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

Size: px
Start display at page:

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

Transcription

1 SECTION CHAPTER 8 Setion 8 1. There re n infinite numer of possile tringles, ll similr, with three given ngles whose sum is If two ngles α nd β of tringle re known, the third ngle n e found immeditely s γ 180 α β. Then the given side will e inluded etween two of the three known ngles.. It is lled the miguous se euse there re two tringles determined y the given informtion. Note: Answers hve een rounded to the numer of signifint digits given in Tle 1; n sign hs een used rther thn. 8. γ 180 ( ) γ α 180 ( ) α β 180 ( ) β α 180 ( ) α 0 sin sin sin sin sin 41 1 sin10 sin 33 1 sin10 14 m 1 m sin sin 9 sin101 sin 3 55 mm sin sin 5 sin sin83 sin m sin sin 43 sin sin105 sin 3 sin 5 5 mi sin sin 77 sin 0 sin 9 sin101 4 mm sin 47 sin m sin 5 sin mi 1. 3 ft, ft, α 30 SSA: h sin α sin tringle; se () 0. ft, ft, α 30 SSA: h sin α sin 30 3 > 0 tringles; se () ft, ft, α 30 SSA: h sin α sin > 3 1 tringle; se (d). 5 ft, ft, α 30 SSA: h sin α sin < 5 < tringles; se ()

2 1 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 4. β 7.5, γ 54.5, 9.7 inhes α 180 ( ) α 98 sin sin 9.7 sin 98 sin in sin sin sin sin in. α 1.7, β 34.4, 18.3 km γ 180 ( ) sin sin sin sin γ.9 sin sin 34.4 sin sin km 1. km 8. α.3, 14.7 inhes, 35. inhes sin sin 14.7 sin sin sin β > 1 No solution 30. h sin α 17 sin ; h < <, so two tringles possile sin sin sin 43.5 sin sin β β 59.1 or β 10.9 Tringle I β 59.1 α + β + γ γ 180 γ 77.4 Tringle II β' 10.9 α + β' + γ' γ' 180 γ' 15. sin sin sin sin sin 43.5 sin m 138 sin 43.5 sin m h sin β 44 sin ; 135 h < <, so two tringles possile sin sin sin sin sin α α 5.0 or α 14.0 Tringle I α 5.0 α + β + γ γ 180 γ 9.7 sin sin sin 7.3 sin m 135 Tringle II α' 14.0 α' + β + γ' γ' 180 γ' 8.7 sin sin sin 7.3 sin 8.7 ' 141 m 135

3 SECTION α 137.3, 13.9 m, 19.1 m sin sin 13.9 sin sin β 8.73 β + α 0.03 > 180 No solution 3. β 33 50', 73 m, 140 m sin sin sin sin sin 3350' sin sin 3350' sin18 30' α 17 40' 1740 m γ 180 (33 50' ') γ 18 30' 38. h sin β 9 sin 30 4 h, so there is one tringle sin sin sin sin sin α 1 α 90 α + β + γ γ 180 γ 0 sin sin sin 30 sin 0 80 in 4 4. From the lw of sines sin sin sin sin nd sin sin sin sin () 40. α 37.3, 4.8 m k 4.8 sin is k suh tht 0 < < k gives no solution; k gives one solution; k < < gives two solutions. (1) sin sin sin sin os sin os sin os sin sin os sin (3) dding (1) nd () ut sin sin 1 (180 γ) sin 90 os similrly, sutrting (1) nd () sin sin os sin sin os sin sin os sin sin sin os sin os sin (4) dividing (4) y (3) ut os os 1 (180 γ) os 90 sin

4 14 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY sin os os sin sin sin os os tn tn ut tn ot 1 tn tn tn tn (B) from 7, 41 α 73 0 β tn 738 tn Answers gree to two signifint digits, the ury of the given informtion. 44. sin 53 sin 830' 10 sin 9830' 8.08 miles from A 4.8 miles from B 4. tn 43 5' h x h x tn 43 5' h h tn h x 000 tn tn 38 + h tn 435' h tn 38 h 1 tn 43 5' 000 tn tn 38 h tn38 1 tn 435' tn 435' h The distne ove se level h 14,490 feet to 4 signifint digits sin 1.7 sin11 sin α 45 γ " α 135 γ " sin(sev) SEV

