Section 11.2: The Law of Sines, from College Trigonometry: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a

Size: px
Start display at page:

Download "Section 11.2: The Law of Sines, from College Trigonometry: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a"

Transcription

1 Setion 11.: The Lw of Sines, from College Trigonometry: Correted Edition y Crl Stitz, Ph.D. nd Jeff Zeger, Ph.D. is ville under Cretive Commons Attriution-NonCommeril-ShreAlike 3.0 liense. 013, Crl Stitz.

2 896 Applitions of Trigonometry 11. The Lw of Sines Trigonometry literlly mens mesuring tringles nd with Chpter 10 under our elts, we re more thn prepred to do just tht. The min gol of this setion nd the next is to develop theorems whih llow us to solve tringles tht is, find the length of eh side of tringle nd the mesure of eh of its ngles. In Setions 10., 10.3 nd 10.6, we ve hd some experiene solving right tringles. The following exmple reviews wht we know. Exmple Given right tringle with hypotenuse of length 7 units nd one leg of length 4 units, find the length of the remining side nd the mesures of the remining ngles. Express the ngles in deiml degrees, rounded to the nerest hundreth of degree. Solution. For definitiveness, we lel the tringle elow. = 7 = 4 To find the length of the missing side, we use the Pythgoren Theorem to get + 4 = 7 whih then yields = 33 units. Now tht ll three sides of the tringle re known, there re severl wys we n find using the inverse trigonometri funtions. To derese the hnes of propgting error, however, we stik to using the dt given to us in the prolem. In this se, the lengths 4 nd 7 were given, so we wnt to relte these to. Aording to Theorem 10.4, os() = 4 7. Sine is n ute ngle, = ros ( 4 7) rdins. Converting to degrees, we find Now tht we hve the mesure of ngle, we ould find the mesure of ngle using the ft tht nd re omplements so + = 90. One gin, we opt to use the dt given to us in the prolem. Aording to Theorem 10.4, we hve tht sin() = 4 7 so = rsin ( 4 7) rdins nd we hve A few remrks out Exmple re in order. First, we dhere to the onvention tht lower se Greek letter denotes n ngle 1 nd the orresponding lowerse English letter represents the side opposite tht ngle. Thus, is the side opposite, is the side opposite nd is the side opposite γ. Tken together, the pirs (, ), (, ) nd (γ, ) re lled ngle-side opposite pirs. Seond, s mentioned erlier, we will strive to solve for quntities using the originl dt given in the prolem whenever possile. While this is not lwys the esiest or fstest wy to proeed, it 1 s well s the mesure of sid ngle s well s the length of sid side

3 11. The Lw of Sines 897 minimizes the hnes of propgted error. 3 Third, sine mny of the pplitions whih require solving tringles in the wild rely on degree mesure, we shll dopt this onvention for the time eing. 4 The Pythgoren Theorem long with Theorems 10.4 nd llow us to esily hndle ny given right tringle prolem, ut wht if the tringle isn t right tringle? In ertin ses, we n use the Lw of Sines to help. Theorem 11.. The Lw of Sines: Given tringle with ngle-side opposite pirs (, ), (, ) nd (γ, ), the following rtios hold or, equivlently, sin() = sin() = sin(γ) sin() = sin() = sin(γ) The proof of the Lw of Sines n e roken into three ses. For our first se, onsider the tringle ABC elow, ll of whose ngles re ute, with ngle-side opposite pirs (, ), (, ) nd (γ, ). If we drop n ltitude from vertex B, we divide the tringle into two right tringles: ABQ nd BCQ. If we ll the length of the ltitude h (for height), we get from Theorem 10.4 tht sin() = h nd sin(γ) = h so tht h = sin() = sin(γ). After some rerrngement of the lst eqution, we get sin() using the tringles ABQ nd ACQ to get sin() B γ = sin(γ). If we drop n ltitude from vertex A, we n proeed s ove = sin(γ), ompleting the proof for this se. A C A C Q A C For our next se onsider the tringle ABC elow with otuse ngle. Extending n ltitude from vertex A gives two right tringles, s in the previous se: ABQ nd ACQ. Proeeding s efore, we get h = sin(γ) nd h = sin() so tht sin() = sin(γ). B h γ h B Q γ B γ B Q h γ A C A C 3 Your Siene tehers should thnk us for this. 4 Don t worry! Rdins will e k efore you know it!

4 898 Applitions of Trigonometry Dropping n ltitude from vertex B lso genertes two right tringles, ABQ nd BCQ. We know tht sin( ) = h so tht h = sin( ). Sine = 180, sin( ) = sin(), so in ft, we hve h = sin(). Proeeding to BCQ, we get sin(γ) = h so h = sin(γ). Putting this together with the previous eqution, we get sin(γ) B = sin(), nd we re finished with this se. h γ Q A C The remining se is when ABC is right tringle. In this se, the Lw of Sines redues to the formuls given in Theorem 10.4 nd is left to the reder. In order to use the Lw of Sines to solve tringle, we need t lest one ngle-side opposite pir. The next exmple showses some of the power, nd the pitflls, of the Lw of Sines. Exmple Solve the following tringles. Give ext nswers nd deiml pproximtions (rounded to hundredths) nd sketh the tringle. 1. = 10, = 7 units, = 45. = 85, = 30, = 5.5 units 3. = 30, = 1 units, = 4 units 4. = 30, = units, = 4 units 5. = 30, = 3 units, = 4 units 6. = 30, = 4 units, = 4 units Solution. 1. Knowing n ngle-side opposite pir, nmely nd, we my proeed in using the Lw of Sines. Sine = 45, we use sin(45 ) = 7 sin(10 ) so = 7 sin(45 ) sin(10 ) = units. Now tht we hve two ngle-side pirs, it is time to find the third. To find γ, we use the ft tht the sum of the mesures of the ngles in tringle is 180. Hene, γ = = 15. To find, we hve no hoie ut to used the derived vlue γ = 15, yet we n minimize the propgtion of error here y using the given ngle-side opposite pir (, ). The Lw of Sines gives us sin(15 ) = 7 sin(10 ) so tht = 7 sin(15 ) sin(10 ).09 units.5. In this exmple, we re not immeditely given n ngle-side opposite pir, ut s we hve the mesures of nd, we n solve for γ sine γ = = 65. As in the previous exmple, we re fored to use derived vlue in our omputtions sine the only 5 The ext vlue of sin(15 ) ould e found using the differene identity for sine or hlf-ngle formul, ut tht eomes unneessrily messy for the disussion t hnd. Thus ext here mens 7 sin(15 ) sin(10 ).

5 11. The Lw of Sines 899 ngle-side pir ville is (γ, ). The Lw of Sines gives sin(85 ) = 5.5 sin(65 ). After the usul rerrngement, we get = 5.5 sin(85 ) sin(65 ) 5.77 units. To find we use the ngle-side pir (γ, ) whih yields sin(30 ) = 5.5 sin(65 ) hene = 5.5 sin(30 ) sin(65 ).90 units. = 30 = 45 = 7 = = 10 γ = 15 = 85 γ = Tringle for numer 1 Tringle for numer 3. Sine we re given (, ) nd, we use the Lw of Sines to find the mesure of γ. We strt with sin(γ) 4 = sin(30 ) 1 nd get sin(γ) = 4 sin (30 ) =. Sine the rnge of the sine funtion is [ 1, 1], there is no rel numer with sin(γ) =. Geometrilly, we see tht side is just too short to mke tringle. The next three exmples keep the sme vlues for the mesure of nd the length of while vrying the length of. We will disuss this se in more detil fter we see wht hppens in those exmples. 4. In this se, we hve the mesure of = 30, = nd = 4. Using the Lw of Sines, we get sin(γ) 4 = sin(30 ) so sin(γ) = sin (30 ) = 1. Now γ is n ngle in tringle whih lso ontins = 30. This mens tht γ must mesure etween 0 nd 150 in order to fit inside the tringle with. The only ngle tht stisfies this requirement nd hs sin(γ) = 1 is γ = 90. In other words, we hve right tringle. We find the mesure of to e = = 60 nd then determine using the Lw of Sines. We find = sin(60 ) sin(30 ) = units. In this se, the side is preisely long enough to form unique right tringle. = 4 = 1 = 4 = 60 = = 30 = Digrm for numer 3 Tringle for numer 4 5. Proeeding s we hve in the previous two exmples, we use the Lw of Sines to find γ. In this se, we hve sin(γ) 4 = sin(30 ) 3 or sin(γ) = 4 sin(30 ) 3 = 3. Sine γ lies in tringle with = 30,

