ sin re of ^ sin re sin sin sin 1. Trcking Stellite The pth of stellite orbiting the erth cuses it to pss directl over two trcking sttions nd, which re 50 mi prt. When the stellite is on one 0 D 8 D 0* 1 5*

41 SETION 6.4 The Lw of Sines 507 side of the two sttions, the ngles of elevtion t nd re mesured to be 87.0 nd 84., respectivel. () How fr is the stellite from sttion? (b) How high is the stellite bove the ground? 6. Rdio ntenn short-wve rdio ntenn is supported b two gu wires, 165 ft nd 180 ft long. Ech wire is ttched to the top of the ntenn nd nchored to the ground, t two nchor points on opposite sides of the ntenn. The shorter wire mkes n ngle of 67 with the ground. How fr prt re the nchor points? 7. Height of Tree tree on hillside csts shdow 15 ft down the hill. If the ngle of inclintion of the hillside is to the horizontl nd the ngle of elevtion of the sun is 5, find the height of the tree. 87.0* 84.*. Flight of Plne pilot is fling over stright highw. He determines the ngles of depression to two mileposts, 5 mi prt, to be nd 48, s shown in the figure. () Find the distnce of the plne from point. (b) Find the elevtion of the plne ft. Distnce cross River To find the distnce cross river, surveor chooses points nd, which re 00 ft prt on one side of the river (see the figure). She then chooses reference point on the opposite side of the river nd finds tht 8 nd 5. pproimte the distnce from to. * 5 mi 00 ft 8* 5* 48* 8. Length of Gu Wire communictions tower is locted t the top of steep hill, s shown. The ngle of inclintion of the hill is 58. gu wire is to be ttched to the top of the tower nd to the ground, 100 m downhill from the bse of the tower. The ngle in the figure is determined to be 1. Find the length of cble required for the gu wire lculting Distnce Observers t P nd Q re locted on the side of hill tht is inclined to the horizontl, s shown. The observer t P determines the ngle of elevtion to hot-ir blloon to be 6. t the sme instnt, the observer t Q mesures the ngle of elevtion to the blloon to be 71. If P is 60 m down the hill from Q, find the distnce from Q to the blloon. å 4. Distnce cross Lke Points nd re seprted b lke. To find the distnce between them, surveor loctes point on lnd such tht He lso mesures s 1 ft nd s 57 ft. Find the distnce between nd. 5. The Lening Tower of Pis The bell tower of the cthedrl in Pis, Itl, lens 5.6 from the verticl. tourist stnds 105 m from its bse, with the tower lening directl towrd her. She mesures the ngle of elevtion to the top of the tower to be 9.. Find the length of the tower to the nerest meter. P * Q 60 m

42 508 HPTER 6 Trigonometric Functions of ngles 40. lculting n ngle wter tower 0 m tll is locted t the top of hill. From distnce of 10 m down the hill, it is observed tht the ngle formed between the top nd bse of the tower is 8. Find the ngle of inclintion of the hill. 8* 0 m hve mesure 60. () Show tht the rdius r of the common fce is given b r b b [Hint: Use the Lw of Sines together with the fct tht n ngle u nd its supplement 180 u hve the sme sine.] (b) Find the rdius of the common fce if the rdii of the bubbles re 4 cm nd cm. (c) Wht shpe does the common fce tke if the two bubbles hve equl rdii? 10 m b r D 41. Distnces to Venus The elongtion of plnet is the ngle formed b the plnet, erth, nd sun (see the figure). It is known tht the distnce from the sun to Venus is 0.7 U (see Eercise 65 in Section 6.). t certin time the elongtion of Venus is found to be 9.4. Find the possible distnces from the erth to Venus t tht time in stronomicl Units (U). Venus sun Discover Discussion 4. Number of Solutions in the mbiguous se We hve seen tht when using the Lw of Sines to solve tringle in the SS cse, there m be two, one, or no solution(s). Sketch tringles like those in Figure 6 to verif the criteri in the tble for the number of solutions if ou re given nd sides nd b. Venus å 1 U erth 4. Sop ubbles When two bubbles cling together in midir, their common surfce is prt of sphere whose center D lies on the line pssing throught the centers of the bubbles (see the figure). lso, ngles nd D ech riterion Number of Solutions b 1 b b sin b sin 1 b sin 0 If 0 nd b 100, use these criteri to find the rnge of vlues of for which the tringle hs two solutions, one solution, or no solution. 6.5 The Lw of osines The Lw of Sines cnnot be used directl to solve tringles if we know two sides nd the ngle between them or if we know ll three sides (these re ses nd 4 of the preceding section). In these two cses, the Lw of osines pplies.