5 SECTION α α 48 β β β + γ γ 180 γ 31 h 157 sin sin h sin sin 48 h 109 ft, to the nerest foot sin 3. R sin 58.4 R 9.73 mm to 3 signifint digits s R mm to 3 signifint digits 5. Let x e the length of the side in the horizontl tringle tht is lso in the vertil tringle with ngle γ. In the horizontl tringle sin(180 (α + β)) sin 180 os(α + β) os 180 sin(α + β) x sin d 0 os(α + β) ( 1) sin(α + β) sin(180 ( )) sin(α + β) x d sin α s(α + β) 1 In the vertil tringle s( ) Setion 8 tn γ h x h x tn γ d sin α s(α + β) tn γ. Use the lw of osines to determine the lrgest ngle (lrgest euse it is opposite the longest side). Then use the lw of sines to determine seond ngle nd the ft tht the sum of the ngles is 180 to determine the third. 4. Sustituting into + os γ gives + os γ whih redues to os γ 1 or γ 0, onfirming tht in n equilterl tringle ll ngles mesure 0.. There is no SSA ongruene theorem, sine the given informtion does not determine tringle uniquely. There is n SSS ongruene theorem sine three sides tht stisfy the tringle inequlity determine the tringle uniquely. 8. A tringle n hve t most one otuse ngle. Sine α 93.5 is otuse oth γ nd β must e ute. [β + γ ; thus oth β nd γ re less thn 90.]

6 1 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 10. β 57.3,.08 m, 5.5 m + os β (.08)(5.5)os m Solve for smllest ngle: 5.5 sin 5.48 sin sin 5.48 sin 57.3 γ 53.7 α 9.0 or α 180 ( ) 1. α ', 8.44 in, 0.3 in + os α (8.44)(0.3)os ' 7.0 in Solve for smllest ngle: sin13550' sin sin 7 sin13550' β 1 30' γ 31 40' or γ 180 (135 50' ') 14. If 1.5 m, 5.3 m, 10.7 m, then sides nd re not long enough to onstrut tringle: + < mi, 0.7 mi, 1. mi Solve for the lrgest ngle: + os β (10.5)(1.)os β os β β sin γ. sin α 180 ( ) m, 9.4 m, 33.7 m Solving for the lrgest ngle: + os γ (31.5)(9.4)os γ os γ γ sin sin α 59.4 β 180 ( ) β + γ > 180 no solution. β 7.3, 13.7 yds, 0.1 yd + os β (13.7)(0.1)os yd 10.1 sin sin α 38.5 γ 180 ( ) 114. (Answers will vry slightly depending on method used.) 4. β 13.4, γ 17.3, 7. km α 180 (β + γ) 180 ( ) 30.3 sin sin sin sin sin13.4 sin km; 7. km. γ.4, 5.5 m, 5.5 m γ β.4 α 180 (γ + β) 180 ((.4 )) sin.4 sin m m, 5.3 m, 8. m 30. α 4.7, 18.1 m,. m

7 SECTION 8-17 Angle opposite longest side: + os α (5.3)(8.)os α α sin sin β 9.7 γ 180 (α + β) 180 ( ) γ γ 47.9, 35. in, 5.5 in sin sin 5.5 sin sin sin β No solution 34. h sin β 98.5 sin h < < There re two tringles. sin sin sin 5.1 sin sin γ γ 51.1 or 18.9 Tringle I γ 51.1 α + β + γ α α sin sin sin 5.1 sin m sin sin 18.1 sin 4.7. β 5.3 or β' sin β 5.3 γ 180 ( ) 8.0 β' γ' 180 ( ) sin 4.7 sin 8.0 ' sin m ' 7.93 m Tringle I: β 5.3, γ 8.0, 3.1 m Tringle II: β' 114.7, γ' 18., ' 7.93 m Tringle II γ' 18.9 α' + β + γ' α' α'.0 sin sin sin 5.1 sin ' 55.5 m 3. h A ,100sq.yds o h o sin 49.3 h 10.1sin sin 49.3 o A 701 sq. m. o h sin 41 h 4sin 41 4 o

8 18 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY A sin 41,900 o sq. ft. 4. o o o o o o o sin sin sin 41 o 9.5 sin o h 9.5sin 41 sin 7 h sin 7 o o 9.5sin 41 sin o sin sin41 o o A sin 7 sq. m. o sin o o s 109 A sq. yds. 4. Given + : + os γ lw of osines + + os γ sustitution os γ 0 os γ 0 γ x + y : os β x x os β : os γ y y os γ x y x os β + os γ os β x + x os β os β x os β ( os γ) os β + os γ os(180 γ) x os γ x x os γ 50. It is wiser to strt y using the lw of osines to find the lrgest ngle, in this se β. Then os 11 so β 10. Then sine, α must equl γ, hene from α + γ we get α γ 30. By finding α first, the inorret onlusion β 30 ws drwn from the eqution sin β we just sw, β ; tully, s 5. AB (85)(73) os m (8.)(8.)os θ (8)(3)os(144 50) 10. ft