6 900 Applitions of Trigonometry we must hve tht 0 < γ < 150. There re two ngles γ tht fll in this rnge nd hve sin(γ) = 3 : γ = rsin ( ) 3 rdins nd γ = π rsin ( 3) rdins At this point, we puse to see if it mkes sense tht we tully hve two vile ses to onsider. As we hve disussed, oth ndidtes for γ re omptile with the given ngle-side pir (, ) = (30, 3) in tht oth hoies for γ n fit in tringle with nd oth hve sine of 3. The only other given piee of informtion is tht = 4 units. Sine >, it must e true tht γ, whih is opposite, hs greter mesure thn whih is opposite. In oth ses, γ >, so oth ndidtes for γ re omptile with this lst piee of given informtion s well. Thus hve two tringles on our hnds. In the se γ = rsin ( 3) rdins 41.81, we find = Using the Lw of Sines with the ngle-side opposite pir (, ) nd, we find 3 sin( ) sin(30 ) 5.70 units. In the se γ = π rsin ( 3) rdins , we repet the ext sme steps nd find nd 1.3 units. 7 tringles re drwn elow. Both = 4 = = 3 γ = 4 = 3 = 30 γ For this lst prolem, we repet the usul Lw of Sines routine to find tht sin(γ) 4 = sin(30 ) 4 so tht sin(γ) = 1. Sine γ must inhit tringle with = 30, we must hve 0 < γ < 150. Sine the mesure of γ must e stritly less thn 150, there is just one ngle whih stisfies oth required onditions, nmely γ = 30. So = = 10 nd, using the Lw of Sines one lst time, = 4 sin(10 ) sin(30 ) = units. = 4 = 10 = 4 = 30 γ = Some remrks out Exmple 11.. re in order. We first note tht if we re given the mesures of two of the ngles in tringle, sy nd, the mesure of the third ngle γ is uniquely 6 To find n ext expression for, we onvert everything k to rdins: = 30 = π rdins, γ = rsin ( ) 6 3 rdins nd 180 = π rdins. Hene, = π π rsin ( ) 6 3 = 5π rsin ( ) 6 3 rdins An ext nswer for in this se is = rsin ( ) 3 π rdins

7 11. The Lw of Sines 901 determined using the eqution γ = 180. Knowing the mesures of ll three ngles of tringle ompletely determines its shpe. If in ddition we re given the length of one of the sides of the tringle, we n then use the Lw of Sines to find the lengths of the remining two sides to determine the size of the tringle. Suh is the se in numers 1 nd ove. In numer 1, the given side is djent to just one of the ngles this is lled the Angle-Angle-Side (AAS) se. 8 In numer, the given side is djent to oth ngles whih mens we re in the so-lled Angle-Side-Angle (ASA) se. If, on the other hnd, we re given the mesure of just one of the ngles in the tringle long with the length of two sides, only one of whih is djent to the given ngle, we re in the Angle-Side-Side (ASS) se. 9 In numer 3, the length of the one given side ws too short to even form tringle; in numer 4, the length of ws just long enough to form right tringle; in 5, ws long enough, ut not too long, so tht two tringles were possile; nd in numer 6, side ws long enough to form tringle ut too long to swing k nd form two. These four ses exemplify ll of the possiilities in the Angle-Side-Side se whih re summrized in the following theorem. Theorem Suppose (, ) nd (γ, ) re intended to e ngle-side pirs in tringle where, nd re given. Let h = sin() ˆ ˆ ˆ ˆ If < h, then no tringle exists whih stisfies the given riteri. If = h, then γ = 90 so extly one (right) tringle exists whih stisfies the riteri. If h < <, then two distint tringles exist whih stisfy the given riteri. If, then γ is ute nd extly one tringle exists whih stisfies the given riteri Theorem 11.3 is proved on se-y-se sis. If < h, then < sin(). If tringle were to exist, the Lw of Sines would hve sin(γ) = sin() so tht sin(γ) = sin() > = 1, whih is impossile. In the figure elow, we see geometrilly why this is the se. h = sin() = h = sin() < h, no tringle = h, γ = 90 Simply put, if < h the side is too short to onnet to form tringle. This mens if h, we re lwys gurnteed to hve t lest one tringle, nd the remining prts of the theorem 8 If this sounds fmilir, it should. From high shool Geometry, we know there re four ongruene onditions for tringles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) nd Side-Side-Side (SSS). If we re given informtion out tringle tht meets one of these four riteri, then we re gurnteed tht extly one tringle exists whih stisfies the given riteri. 9 In more reputle ooks, this is lled the Side-Side-Angle or SSA se.

8 90 Applitions of Trigonometry tell us wht kind nd how mny tringles to expet in eh se. If = h, then = sin() nd the Lw of Sines gives sin() = sin(γ) so tht sin(γ) = sin() = = 1. Here, γ = 90 s required. Moving long, now suppose h < <. As efore, the Lw of Sines 10 gives sin(γ) = sin(). Sine h <, sin() < or sin() < 1 whih mens there re two solutions to sin(γ) = sin() : n ute ngle whih we ll ll γ 0, nd its supplement, 180 γ 0. We need to rgue tht eh of these ngles fit into tringle with. Sine (, ) nd (γ 0, ) re ngle-side opposite pirs, the ssumption > in this se gives us γ 0 >. Sine γ 0 is ute, we must hve tht is ute s well. This mens one tringle n ontin oth nd γ 0, giving us one of the tringles promised in the theorem. If we mnipulte the inequlity γ 0 > it, we hve 180 γ 0 < 180 whih gives (180 γ 0 ) + < 180. This proves tringle n ontin oth of the ngles nd (180 γ 0 ), giving us the seond tringle predited in the theorem. To prove the lst se in the theorem, we ssume. Then γ, whih fores γ to e n ute ngle. Hene, we get only one tringle in this se, ompleting the proof. γ 0 γ 0 h h < <, two tringles h γ, one tringle One lst omment efore we use the Lw of Sines to solve n pplition prolem. In the Angle- Side-Side se, if you re given n otuse ngle to egin with then it is impossile to hve the two tringle se. Think out this efore reding further. Exmple Ssquth Islnd lies off the ost of Ippizuti Lke. Two sightings, tken 5 miles prt, re mde to the islnd. The ngle etween the shore nd the islnd t the first oservtion point is 30 nd t the seond point the ngle is 45. Assuming stright ostline, find the distne from the seond oservtion point to the islnd. Wht point on the shore is losest to the islnd? How fr is the islnd from this point? Solution. We sketh the prolem elow with the first oservtion point leled s P nd the seond s Q. In order to use the Lw of Sines to find the distne d from Q to the islnd, we first need to find the mesure of whih is the ngle opposite the side of length 5 miles. To tht end, we note tht the ngles γ nd 45 re supplementl, so tht γ = = 135. We n now find = γ = = 15. By the Lw of Sines, we hve d sin(30 ) = 5 sin(15 ) whih gives d = 5 sin(30 ) sin(15 ) 9.66 miles. Next, to find the point on the ost losest to the islnd, whih we ve leled s C, we need to find the perpendiulr distne from the islnd to the ost Rememer, we hve lredy rgued tht tringle exists in this se! 11 Do you see why C must lie to the right of Q?