43 SETION 6.5 The Lw of osines 509 The Lw of osines b In n tringle (see Figure 1), we hve c b c bc cos b c c cos Figure 1 c b b cos (b ç, b ß ) b (0, 0) c (c, 0) Figure Proof To prove the Lw of osines, plce tringle so tht is t the origin, s shown in Figure. The coordintes of the vertices nd re 1c, 0 nd (b cos, b sin ), respectivel. (You should check tht the coordintes of these points will be the sme if we drw ngle s n cute ngle.) Using the Distnce Formul, we get 1b cos c 1b sin 0 b cos bc cos c b sin b 1cos sin bc cos c b c bc cos ecuse sin cos 1 This proves the first formul. The other two formuls re obtined in the sme w b plcing ech of the other vertices of the tringle t the origin nd repeting the preceding rgument. In words, the Lw of osines ss tht the squre of n side of tringle is equl to the sum of the squres of the other two sides, minus twice the product of those two sides times the cosine of the included ngle. If one of the ngles of tringle, s, is right ngle, then cos 0 nd the Lw of osines reduces to the Pthgoren Theorem, c b. Thus, the Pthgoren Theorem is specil cse of the Lw of osines. Emple 1 Length of Tunnel tunnel is to be built through mountin. To estimte the length of the tunnel, surveor mkes the mesurements shown in Figure. Use the surveor s dt to pproimte the length of the tunnel. Figure 88 ft 8.4* 1 ft Solution To pproimte the length c of the tunnel, we use the Lw of osines: c b b cos Lw of osines cos 8.4 Substitute Use clcultor c Tke squre roots Thus, the tunnel will be pproimtel 417 ft long.

44 510 HPTER 6 Trigonometric Functions of ngles b=8 c=1 =5 Emple SSS, the Lw of osines The sides of tringle re 5, b 8, nd c 1 (see Figure 4). Find the ngles of the tringle. Figure 4 Solution We first find. From the Lw of osines, we hve b c bc cos. Solving for cos, we get cos b c bc OS 1 Using clcultor, we find tht 18. In the sme w the equtions OR INV OS OR R OS cos c b c cos b c b give 9 nd 1. Of course, once two ngles re clculted, the third cn more esil be found from the fct tht the sum of the ngles of tringle is 180. However, it s good ide to clculte ll three ngles using the Lw of osines nd dd the three ngles s check on our computtions. Emple SS, the Lw of osines Solve tringle, where 46.5, b 10.5, nd c Solution We cn find using the Lw of osines. b c bc cos cos Thus, We lso use the Lw of osines to find nd, s in Emple. b= * Å * 5.* c=18.0 Figure 5 cos c b c cos b c b Using clcultor, we find tht 5. nd 98.. To summrize: 5., 98., nd 1.. (See Figure 5.) We could hve used the Lw of Sines to find nd in Emple, since we knew ll three sides nd n ngle in the tringle. ut knowing the sine of n ngle does not uniquel specif the ngle, since n ngle u nd its supplement 180 u both hve the sme sine. Thus we would need to decide which of the two ngles is the correct choice. This mbiguit does not rise when we use the Lw of osines, becuse ever ngle between 0 nd 180 hs unique cosine. So using onl the Lw of osines is preferble in problems like Emple.

45 SETION 6.5 The Lw of osines 511 Nvigtion: Heding nd ering In nvigtion direction is often given s bering, tht is, s n cute ngle mesured from due north or due south. The bering N 0 E, for emple, indictes direction tht points 0 to the est of due north (see Figure 6). N N N N W 0 E W 60 E W E W E S N 0 E S N 60 W S S 70 W S S 50 E Figure 6 Emple 4 Nvigtion 00 mi 40* 100 mi pilot sets out from n irport nd heds in the direction N 0 E, fling t 00 mi/h. fter one hour, he mkes course correction nd heds in the direction N 40 E. Hlf n hour fter tht, engine trouble forces him to mke n emergenc lnding. () Find the distnce between the irport nd his finl lnding point. (b) Find the bering from the irport to his finl lnding point. 0* Figure 7 nother ngle with sine is ut this is clerl too lrge to be in. Solution () In one hour the plne trvels 00 mi, nd in hlf n hour it trvels 100 mi, so we cn plot the pilot s course s in Figure 7. When he mkes his course correction, he turns 0 to the right, so the ngle between the two legs of his trip is So b the Lw of osines we hve b # 00 # 100 cos , Thus, b The pilot lnds bout 96 mi from his strting point. (b) We first use the Lw of Sines to find. sin 100 sin sin 100 # Using the ke on clcultor, we find tht From Figure 7 we see tht the line from the irport to the finl lnding site points in the direction est of due north. Thus, the bering is bout N 6.6 E. SIN sin

46 51 HPTER 6 Trigonometric Functions of ngles The re of Tringle n interesting ppliction of the Lw of osines involves formul for finding the re of tringle from the lengths of its three sides (see Figure 8). c Heron s Formul The re of tringle is given b 1s1s 1s b1s c Figure 8 b where s 1 1 b c is the semiperimeter of the tringle; tht is, s is hlf the perimeter. Proof We strt with the formul 1 b sin from Section 6.. Thus 1 4 b sin 1 4 b 11 cos Pthgoren identit Fctor Net, we write the epressions 1 cos nd 1 cos in terms of, b nd c. the Lw of osines we hve Similrl 1 4 b 11 cos 11 cos cos b c b 1 cos 1 b c b b b c b 1 b c b 1 b c1 b c b Lw of osines dd 1 ommon denomintor Fctor 1c b1c b 1 cos b Difference of squres Substituting these epressions in the formul we obtined for gives To see tht the fctors in the lst two products re equl, note for emple tht b c b c s c c 1 b c1 b c 1 4 b b 1 b c 1 b c s1s c1s b1s Heron s Formul now follows b tking the squre root of ech side. 1c b1c b 1c b b 1c b

47 SETION 6.5 The Lw of osines 51 Figure 9 15 ft 15 ft 80 ft Emple 5 re of Lot businessmn wishes to bu tringulr lot in bus downtown loction (see Figure 9). The lot frontges on the three djcent streets re 15, 80, nd 15 ft. Find the re of the lot. Solution The semiperimeter of the lot is s 60 Heron s Formul the re is ,451.6 Thus, the re is pproimtel 17,45 ft. 6.5 Eercises 1 8 Use the Lw of osines to determine the indicted side or ngle u * * * * * Solve tringle * 11..0, b 4.0, 5 1. b 60, c 0, , b 5, c , b 1, c b 15, c 16, , c 50, , b 65, , 61, Find the indicted side or ngle u. (Use either the Lw of Sines or the Lw of osines, s pproprite.) * 85* 1.. 0* * *