9 11. The Lw of Sines 903 Let x denote the distne from the seond oservtion point Q to the point C nd let y denote the distne from C to the islnd. ( Using Theorem 10.4, we get sin (45 ) = y d. After some rerrnging, ) we find y = d sin (45 ) miles. Hene, the islnd is pproximtely 6.83 miles from the ost. To find the distne from Q to C, we note tht = = 45 so y symmetry, 1 we get x = y 6.83 miles. Hene, the point on the shore losest to the islnd is pproximtely 6.83 miles down the ost from the seond oservtion point. Ssquth Islnd Ssquth Islnd d 9.66 miles d 9.66 miles y miles P 30 γ Q 45 Shoreline Q 45 C 5 miles x miles We lose this setion with new formul to ompute the re enlosed y tringle. Its proof uses the sme ses nd digrms s the proof of the Lw of Sines nd is left s n exerise. Theorem Suppose (, ), (, ) nd (γ, ) re the ngle-side opposite pirs of tringle. Then the re A enlosed y the tringle is given y A = 1 sin() = 1 sin() = 1 sin(γ) Exmple Find the re of the tringle in Exmple 11.. numer 1. Solution. From our work in Exmple 11.. numer 1, we hve ll three ngles nd ll three sides to work with. However, to minimize propgted error, we hoose A = 1 sin() from Theorem 11.4 euse it uses the most piees of given informtion. We re ( given ) = 7 nd = 45, nd we lulted = 7 sin(15 ) sin(10 ). Using these vlues, we find A = 1 (7) 7 sin(15 ) sin(10 ) sin (45 ) = 5.18 squre units. The reder is enourged to hek this nswer ginst the results otined using the other formuls in Theorem Or y Theorem 10.4 gin...

10 904 Applitions of Trigonometry Exerises In Exerises 1-0, solve for the remining side(s) nd ngle(s) if possile. As in the text, (, ), (, ) nd (γ, ) re ngle-side opposite pirs. 1. = 13, = 17, = 5. = 73., = 54.1, = = 95, = 85, = = 95, = 6, = = 117, = 35, = 4 6. = 117, = 45, = 4 7. = 68.7, = 88, = 9 8. = 4, = 17, = = 68.7, = 70, = = 30, = 7, = = 4, = 39, = γ = 53, = 53, = = 6, = 57, = γ = 74.6, = 3, = = 10, = 16.75, = = 10, = 16.75, = = 10, γ = 35, = = 9.13, γ = 83.95, = γ = 10, = 61, = 4 0. = 50, = 5, = Find the re of the tringles given in Exerises 1, 1 nd 0 ove. (Another Clssi Applition: Grde of Rod) The grde of rod is muh like the pith of roof (See Exmple ) in tht it expresses the rtio of rise/run. In the se of rod, this rtio is lwys positive euse it is mesured going uphill nd it is usully given s perentge. For exmple, rod whih rises 7 feet for every 100 feet of (horizontl) forwrd progress is sid to hve 7% grde. However, if we wnt to pply ny Trigonometry to story prolem involving rods going uphill or downhill, we need to view the grde s n ngle with respet to the horizontl. In Exerises - 4, we first hve you hnge rod grdes into ngles nd then use the Lw of Sines in n pplition.. Using right tringle with horizontl leg of length 100 nd vertil leg with length 7, show tht 7% grde mens tht the rod (hypotenuse) mkes out 4 ngle with the horizontl. (It will not e extly 4, ut it s pretty lose.) 3. Wht grde is given y 9.65 ngle mde y the rod nd the horizontl? I hve friends who live in Pifi, CA nd their rod is tully this steep. It s not nie rod to drive.

11 11. The Lw of Sines Along long, stright streth of mountin rod with 7% grde, you see tll tree stnding perfetly plum longside the rod. 14 From point 500 feet downhill from the tree, the ngle of inlintion from the rod to the top of the tree is 6. Use the Lw of Sines to find the height of the tree. (Hint: First show tht the tree mkes 94 ngle with the rod.) (Another Clssi Applition: Berings) In the next severl exerises we introdue nd work with the nvigtion tool known s erings. Simply put, ering is the diretion you re heding ording to ompss. The lssi nomenlture for erings, however, is not given s n ngle in stndrd position, so we must first understnd the nottion. A ering is given s n ute ngle of rottion (to the est or to the west) wy from the north-south (up nd down) line of ompss rose. For exmple, N40 E (red 40 est of north ) is ering whih is rotted lokwise 40 from due north. If we imgine stnding t the origin in the Crtesin Plne, this ering would hve us heding into Qudrnt I long the terminl side of θ = 50. Similrly, S50 W would point into Qudrnt III long the terminl side of θ = 0 euse we strted out pointing due south (long θ = 70 ) nd rotted lokwise 50 k to 0. Counter-lokwise rottions would e found in the erings N60 W (whih is on the terminl side of θ = 150 ) nd S7 E (whih lies long the terminl side of θ = 97 ). These four erings re drwn in the plne elow. N60 W 60 N 40 N40 E W E S50 W 50 S 7 S7 E The rdinl diretions north, south, est nd west re usully not given s erings in the fshion desried ove, ut rther, one just refers to them s due north, due south, due est nd due west, respetively, nd it is ssumed tht you know whih qudrntl ngle goes with eh rdinl diretion. (Hint: Look t the digrm ove.) 5. Find the ngle θ in stndrd position with 0 θ < 360 whih orresponds to eh of the erings given elow. () due west () S83 E () N5.5 E (d) due south 14 The word plum here mens tht the tree is perpendiulr to the horizontl.

12 906 Applitions of Trigonometry (e) N31.5 W (f) S W 15 (g) N45 E (h) S45 W 6. The Colonel spots mpfire t of ering N4 E from his urrent position. Srge, who is positioned 3000 feet due est of the Colonel, rekons the ering to the fire to e N0 W from his urrent position. Determine the distne from the mpfire to eh mn, rounded to the nerest foot. 7. A hiker strts wlking due west from Ssquth Point nd gets to the Chupr Trilhed efore she relizes tht she hsn t reset her pedometer. From the Chupr Trilhed she hikes for 5 miles long ering of N53 W whih rings her to the Muffin Ridge Oservtory. From there, she knows ering of S65 E will tke her stright k to Ssquth Point. How fr will she hve to wlk to get from the Muffin Ridge Oservtory to Ssquh Point? Wht is the distne etween Ssquth Point nd the Chupr Trilhed? 8. The ptin of the SS Bigfoot sees signl flre t ering of N15 E from her urrent lotion. From his position, the ptin of the HMS Ssquth finds the signl flre to e t ering of N75 W. If the SS Bigfoot is 5 miles from the HMS Ssquth nd the ering from the SS Bigfoot to the HMS Ssquth is N50 E, find the distnes from the flre to eh vessel, rounded to the nerest tenth of mile. 9. Crl spies potentil Ssquth nest t ering of N10 E nd rdios Jeff, who is t ering of N50 E from Crl s position. From Jeff s position, the nest is t ering of S70 W. If Jeff nd Crl re 500 feet prt, how fr is Jeff from the Ssquth nest? Round your nswer to the nerest foot. 30. A hiker determines the ering to lodge from her urrent position is S40 W. She proeeds to hike miles t ering of S0 E t whih point she determines the ering to the lodge is S75 W. How fr is she from the lodge t this point? Round your nswer to the nerest hundredth of mile. 31. A wthtower spots ship off shore t ering of N70 E. A seond tower, whih is 50 miles from the first t ering of S80 E from the first tower, determines the ering to the ship to e N5 W. How fr is the ot from the seond tower? Round your nswer to the nerest tenth of mile. 3. Skippy nd Slly deide to hunt UFOs. One night, they position themselves miles prt on n ndoned streth of desert runwy. An hour into their investigtion, Skippy spies UFO hovering over spot on the runwy diretly etween him nd Slly. He reords the ngle of inlintion from the ground to the rft to e 75 nd rdios Slly immeditely to find the ngle of inlintion from her position to the rft is 50. How high off the ground is the UFO t this point? Round your nswer to the nerest foot. (Rell: 1 mile is 580 feet.) 15 See Exmple in Setion 10.1 for review of the DMS system.