48 514 HPTER 6 Trigonometric Functions of ngles * Find the re of the tringle whose sides hve the given lengths. 7. 9, b 1, c , b, c 9. 7, b 8, c , b 100, c Find the re of the shded figure, correct to two decimls * * * Three circles of rdii 4, 5, nd 6 cm re mutull tngent. Find the shded re enclosed between the circles * 10 5* 60* Prove tht in tringle These re clled the Projection Lws. [Hint: To get the first eqution, dd the second nd third equtions in the Lw of osines nd solve for.] pplictions 7. Surveing To find the distnce cross smll lke, surveor hs tken the mesurements shown. Find the distnce cross the lke using this informtion. b cos c cos.8 mi b c cos cos c cos b cos 40.*.56 mi 8. Geometr prllelogrm hs sides of lengths nd 5, nd one ngle is 50. Find the lengths of the digonls. 9. lculting Distnce Two stright rods diverge t n ngle of 65. Two crs leve the intersection t :00 P.M., one trveling t 50 mi/h nd the other t 0 mi/h. How fr prt re the crs t :0 P.M.? 40. lculting Distnce cr trvels long stright rod, heding est for 1 h, then trveling for 0 min on nother rod tht leds northest. If the cr hs mintined constnt speed of 40 mi/h, how fr is it from its strting position? 41. Ded Reckoning pilot flies in stright pth for 1 h 0 min. She then mkes course correction, heding 10 to the right of her originl course, nd flies h in the new direction. If she mintins constnt speed of 65 mi/h, how fr is she from her strting position? 4. Nvigtion Two bots leve the sme port t the sme time. One trvels t speed of 0 mi/h in the direction N 50 E nd the other trvels t speed of 6 mi/h in direction S 70 E (see the figure). How fr prt re the two bots fter one hour? N N 50 E W 50 S 70 E S 70 E

49 SETION 6.5 The Lw of osines Nvigtion fishermn leves his home port nd heds in the direction N 70 W. He trvels 0 mi nd reches Egg Islnd. The net d he sils N 10 E for 50 mi, reching Forrest Islnd. () Find the distnce between the fishermn s home port nd Forrest Islnd. (b) Find the bering from Forrest Islnd bck to his home port. Forrest Islnd 47. Fling Kites bo is fling two kites t the sme time. He hs 80 ft of line out to one kite nd 40 ft to the other. He estimtes the ngle between the two lines to be 0. pproimte the distnce between the kites. 80 ft 0 40 ft mi Egg Islnd 0 mi 70 Home port 48. Securing Tower 15-ft tower is locted on the side of mountin tht is inclined to the horizontl. gu wire is to be ttched to the top of the tower nd nchored t point 55 ft downhill from the bse of the tower. Find the shortest length of wire needed. 44. Nvigtion irport is 00 mi from irport t bering N 50 E (see the figure). pilot wishing to fl from to mistkenl flies due est t 00 mi/h for 0 minutes, when he notices his error. () How fr is the pilot from his destintion t the time he notices the error? (b) Wht bering should he hed his plne in order to rrive t irport? irport 55 ft 15 ft mi 49. ble r steep mountin is inclined 74 to the horizontl nd rises 400 ft bove the surrounding plin. cble cr is to be instlled from point 800 ft from the bse to the top of the mountin, s shown. Find the shortest length of cble needed. irport 45. Tringulr Field tringulr field hs sides of lengths, 6, nd 44 d. Find the lrgest ngle. 46. Towing rge Two tugbots tht re 10 ft prt pull brge, s shown. If the length of one cble is 1 ft nd the length of the other is 0 ft, find the ngle formed b the two cbles. 800 ft 74* 400 ft 1 ft 0 ft 10 ft 50. N Tower The N Tower in Toronto, nd, is the tllest free-stnding structure in the world. womn on the observtion deck, 1150 ft bove the ground, wnts to determine the distnce between two lndmrks on the ground below. She observes tht the ngle formed b the lines of sight to these two lndmrks is 4. She lso observes tht the ngle between the verticl nd the line of sight to one of the

50 516 HPTER 6 Trigonometric Functions of ngles lndmrks is 6 nd to the other lndmrk is 54. Find the distnce between the two lndmrks. 51. Lnd Vlue Lnd in downtown olumbi is vlued t $0 squre foot. Wht is the vlue of tringulr lot with sides of lengths 11, 148, nd 190 ft? Discover Discussion 5. Solving for the ngles in Tringle The prgrph tht follows the solution of Emple on pge 510 eplins n lterntive method for finding nd, using the Lw of Sines. Use this method to solve the tringle in the emple, finding first nd then. Eplin how ou chose the pproprite vlue for the mesure of. Which method do ou prefer for solving n SS tringle problem, the one eplined in Emple or the one ou used in this eercise? 6 Review oncept heck 1. () Eplin the difference between positive ngle nd negtive ngle. (b) How is n ngle of mesure 1 degree formed? (c) How is n ngle of mesure 1 rdin formed? (d) How is the rdin mesure of n ngle u defined? (e) How do ou convert from degrees to rdins? (f) How do ou convert from rdins to degrees?. () When is n ngle in stndrd position? (b) When re two ngles coterminl?. () Wht is the length s of n rc of circle with rdius r tht subtends centrl ngle of u rdins? (b) Wht is the re of sector of circle with rdius r nd centrl ngle u rdins? 4. If u is n cute ngle in right tringle, define the si trigonometric rtios in terms of the djcent nd opposite sides nd the hpotenuse. 5. Wht does it men to solve tringle? 6. If u is n ngle in stndrd position, P1, is point on the terminl side, nd r is the distnce from the origin to P, write epressions for the si trigonometric functions of u. 7. Which trigonometric functions re positive in qudrnts I, II, III, nd IV? 8. If u is n ngle in stndrd position, wht is its reference ngle u? 9. () Stte the reciprocl identities. (b) Stte the Pthgoren identities. 10. () Wht is the re of tringle with sides of length nd b nd with included ngle u? (b) Wht is the re of tringle with sides of length, b, nd c? 11. () Stte the Lw of Sines. (b) Stte the Lw of osines. 1. Eplin the mbiguous cse in the Lw of Sines. Eercises 1 Find the rdin mesure tht corresponds to the given degree mesure. 1. () 60 (b) 0 (c) 15 (d) 90. () 4 (b) 0 (c) 750 (d) 5 4 Find the degree mesure tht corresponds to the given rdin mesure. 5p 9p. () (b) p (c) (d).1 6 4

51 HPTER 6 Review p 4. () 8 (b) 5 (c) (d) 6 5. Find the length of n rc of circle of rdius 8 m if the rc subtends centrl ngle of 1 rd. 6. Find the mesure of centrl ngle u in circle of rdius 5 ft if the ngle is subtended b n rc of length 7 ft. 7. circulr rc of length 100 ft subtends centrl ngle of 70. Find the rdius of the circle. 8. How mn revolutions will cr wheel of dimeter 8 in. mke over period of hlf n hour if the cr is trveling t 60 mi/h? 9. New York nd Los ngeles re 450 mi prt. Find the ngle tht the rc between these two cities subtends t the center of the erth. (The rdius of the erth is 960 mi.) 10. Find the re of sector with centrl ngle rd in circle of rdius 5 m. 11. Find the re of sector with centrl ngle 5 in circle of rdius 00 ft. 1. sector in circle of rdius 5 ft hs n re of 15 ft. Find the centrl ngle of the sector. 1. potter s wheel with rdius 8 in. spins t 150 rpm. Find the ngulr nd liner speeds of point on the rim of the wheel. p 5 (b) Find the ngulr speed of the wheels. (c) How fst (in mi/h) is the cr trveling? Find the vlues of the si trigonometric rtios of u Find the sides lbeled nd, correct to two deciml plces * 7 5 Ger 5 Rtio 1st 4.1 nd.0 rd 1.6 4th 0.9 5th 0.7 5* in. 0* 0* 0* In n utomobile trnsmission ger rtio g is the rtio ngulr speed of engine g ngulr speed of wheels 1 1 Solve the tringle. 1.. The ngulr speed of the engine is shown on the tchometer (in rpm). certin sports cr hs wheels with rdius 11 in. Its ger rtios re shown in the following tble. Suppose the cr is in fourth ger nd the tchometer reds 500 rpm. () Find the ngulr speed of the engine. 0* 60* 0