13 11. The Lw of Sines The ngle of depression from n oserver in n prtment omplex to grgoyle on the uilding next door is 55. From point five stories elow the originl oserver, the ngle of inlintion to the grgoyle is 0. Find the distne from eh oserver to the grgoyle nd the distne from the grgoyle to the prtment omplex. Round your nswers to the nerest foot. (Use the rule of thum tht one story of uilding is 9 feet.) 34. Prove tht the Lw of Sines holds when ABC is right tringle. 35. Disuss with your lssmtes why knowing only the three ngles of tringle is not enough to determine ny of the sides. 36. Disuss with your lssmtes why the Lw of Sines nnot e used to find the ngles in the tringle when only the three sides re given. Also disuss wht hppens if only two sides nd the ngle etween them re given. (Sid nother wy, explin why the Lw of Sines nnot e used in the SSS nd SAS ses.) 37. Given = 30 nd = 10, hoose four different vlues for so tht () the informtion yields no tringle () the informtion yields extly one right tringle () the informtion yields two distint tringles (d) the informtion yields extly one otuse tringle Explin why you nnot hoose in suh wy s to hve = 30, = 10 nd your hoie of yield only one tringle where tht unique tringle hs three ute ngles. 38. Use the ses nd digrms in the proof of the Lw of Sines (Theorem 11.) to prove the re formuls given in Theorem Why do those formuls yield squre units when four quntities re eing multiplied together?

14 908 Applitions of Trigonometry 11.. Answers 1. = 13 = 17 γ = 150 = = 73. = 54.1 γ = 5.7 = Informtion does not produe tringle 4. = 95 = 6 γ = 3 = Informtion does not produe tringle 6. = γ 6.7 = 45 = = γ 34.4 = 88 = = γ = 17 = = γ 8. = 88 = = γ 5.66 = 17 = Informtion does not produe tringle 10. = 30 = 90 γ = 60 = 7 = 14 = = γ 114. = 39 = = 53 = 74 γ = 53 = = = γ 4.57 = 57 = γ = 74.6 = = 3 = γ = 57 = γ = 74.6 = = = 10 γ = = Informtion does not produe tringle 17. = 43 = 10 γ = = = 66.9 = 9.13 γ = = Informtion does not produe tringle 0. = 50.5 γ = 5 = The re of the tringle from Exerise 1 is out 8.1 squre units. The re of the tringle from Exerise 1 is out squre units. The re of the tringle from Exerise 0 is out 149 squre units.. rtn ( ) rdins, whih is equivlent to Aout 17% 4. Aout 53 feet

15 11. The Lw of Sines () θ = 180 () θ = 353 () θ = 84.5 (d) θ = 70 (e) θ = 11.5 (f) θ = (g) θ = 45 (h) θ = 5 6. The Colonel is out 3193 feet from the mpfire. Srge is out 55 feet to the mpfire. 7. The distne from the Muffin Ridge Oservtory to Ssquh Point is out 7.1 miles. The distne from Ssquth Point to the Chupr Trilhed is out.46 miles. 8. The SS Bigfoot is out 4.1 miles from the flre. The HMS Ssquth is out.9 miles from the flre. 9. Jeff is out 371 feet from the nest. 30. She is out 3.0 miles from the lodge 31. The ot is out 5.1 miles from the seond tower. 3. The UFO is hovering out 9539 feet ove the ground. 33. The grgoyle is out 44 feet from the oserver on the upper floor. The grgoyle is out 7 feet from the oserver on the lower floor. The grgoyle is out 5 feet from the other uilding.

PYTHAGORAS THEOREM WHAT S IN CHAPTER 1? IN THIS CHAPTER YOU WILL:

PYTHAGORAS THEOREM WHAT S IN CHAPTER 1? IN THIS CHAPTER YOU WILL: PYTHAGORAS THEOREM 1 WHAT S IN CHAPTER 1? 1 01 Squres, squre roots nd surds 1 02 Pythgors theorem 1 03 Finding the hypotenuse 1 04 Finding shorter side 1 05 Mixed prolems 1 06 Testing for right-ngled tringles

More information

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the

More information

Section 1.3 Triangles

Section 1.3 Triangles Se 1.3 Tringles 21 Setion 1.3 Tringles LELING TRINGLE The line segments tht form tringle re lled the sides of the tringle. Eh pir of sides forms n ngle, lled n interior ngle, nd eh tringle hs three interior

More information

Date Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( )

Date Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( ) UNIT 5 TRIGONOMETRI RTIOS Dte Lesson Text TOPI Homework pr. 4 5.1 (48) Trigonometry Review WS 5.1 # 3 5, 9 11, (1, 13)doso pr. 6 5. (49) Relted ngles omplete lesson shell & WS 5. pr. 30 5.3 (50) 5.3 5.4

More information

Non Right Angled Triangles

Non Right Angled Triangles Non Right ngled Tringles Non Right ngled Tringles urriulum Redy www.mthletis.om Non Right ngled Tringles NON RIGHT NGLED TRINGLES sin i, os i nd tn i re lso useful in non-right ngled tringles. This unit

More information

Lesson 2: The Pythagorean Theorem and Similar Triangles. A Brief Review of the Pythagorean Theorem.

Lesson 2: The Pythagorean Theorem and Similar Triangles. A Brief Review of the Pythagorean Theorem. 27 Lesson 2: The Pythgoren Theorem nd Similr Tringles A Brief Review of the Pythgoren Theorem. Rell tht n ngle whih mesures 90º is lled right ngle. If one of the ngles of tringle is right ngle, then we

More information

Comparing the Pre-image and Image of a Dilation

Comparing the Pre-image and Image of a Dilation hpter Summry Key Terms Postultes nd Theorems similr tringles (.1) inluded ngle (.2) inluded side (.2) geometri men (.) indiret mesurement (.6) ngle-ngle Similrity Theorem (.2) Side-Side-Side Similrity

More information

2.1 ANGLES AND THEIR MEASURE. y I

2.1 ANGLES AND THEIR MEASURE. y I .1 ANGLES AND THEIR MEASURE Given two interseting lines or line segments, the mount of rottion out the point of intersetion (the vertex) required to ring one into orrespondene with the other is lled the

More information

Maintaining Mathematical Proficiency

Maintaining Mathematical Proficiency Nme Dte hpter 9 Mintining Mthemtil Profiieny Simplify the epression. 1. 500. 189 3. 5 4. 4 3 5. 11 5 6. 8 Solve the proportion. 9 3 14 7. = 8. = 9. 1 7 5 4 = 4 10. 0 6 = 11. 7 4 10 = 1. 5 9 15 3 = 5 +

More information

Activities. 4.1 Pythagoras' Theorem 4.2 Spirals 4.3 Clinometers 4.4 Radar 4.5 Posting Parcels 4.6 Interlocking Pipes 4.7 Sine Rule Notes and Solutions