52 518 HPTER 6 Trigonometric Functions of ngles. Epress the lengths nd b in the figure in terms of the trigonometric rtios of u. b pilot mesures the ngles of depression to two ships to be 40 nd 5 (see the figure). If the pilot is fling t n elevtion of 5,000 ft, find the distnce between the two ships. 40* 5* 4. The highest free-stnding tower in the world is the N Tower in Toronto, nd. From distnce of 1 km from its bse, the ngle of elevtion to the top of the tower is Find the height of the tower. 5. Find the perimeter of regulr hegon tht is inscribed in circle of rdius 8 m. 6. The pistons in cr engine move up nd down repetedl to turn the crnkshft, s shown. Find the height of the point P bove the center O of the crnkshft in terms of the ngle u. P O 7. s viewed from the erth, the ngle subtended b the full moon is Use this informtion nd the fct tht the distnce from the erth to the moon is 6,900 mi to find the rdius of the moon. Q 8 in Find the ect vlue. 9. sin cot p 4. sin 405 b 5. cos csc 8p cot csc 9p 4 1. tn1 15. cos 5p 6 sec p sec 1p 6 tn p Find the vlues of the si trigonometric rtios of the ngle u in stndrd position if the point 1 5, 1 is on the terminl side of u. 4. Find sin u if u is in stndrd position nd its terminl side intersects the circle of rdius 1 centered t the origin t the point 1 1/, Find the cute ngle tht is formed b the line nd the -is. 44. Find the si trigonometric rtios of the ngle u in stndrd position if its terminl side is in qudrnt III nd is prllel to the line Write the first epression in terms of the second, for u in the given qudrnt. 45. tn u, cos u; u in qudrnt II 46. sec u, sin u; u in qudrnt III 47. tn u, sin u; u in n qudrnt 48. csc u cos u, sin u; u in n qudrnt

53 HPTER 6 Review Find the vlues of the si trigonometric functions of u from the informtion given. 49. tn u 17/, sec u sec u 41, csc u sin u, cos u 0 5. sec u 1 5 5, tn u 0 5. If tn u 1 for u in qudrnt II, find sin u cos u. 54. If sin u 1 for u in qudrnt I, find tn u sec u. 55. If tn u 1, find sin u cos u. 56. If cos u 1/ nd p/ u p, find sin u Find the side lbeled * 80* 45* 105* 64. From point on the ground, the ngle of elevtion to the top of tll building is 4.1. From point, which is 600 ft closer to the building, the ngle of elevtion is mesured to be 0.. Find the height of the building. 65. Find the distnce between points nd on opposite sides of lke from the informtion shown. 4.1* 600 ft. mi 4* 0.* * * * Two ships leve port t the sme time. One trvels t 0 mi/h in direction N E, nd the other trvels t 8 mi/h in direction S 4 E (see the figure). How fr prt re the two ships fter h? * 5.6 mi 66. bot is cruising the ocen off stright shoreline. Points nd re 10 mi prt on the shore, s shown. It is found tht 4. nd Find the shortest distnce from the bot to the shore. Shoreline 4.* 10 mi N * N E 68.9* W E S 4* S 4 E 67. Find the re of tringle with sides of length 8 nd 14 nd included ngle Find the re of tringle with sides of length 5, 6, nd 8.

54 50 HPTER 6 Trigonometric Functions of ngles 6 Test 1. Find the rdin mesures tht correspond to the degree mesures 0 nd 15. 4p. Find the degree mesures tht correspond to the rdin mesures nd 1... The rotor bldes of helicopter re 16 ft long nd re rotting t 10 rpm. () Find the ngulr speed of the rotor. (b) Find the liner speed of point on the tip of blde. 4. Find the ect vlue of ech of the following. () sin 405 (b) tn1 150 (c) sec 5p (d) csc 5p 5. Find tn u sin u for the ngle u shown. 6. Epress the lengths nd b shown in the figure in terms of u. 4 b 7. If cos u 1 nd u is in qudrnt III, find tn u cot u csc u. 8. If sin u 5 nd tn u 5 1 1, find sec u. 9. Epress tn u in terms of sec u for u in qudrnt II. 10. The bse of the ldder in the figure is 6 ft from the building, nd the ngle formed b the ldder nd the ground is 7. How high up the building does the ldder touch? 7* 6 ft