Activities. 4.1 Pythagoras' Theorem 4.2 Spirals 4.3 Clinometers 4.4 Radar 4.5 Posting Parcels 4.6 Interlocking Pipes 4.7 Sine Rule Notes and Solutions MEP: Demonstrtion Projet UNIT 4: Trigonometry UNIT 4 Trigonometry tivities tivities 4. Pythgors' Theorem 4.2 Spirls 4.3 linometers 4.4 Rdr 4.5 Posting Prels 4.6 Interloking Pipes 4.7 Sine Rule Notes nd

More information

Something found at a salad bar

Something found at a salad bar Nme PP Something found t sld r 4.7 Notes RIGHT TRINGLE hs extly one right ngle. To solve right tringle, you n use things like SOH-H-TO nd the Pythgoren Theorem. n OLIQUE TRINGLE hs no right ngles. To solve

More information

PYTHAGORAS THEOREM,TRIGONOMETRY,BEARINGS AND THREE DIMENSIONAL PROBLEMS

PYTHAGORAS THEOREM,TRIGONOMETRY,BEARINGS AND THREE DIMENSIONAL PROBLEMS PYTHGORS THEOREM,TRIGONOMETRY,ERINGS ND THREE DIMENSIONL PROLEMS 1.1 PYTHGORS THEOREM: 1. The Pythgors Theorem sttes tht the squre of the hypotenuse is equl to the sum of the squres of the other two sides

More information

2. There are an infinite number of possible triangles, all similar, with three given angles whose sum is 180.

2. There are an infinite number of possible triangles, all similar, with three given angles whose sum is 180. SECTION 8-1 11 CHAPTER 8 Setion 8 1. There re n infinite numer of possile tringles, ll similr, with three given ngles whose sum is 180. 4. If two ngles α nd β of tringle re known, the third ngle n e found

More information

Trigonometry Revision Sheet Q5 of Paper 2

Trigonometry Revision Sheet Q5 of Paper 2 Trigonometry Revision Sheet Q of Pper The Bsis - The Trigonometry setion is ll out tringles. We will normlly e given some of the sides or ngles of tringle nd we use formule nd rules to find the others.

More information

Math Lesson 4-5 The Law of Cosines

Math Lesson 4-5 The Law of Cosines Mth-1060 Lesson 4-5 The Lw of osines Solve using Lw of Sines. 1 17 11 5 15 13 SS SSS Every pir of loops will hve unknowns. Every pir of loops will hve unknowns. We need nother eqution. h Drop nd ltitude

More information

THE PYTHAGOREAN THEOREM

THE PYTHAGOREAN THEOREM THE PYTHAGOREAN THEOREM The Pythgoren Theorem is one of the most well-known nd widely used theorems in mthemtis. We will first look t n informl investigtion of the Pythgoren Theorem, nd then pply this

More information

Lesson 8.1 Graphing Parametric Equations

Lesson 8.1 Graphing Parametric Equations Lesson 8.1 Grphing Prmetric Equtions 1. rete tle for ech pir of prmetric equtions with the given vlues of t.. x t 5. x t 3 c. x t 1 y t 1 y t 3 y t t t {, 1, 0, 1, } t {4,, 0,, 4} t {4, 0,, 4, 8}. Find

More information

Proving the Pythagorean Theorem

Proving the Pythagorean Theorem Proving the Pythgoren Theorem W. Bline Dowler June 30, 2010 Astrt Most people re fmilir with the formul 2 + 2 = 2. However, in most ses, this ws presented in lssroom s n solute with no ttempt t proof or

More information

Trigonometry and Constructive Geometry

Trigonometry and Constructive Geometry Trigonometry nd Construtive Geometry Trining prolems for M2 2018 term 1 Ted Szylowie tedszy@gmil.om 1 Leling geometril figures 1. Prtie writing Greek letters. αβγδɛθλµπψ 2. Lel the sides, ngles nd verties

More information

Intermediate Math Circles Wednesday 17 October 2012 Geometry II: Side Lengths

Intermediate Math Circles Wednesday 17 October 2012 Geometry II: Side Lengths Intermedite Mth Cirles Wednesdy 17 Otoer 01 Geometry II: Side Lengths Lst week we disussed vrious ngle properties. As we progressed through the evening, we proved mny results. This week, we will look t

More information

A Study on the Properties of Rational Triangles

A Study on the Properties of Rational Triangles Interntionl Journl of Mthemtis Reserh. ISSN 0976-5840 Volume 6, Numer (04), pp. 8-9 Interntionl Reserh Pulition House http://www.irphouse.om Study on the Properties of Rtionl Tringles M. Q. lm, M.R. Hssn

More information

GM1 Consolidation Worksheet

GM1 Consolidation Worksheet Cmridge Essentils Mthemtis Core 8 GM1 Consolidtion Worksheet GM1 Consolidtion Worksheet 1 Clulte the size of eh ngle mrked y letter. Give resons for your nswers. or exmple, ngles on stright line dd up

More information

LESSON 11: TRIANGLE FORMULAE

LESSON 11: TRIANGLE FORMULAE . THE SEMIPERIMETER OF TRINGLE LESSON : TRINGLE FORMULE In wht follows, will hve sides, nd, nd these will e opposite ngles, nd respetively. y the tringle inequlity, nd..() So ll of, & re positive rel numers.

More information

Part I: Study the theorem statement.

Part I: Study the theorem statement. Nme 1 Nme 2 Nme 3 A STUDY OF PYTHAGORAS THEOREM Instrutions: Together in groups of 2 or 3, fill out the following worksheet. You my lift nswers from the reding, or nswer on your own. Turn in one pket for

More information

12.4 Similarity in Right Triangles

12.4 Similarity in Right Triangles Nme lss Dte 12.4 Similrit in Right Tringles Essentil Question: How does the ltitude to the hpotenuse of right tringle help ou use similr right tringles to solve prolems? Eplore Identifing Similrit in Right

More information

Project 6: Minigoals Towards Simplifying and Rewriting Expressions

Project 6: Minigoals Towards Simplifying and Rewriting Expressions MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy

More information

Similar Right Triangles

Similar Right Triangles Geometry V1.noteook Ferury 09, 2012 Similr Right Tringles Cn I identify similr tringles in right tringle with the ltitude? Cn I identify the proportions in right tringles? Cn I use the geometri mens theorems

More information

3.1 Review of Sine, Cosine and Tangent for Right Angles

3.1 Review of Sine, Cosine and Tangent for Right Angles Foundtions of Mth 11 Section 3.1 Review of Sine, osine nd Tngent for Right Tringles 125 3.1 Review of Sine, osine nd Tngent for Right ngles The word trigonometry is derived from the Greek words trigon,

More information

Ellipses. The second type of conic is called an ellipse.

Ellipses. The second type of conic is called an ellipse. Ellipses The seond type of oni is lled n ellipse. Definition of Ellipse An ellipse is the set of ll points (, y) in plne, the sum of whose distnes from two distint fied points (foi) is onstnt. (, y) d

More information

Introduction to Olympiad Inequalities

Introduction to Olympiad Inequalities Introdution to Olympid Inequlities Edutionl Studies Progrm HSSP Msshusetts Institute of Tehnology Snj Simonovikj Spring 207 Contents Wrm up nd Am-Gm inequlity 2. Elementry inequlities......................

More information

HS Pre-Algebra Notes Unit 9: Roots, Real Numbers and The Pythagorean Theorem

HS Pre-Algebra Notes Unit 9: Roots, Real Numbers and The Pythagorean Theorem HS Pre-Alger Notes Unit 9: Roots, Rel Numers nd The Pythgoren Theorem Roots nd Cue Roots Syllus Ojetive 5.4: The student will find or pproximte squre roots of numers to 4. CCSS 8.EE.-: Evlute squre roots

More information

1.3 SCALARS AND VECTORS

1.3 SCALARS AND VECTORS Bridge Course Phy I PUC 24 1.3 SCLRS ND VECTORS Introdution: Physis is the study of nturl phenomen. The study of ny nturl phenomenon involves mesurements. For exmple, the distne etween the plnet erth nd

More information

3 Angle Geometry. 3.1 Measuring Angles. 1. Using a protractor, measure the marked angles.

3 Angle Geometry. 3.1 Measuring Angles. 1. Using a protractor, measure the marked angles. 3 ngle Geometry MEP Prtie ook S3 3.1 Mesuring ngles 1. Using protrtor, mesure the mrked ngles. () () (d) (e) (f) 2. Drw ngles with the following sizes. () 22 () 75 120 (d) 90 (e) 153 (f) 45 (g) 180 (h)

More information

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle. Mth 3329-Uniform Geometries Leture 06 1. Review of trigonometry While we re looking t Eulid s Elements, I d like to look t some si trigonometry. Figure 1. The Pythgoren theorem sttes tht if = 90, then

More information

CHENG Chun Chor Litwin The Hong Kong Institute of Education

CHENG Chun Chor Litwin The Hong Kong Institute of Education PE-hing Mi terntionl onferene IV: novtion of Mthemtis Tehing nd Lerning through Lesson Study- onnetion etween ssessment nd Sujet Mtter HENG hun hor Litwin The Hong Kong stitute of Edution Report on using

More information

Geometry of the Circle - Chords and Angles. Geometry of the Circle. Chord and Angles. Curriculum Ready ACMMG: 272.

Geometry of the Circle - Chords and Angles. Geometry of the Circle. Chord and Angles. Curriculum Ready ACMMG: 272. Geometry of the irle - hords nd ngles Geometry of the irle hord nd ngles urriulum Redy MMG: 272 www.mthletis.om hords nd ngles HRS N NGLES The irle is si shpe nd so it n e found lmost nywhere. This setion

More information

= x x 2 = 25 2

= x x 2 = 25 2 9.1 Wrm Up Solve the eqution. 1. 4 2 + 3 2 = x 2 2. 13 2 + x 2 = 25 2 3. 3 2 2 + x 2 = 5 2 2 4. 5 2 + x 2 = 12 2 Mrh 7, 2016 Geometry 9.1 The Pythgoren Theorem 1 Geometry 9.1 The Pythgoren Theorem 9.1

More information

INTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable

INTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable INTEGRATION NOTE: These notes re supposed to supplement Chpter 4 of the online textbook. 1 Integrls of Complex Vlued funtions of REAL vrible If I is n intervl in R (for exmple I = [, b] or I = (, b)) nd

More information

Discrete Structures Lecture 11

Discrete Structures Lecture 11 Introdution Good morning. In this setion we study funtions. A funtion is mpping from one set to nother set or, perhps, from one set to itself. We study the properties of funtions. A mpping my not e funtion.

More information

Naming the sides of a right-angled triangle

Naming the sides of a right-angled triangle 6.2 Wht is trigonometry? The word trigonometry is derived from the Greek words trigonon (tringle) nd metron (mesurement). Thus, it literlly mens to mesure tringle. Trigonometry dels with the reltionship

More information

9.1 Day 1 Warm Up. Solve the equation = x x 2 = March 1, 2017 Geometry 9.1 The Pythagorean Theorem 1

9.1 Day 1 Warm Up. Solve the equation = x x 2 = March 1, 2017 Geometry 9.1 The Pythagorean Theorem 1 9.1 Dy 1 Wrm Up Solve the eqution. 1. 4 2 + 3 2 = x 2 2. 13 2 + x 2 = 25 2 3. 3 2 2 + x 2 = 5 2 2 4. 5 2 + x 2 = 12 2 Mrh 1, 2017 Geometry 9.1 The Pythgoren Theorem 1 9.1 Dy 2 Wrm Up Use the Pythgoren

More information

MATHEMATICS AND STATISTICS 1.6

MATHEMATICS AND STATISTICS 1.6 MTHMTIS N STTISTIS 1.6 pply geometri resoning in solving prolems ternlly ssessed 4 redits S 91031 inding unknown ngles When finding the size of unknown ngles in figure, t lest two steps of resoning will

More information

Pythagoras Theorem. The area of the square on the hypotenuse is equal to the sum of the squares on the other two sides

Pythagoras Theorem. The area of the square on the hypotenuse is equal to the sum of the squares on the other two sides Pythgors theorem nd trigonometry Pythgors Theorem The hypotenuse of right-ngled tringle is the longest side The hypotenuse is lwys opposite the right-ngle 2 = 2 + 2 or 2 = 2-2 or 2 = 2-2 The re of the

More information

Part 4. Integration (with Proofs)

Part 4. Integration (with Proofs) Prt 4. Integrtion (with Proofs) 4.1 Definition Definition A prtition P of [, b] is finite set of points {x 0, x 1,..., x n } with = x 0 < x 1

More information

April 8, 2017 Math 9. Geometry. Solving vector problems. Problem. Prove that if vectors and satisfy, then.

April 8, 2017 Math 9. Geometry. Solving vector problems. Problem. Prove that if vectors and satisfy, then. pril 8, 2017 Mth 9 Geometry Solving vetor prolems Prolem Prove tht if vetors nd stisfy, then Solution 1 onsider the vetor ddition prllelogrm shown in the Figure Sine its digonls hve equl length,, the prllelogrm

More information

SECTION A STUDENT MATERIAL. Part 1. What and Why.?

SECTION A STUDENT MATERIAL. Part 1. What and Why.? SECTION A STUDENT MATERIAL Prt Wht nd Wh.? Student Mteril Prt Prolem n > 0 n > 0 Is the onverse true? Prolem If n is even then n is even. If n is even then n is even. Wht nd Wh? Eploring Pure Mths Are

More information

m A 1 1 A ! and AC 6

m A 1 1 A ! and AC 6 REVIEW SET A Using sle of m represents units, sketh vetor to represent: NON-CALCULATOR n eroplne tking off t n ngle of 8 ± to runw with speed of 6 ms displement of m in north-esterl diretion. Simplif:

More information

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into

More information

Matrices SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics (c) 1. Definition of a Matrix

Matrices SCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics (c) 1. Definition of a Matrix tries Definition of tri mtri is regulr rry of numers enlosed inside rkets SCHOOL OF ENGINEERING & UIL ENVIRONEN Emple he following re ll mtries: ), ) 9, themtis ), d) tries Definition of tri Size of tri

More information

Applications of Trigonometry: Triangles and Vectors

Applications of Trigonometry: Triangles and Vectors 7 Applitions of Trigonometry: Tringles nd Vetors Norfolk, Virgini Atlnti Oen Bermud Bermud In reent dedes, mny people hve ome to elieve tht n imginry re lled the Bermud Tringle, loted off the southestern

More information

p-adic Egyptian Fractions

p-adic Egyptian Fractions p-adic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Set-up 3 4 p-greedy Algorithm 5 5 p-egyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction

More information

1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE

1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check

More information

Probability. b a b. a b 32.

Probability. b a b. a b 32. Proility If n event n hppen in '' wys nd fil in '' wys, nd eh of these wys is eqully likely, then proility or the hne, or its hppening is, nd tht of its filing is eg, If in lottery there re prizes nd lnks,

More information

Symmetrical Components 1

Symmetrical Components 1 Symmetril Components. Introdution These notes should e red together with Setion. of your text. When performing stedy-stte nlysis of high voltge trnsmission systems, we mke use of the per-phse equivlent

More information

Tutorial Worksheet. 1. Find all solutions to the linear system by following the given steps. x + 2y + 3z = 2 2x + 3y + z = 4.