55 HPTER 6 Test Find the side lbeled Refer to the figure below. () Find the re of the shded region. (b) Find the perimeter of the shded region m 16. Refer to the figure below. () Find the ngle opposite the longest side. (b) Find the re of the tringle Two wires tether blloon to the ground, s shown. How high is the blloon bove the ground? h 75* 85* 100 ft

56 Focus on Modeling Surveing How cn we mesure the height of mountin, or the distnce cross lke? Obviousl it m be difficult, inconvenient, or impossible to mesure these distnces directl (tht is, using tpe mesure or rd stick). On the other hnd, it is es to mesure ngles to distnt objects. Tht s where trigonometr comes in the trigonometric rtios relte ngles to distnces, so the cn be used to clculte distnces from the mesured ngles. In this Focus we emine how trigonometr is used to mp town. Modern mp mking methods use stellites nd the Globl Positioning Sstem, but mthemtics remins t the core of the process. Mpping Town student wnts to drw mp of his hometown. To construct n ccurte mp (or scle model), he needs to find distnces between vrious lndmrks in the town. The student mkes the mesurements shown in Figure 1. Note tht onl one distnce is mesured, between it Hll nd the first bridge. ll other mesurements re ngles. Figure 1 The distnces between other lndmrks cn now be found using the Lw of Sines. For emple, the distnce from the bnk to the first bridge is clculted b ppling the Lw of Sines to the tringle with vertices t it Hll, the bnk, nd the first bridge: 0.86 sin 50 sin sin 50 sin 0 1. mi Lw of Sines Solve for lcultor 5

57 Surveing 5 So the distnce between the bnk nd the first bridge is 1. mi. The distnce we just found cn now be used to find other distnces. For instnce, we find the distnce between the bnk nd the cliff s follows: sin sin sin 64 sin mi Lw of Sines Solve for lcultor ontinuing in this fshion, we cn clculte ll the distnces between the lndmrks shown in the rough sketch in Figure 1. We cn use this informtion to drw the mp shown in Figure. it Hll nk hurch N Fire Hll School 0 1/4 1/ /4 1 mile Figure To mke topogrphic mp, we need to mesure elevtion. This concept is eplored in Problems 4 6. Problems 1. ompleting the Mp Find the distnce between the church nd it Hll.. ompleting the Mp Find the distnce between the fire hll nd the school. (You will need to find other distnces first.)

58 54 Focus on Modeling. Determining Distnce surveor on one side of river wishes to find the distnce between points nd on the opposite side of the river. On her side, she chooses points nd D, which re 0 m prt, nd mesures the ngles shown in the figure below. Find the distnce between nd. 50* 40* 0* 0 m 45* D.1* 69.4* 00 m 51.6* 4. Height of liff To mesure the height of n inccessible cliff on the opposite side of river, surveor mkes the mesurements shown in the figure t the left. Find the height of the cliff. 5. Height of Mountin To clculte the height h of mountin, ngle, b, nd distnce d re mesured, s shown in the figure below. () Show tht h d cot cot b (b) Show tht sin sin b h d sin1b (c) Use the formuls from prts () nd (b) to find the height of mountin if 5, b 9, nd d 800 ft. Do ou get the sme nswer from ech formul? h å 40 ft d 68* 4* 9* 6. Determining Distnce surveor hs determined tht mountin is 40 ft high. From the top of the mountin he mesures the ngles of depression to two lndmrks t the bse of the mountin, nd finds them to be 4 nd 9. (Observe tht these re the sme s the ngles of elevtion from the lndmrks s shown in the figure t the left.) The ngle between the lines of sight to the lndmrks is 68. lculte the distnce between the two lndmrks.

59 Surveing Surveing uilding Lots surveor surves two djcent lots nd mkes the following rough sketch showing his mesurements. lculte ll the distnces shown in the figure nd use our result to drw n ccurte mp of the two lots. 8. Gret Surve of Indi The Gret Trigonometric Surve of Indi ws one of the most mssive mpping projects ever undertken (see the mrgin note on pge 504). Do some reserch t our librr or on the Internet to lern more bout the Surve, nd write report on our findings. ritish Librr