Tutorial Worksheet. 1. Find all solutions to the linear system by following the given steps. x + 2y + 3z = 2 2x + 3y + z = 4. Mth 5 Tutoril Week 1 - Jnury 1 1 Nme Setion Tutoril Worksheet 1. Find ll solutions to the liner system by following the given steps x + y + z = x + y + z = 4. y + z = Step 1. Write down the rgumented mtrix

More information

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3 2 The Prllel Circuit Electric Circuits: Figure 2- elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is

More information

Precalculus Notes: Unit 6 Law of Sines & Cosines, Vectors, & Complex Numbers. A can be rewritten as

Precalculus Notes: Unit 6 Law of Sines & Cosines, Vectors, & Complex Numbers. A can be rewritten as Dte: 6.1 Lw of Sines Syllus Ojetie: 3.5 Te student will sole pplition prolems inoling tringles (Lw of Sines). Deriing te Lw of Sines: Consider te two tringles. C C In te ute tringle, sin In te otuse tringle,

More information

The University of Nottingham SCHOOL OF COMPUTER SCIENCE A LEVEL 2 MODULE, SPRING SEMESTER MACHINES AND THEIR LANGUAGES ANSWERS

The University of Nottingham SCHOOL OF COMPUTER SCIENCE A LEVEL 2 MODULE, SPRING SEMESTER MACHINES AND THEIR LANGUAGES ANSWERS The University of ottinghm SCHOOL OF COMPUTR SCIC A LVL 2 MODUL, SPRIG SMSTR 2015 2016 MACHIS AD THIR LAGUAGS ASWRS Time llowed TWO hours Cndidtes my omplete the front over of their nswer ook nd sign their

More information

Plotting Ordered Pairs Using Integers

Plotting Ordered Pairs Using Integers SAMPLE Plotting Ordered Pirs Using Integers Ple two elsti nds on geoord to form oordinte xes shown on the right to help you solve these prolems.. Wht letter of the lphet does eh set of pirs nme?. (, )

More information

Applications of trigonometry

Applications of trigonometry 3 3 3 3 3D 3E 3F 3G 3H Review of right-ngled tringles erings Using the sine rule to find side lengths Using the sine rule to find ngles re of tringle Using the osine rule to find side lengths Using the

More information

Section 4.4. Green s Theorem

Section 4.4. Green s Theorem The Clulus of Funtions of Severl Vriles Setion 4.4 Green s Theorem Green s theorem is n exmple from fmily of theorems whih onnet line integrls (nd their higher-dimensionl nlogues) with the definite integrls

More information

Arrow s Impossibility Theorem

Arrow s Impossibility Theorem Rep Voting Prdoxes Properties Arrow s Theorem Arrow s Impossiility Theorem Leture 12 Arrow s Impossiility Theorem Leture 12, Slide 1 Rep Voting Prdoxes Properties Arrow s Theorem Leture Overview 1 Rep

More information

Review Topic 14: Relationships between two numerical variables

Review Topic 14: Relationships between two numerical variables Review Topi 14: Reltionships etween two numeril vriles Multiple hoie 1. Whih of the following stterplots est demonstrtes line of est fit? A B C D E 2. The regression line eqution for the following grph

More information

Pythagoras Theorem. Pythagoras Theorem. Curriculum Ready ACMMG: 222, 245.

Pythagoras Theorem. Pythagoras Theorem. Curriculum Ready ACMMG: 222, 245. Pythgors Theorem Pythgors Theorem Curriulum Redy ACMMG:, 45 www.mthletis.om Fill in these spes with ny other interesting fts you n find out Pythgors. In the world of Mthemtis, Pythgors is legend. He lived

More information

CS 573 Automata Theory and Formal Languages

CS 573 Automata Theory and Formal Languages Non-determinism Automt Theory nd Forml Lnguges Professor Leslie Lnder Leture # 3 Septemer 6, 2 To hieve our gol, we need the onept of Non-deterministi Finite Automton with -moves (NFA) An NFA is tuple

More information

Linear Algebra Introduction

Linear Algebra Introduction Introdution Wht is Liner Alger out? Liner Alger is rnh of mthemtis whih emerged yers k nd ws one of the pioneer rnhes of mthemtis Though, initilly it strted with solving of the simple liner eqution x +

More information

The Ellipse. is larger than the other.

The Ellipse. is larger than the other. The Ellipse Appolonius of Perg (5 B.C.) disovered tht interseting right irulr one ll the w through with plne slnted ut is not perpendiulr to the is, the intersetion provides resulting urve (oni setion)

More information

1 cos. cos cos cos cos MAT 126H Solutions Take-Home Exam 4. Problem 1

1 cos. cos cos cos cos MAT 126H Solutions Take-Home Exam 4. Problem 1 MAT 16H Solutions Tke-Home Exm 4 Problem 1 ) & b) Using the hlf-ngle formul for cosine, we get: 1 cos 1 4 4 cos cos 8 4 nd 1 8 cos cos 16 4 c) Using the hlf-ngle formul for tngent, we get: cot ( 3π 1 )

More information

Chapter 4 State-Space Planning

Chapter 4 State-Space Planning Leture slides for Automted Plnning: Theory nd Prtie Chpter 4 Stte-Spe Plnning Dn S. Nu CMSC 722, AI Plnning University of Mrylnd, Spring 2008 1 Motivtion Nerly ll plnning proedures re serh proedures Different

More information

Spacetime and the Quantum World Questions Fall 2010

Spacetime and the Quantum World Questions Fall 2010 Spetime nd the Quntum World Questions Fll 2010 1. Cliker Questions from Clss: (1) In toss of two die, wht is the proility tht the sum of the outomes is 6? () P (x 1 + x 2 = 6) = 1 36 - out 3% () P (x 1

More information

Alg 3 Ch 7.2, 8 1. C 2) If A = 30, and C = 45, a = 1 find b and c A

Alg 3 Ch 7.2, 8 1. C 2) If A = 30, and C = 45, a = 1 find b and c A lg 3 h 7.2, 8 1 7.2 Right Tringle Trig ) Use of clcultor sin 10 = sin x =.4741 c ) rete right tringles π 1) If = nd = 25, find 6 c 2) If = 30, nd = 45, = 1 find nd c 3) If in right, with right ngle t,

More information

03. Early Greeks & Aristotle

03. Early Greeks & Aristotle 03. Erly Greeks & Aristotle I. Erly Greeks Topis I. Erly Greeks II. The Method of Exhustion III. Aristotle. Anximnder (. 60 B.C.) to peiron - the unlimited, unounded - fundmentl sustne of relity - underlying

More information

6.5 Improper integrals

6.5 Improper integrals Eerpt from "Clulus" 3 AoPS In. www.rtofprolemsolving.om 6.5. IMPROPER INTEGRALS 6.5 Improper integrls As we ve seen, we use the definite integrl R f to ompute the re of the region under the grph of y =

More information

QUADRATIC EQUATION. Contents

QUADRATIC EQUATION. Contents QUADRATIC EQUATION Contents Topi Pge No. Theory 0-04 Exerise - 05-09 Exerise - 09-3 Exerise - 3 4-5 Exerise - 4 6 Answer Key 7-8 Syllus Qudrti equtions with rel oeffiients, reltions etween roots nd oeffiients,

More information

Instructions to students: Use your Text Book and attempt these questions.

Instructions to students: Use your Text Book and attempt these questions. Instrutions to students: Use your Text Book nd ttempt these questions. Due Dte: 16-09-2018 Unit 2 Chpter 8 Test Slrs nd vetors Totl mrks 50 Nme: Clss: Dte: Setion A Selet the est nswer for eh question.

More information

8 THREE PHASE A.C. CIRCUITS

8 THREE PHASE A.C. CIRCUITS 8 THREE PHSE.. IRUITS The signls in hpter 7 were sinusoidl lternting voltges nd urrents of the so-lled single se type. n emf of suh type n e esily generted y rotting single loop of ondutor (or single winding),

More information

Lecture 6: Coding theory

Lecture 6: Coding theory Leture 6: Coing theory Biology 429 Crl Bergstrom Ferury 4, 2008 Soures: This leture loosely follows Cover n Thoms Chpter 5 n Yeung Chpter 3. As usul, some of the text n equtions re tken iretly from those

More information

Green s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e

Green s Theorem. (2x e y ) da. (2x e y ) dx dy. x 2 xe y. (1 e y ) dy. y=1. = y e y. y=0. = 2 e Green s Theorem. Let be the boundry of the unit squre, y, oriented ounterlokwise, nd let F be the vetor field F, y e y +, 2 y. Find F d r. Solution. Let s write P, y e y + nd Q, y 2 y, so tht F P, Q. Let

More information

Arrow s Impossibility Theorem

Arrow s Impossibility Theorem Rep Fun Gme Properties Arrow s Theorem Arrow s Impossiility Theorem Leture 12 Arrow s Impossiility Theorem Leture 12, Slide 1 Rep Fun Gme Properties Arrow s Theorem Leture Overview 1 Rep 2 Fun Gme 3 Properties

More information

Section 6: Area, Volume, and Average Value

Section 6: Area, Volume, and Average Value Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find

More information

A study of Pythagoras Theorem

A study of Pythagoras Theorem CHAPTER 19 A study of Pythgors Theorem Reson is immortl, ll else mortl. Pythgors, Diogenes Lertius (Lives of Eminent Philosophers) Pythgors Theorem is proly the est-known mthemticl theorem. Even most nonmthemticins

More information

CS311 Computational Structures Regular Languages and Regular Grammars. Lecture 6

CS311 Computational Structures Regular Languages and Regular Grammars. Lecture 6 CS311 Computtionl Strutures Regulr Lnguges nd Regulr Grmmrs Leture 6 1 Wht we know so fr: RLs re losed under produt, union nd * Every RL n e written s RE, nd every RE represents RL Every RL n e reognized

More information

VECTOR ALGEBRA. Syllabus :

VECTOR ALGEBRA. Syllabus : MV VECTOR ALGEBRA Syllus : Vetors nd Slrs, ddition of vetors, omponent of vetor, omponents of vetor in two dimensions nd three dimensionl spe, slr nd vetor produts, slr nd vetor triple produt. Einstein

More information

Vectors. a Write down the vector AB as a column vector ( x y ). A (3, 2) x point C such that BC = 3. . Go to a OA = a

Vectors. a Write down the vector AB as a column vector ( x y ). A (3, 2) x point C such that BC = 3. . Go to a OA = a Streth lesson: Vetors Streth ojetives efore you strt this hpter, mrk how onfident you feel out eh of the sttements elow: I n lulte using olumn vetors nd represent the sum nd differene of two vetors grphilly.

More information

(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.

(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P. Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time

More information

Algorithm Design and Analysis

Algorithm Design and Analysis Algorithm Design nd Anlysis LECTURE 5 Supplement Greedy Algorithms Cont d Minimizing lteness Ching (NOT overed in leture) Adm Smith 9/8/10 A. Smith; sed on slides y E. Demine, C. Leiserson, S. Rskhodnikov,

More information

Math 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1)

Math 32B Discussion Session Week 8 Notes February 28 and March 2, f(b) f(a) = f (t)dt (1) Green s Theorem Mth 3B isussion Session Week 8 Notes Februry 8 nd Mrh, 7 Very shortly fter you lerned how to integrte single-vrible funtions, you lerned the Fundmentl Theorem of lulus the wy most integrtion

More information

SAMPLE. Trigonometry. Naming the sides of a right-angled triangle

SAMPLE. Trigonometry. Naming the sides of a right-angled triangle H P T E R 7 Trigonometry How re sin, os nd tn defined using right-ngled tringle? How n the trigonometri rtios e used to find the side lengths or ngles in right-ngled tringles? Wht is ment y n ngle of elevtion

More information

Unit 6 Solving Oblique Triangles - Classwork

Unit 6 Solving Oblique Triangles - Classwork Unit 6 Solving Oblique Tringles - Clsswork A. The Lw of Sines ASA nd AAS In geometry, we lerned to prove congruence of tringles tht is when two tringles re exctly the sme. We used severl rules to prove

More information

For a, b, c, d positive if a b and. ac bd. Reciprocal relations for a and b positive. If a > b then a ab > b. then

For a, b, c, d positive if a b and. ac bd. Reciprocal relations for a and b positive. If a > b then a ab > b. then Slrs-7.2-ADV-.7 Improper Definite Integrls 27.. D.dox Pge of Improper Definite Integrls Before we strt the min topi we present relevnt lger nd it review. See Appendix J for more lger review. Inequlities:

More information

5Trigonometric UNCORRECTED PAGE PROOFS. ratios and their applications

5Trigonometric UNCORRECTED PAGE PROOFS. ratios and their applications 5Trigonometri rtios nd their pplitions 5.1 Kik off with CS 5.2 Trigonometry of right-ngled tringles 5.3 Elevtion, depression nd erings 5.4 The sine rule 5.5 The osine rule 5.6 rs, setors nd segments 5.7

More information

TOPIC: LINEAR ALGEBRA MATRICES

TOPIC: LINEAR ALGEBRA MATRICES Interntionl Blurete LECTUE NOTES for FUTHE MATHEMATICS Dr TOPIC: LINEA ALGEBA MATICES. DEFINITION OF A MATIX MATIX OPEATIONS.. THE DETEMINANT deta THE INVESE A -... SYSTEMS OF LINEA EQUATIONS. 8. THE AUGMENTED

More information

Exercise sheet 6: Solutions

Exercise sheet 6: Solutions Eerise sheet 6: Solutions Cvet emptor: These re merel etended hints, rther thn omplete solutions. 1. If grph G hs hromti numer k > 1, prove tht its verte set n e prtitioned into two nonempt sets V 1 nd

More information

What else can you do?

What else can you do? Wht else cn you do? ngle sums The size of specil ngle types lernt erlier cn e used to find unknown ngles. tht form stright line dd to 180c. lculte the size of + M, if L is stright line M + L = 180c( stright

More information

set is not closed under matrix [ multiplication, ] and does not form a group.

set is not closed under matrix [ multiplication, ] and does not form a group. Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed

More information

Designing Information Devices and Systems I Spring 2018 Homework 7

Designing Information Devices and Systems I Spring 2018 Homework 7 EECS 16A Designing Informtion Devices nd Systems I Spring 2018 omework 7 This homework is due Mrch 12, 2018, t 23:59. Self-grdes re due Mrch 15, 2018, t 23:59. Sumission Formt Your homework sumission should

More information

2 Calculate the size of each angle marked by a letter in these triangles.

2 Calculate the size of each angle marked by a letter in these triangles. Cmridge Essentils Mthemtics Support 8 GM1.1 GM1.1 1 Clculte the size of ech ngle mrked y letter. c 2 Clculte the size of ech ngle mrked y letter in these tringles. c d 3 Clculte the size of ech ngle mrked

More information

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus 7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e

More information

NON-DETERMINISTIC FSA

NON-DETERMINISTIC FSA Tw o types of non-determinism: NON-DETERMINISTIC FS () Multiple strt-sttes; strt-sttes S Q. The lnguge L(M) ={x:x tkes M from some strt-stte to some finl-stte nd ll of x is proessed}. The string x = is

More information

5. Every rational number have either terminating or repeating (recurring) decimal representation.

5. Every rational number have either terminating or repeating (recurring) decimal representation. CHAPTER NUMBER SYSTEMS Points to Rememer :. Numer used for ounting,,,,... re known s Nturl numers.. All nturl numers together with zero i.e. 0,,,,,... re known s whole numers.. All nturl numers, zero nd

More information