9 SECTION θ After hours, Plne A hs trveled 800 miles, Plne B 1000 miles. The ngle etween them is (1000)(800) os mi 0. The ngle t the enter is An isoseles tringle is formed, so the other two ngles re Let x e the hord formed in the irle. Using the lw of sines to solve, 7.09 sin 70 x sin 40 x The perimeter, to 3 signifint digits, is 9x 43. m.. OA OB AB AB OA + OB (OA)(OB)os θ 5 1 ( 5 ) 5 + ( ) (5)( )os θ (4 5) (3 1) 5 θ 0.44 rdin 4. The sides of the tringle hve lengths of 5 + 7, , nd Find ngle γ first (ngle opposite longest side): (10)(7)os γ γ ' 13 7 Angle α: sin 9810 sin α ' Angle β: β 180 (α + γ) 180 (98 10' ') 49 40'. AB AC BC (AB) (AC) + (CB) (AC)(BC)os(ACB) os(acb) Angle ACB (CS) R + (ST) (R)(ST) os 1.4 (CS) (394)(1034) os 1.4 CS 40 height CS R miles Setion 8 3. Answers will vry. 4. A vetor tht hs mgnitude 1 is lled unit vetor.. The differene is the vetor tht represents the wind or urrent! Tht is, the pprent veloity is the veloity reltive to the ir, while the tul veloity is the resultnt of the pprent veloity nd the wind veloity, tht is, the veloity reltive to the ground. 8. The oordintes of P re given y (x p, y p ) (x x, y y ) (3, 15 7) (1, 8) Hene OP 1, 8 1. The oordintes of P re given y (x p, y p ) (x x, y y ) (0 9, 0 ( 7)) ( 9, 7) Hene OP 9,7 10. The oordintes of P re given y (x p, y p ) (x x, y y ) (8 ( 5), ( 1) ) (13, 3) Hene OP 13, OP AB 7,1 1. v v ( 4) 3 5 5

10 0 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 0. v 10 ( 10) v ( 1) 5 4. u + v u v u v os(180 ). u + v u v u v os(180 ) (10)(84)os(18044 ) 190 gm 190 sin13 84 sin α (8.0)(.0) os(1804 ) 9.1 knots.0 sin 9.1 sin11 α sin137 v sin17 u v 14 kg; sin 33 sin137 u 1 kg sin19.5 u sin 3.7 v sin17.8 u 30 mph nd v 173 mph 3. (A) u + v 1, 3, 1 3, ( ), 0 (B) u v 1, 3, 1 3, ( ) 4, 4 (C) u v + 3w 1, 3, 3 0, 30,4 5,0 34. (A) u + v 3,, 3 ( ), 5,4 (B) u v 3,, 3 ( ), 1,0 (C) u v + 3w 3,, 3 3,0 9,40 13, 3., 5,0 0, 5 1,0 5 0,1 i 5 j 38. 0, 7 7 0,1 7 j uuur 40. AB 0 ( ), ( 1),3,0 0,3 1,0 3 0,1 i 3j 4. u v 3i j (i + 4j) 3i j i 4j i j 44. 3u + v 3(3i j) + (i + 4j) 9i j + 4i + 8j 13i + j 4. u 3v + w 3i j 3(i + 4j) + (i) 3i j i 1j + 4i i 14j 48. v u 1 v v ,1 5 1, v ( 3) 13 u 1 v v 1 13, 3 3, v 0 ( 17) u 1 v v , 17 0, v ( ) ( 7) 9 3 u 1 v v 1 ( i 7 j) 3 5. True. The stndrd vetor is n ext representtion of the vetor. 58. Flse. i is nother unit vetor, nd there re n infinite numer. 3 i 7 3 j 7, 3 3

11 SECTION True. v + v v hs the sme diretion s v nd twie the mgnitude.. Flse. i + j is not unit vetor. 4. u + v, d,, d. u + ( u),,,,, d d,, v + u ( ), ( ) 0, m(u + v) m,, d m, d 70. 1u 1, 1,1, u 7. v m m, m md m, m m, md m, m, d mu + mv (15)(3.9) os (5 45 ) 14 mph 3.9 sin 14 sin(70 ) θ 15 heding mph t R (300)(900) os sin sin α The third ngle: 180 ( ) 17 diretion v sin θ 4 55 θ mph t 349. ( ) 78. (A) prllel fore 500 sin l (B) fore perpendiulr 500 os l 80. Left: 41 sin 31 1; Right: 31 sin > 0 slide left 8. Let the left tension e represented y T L nd the right tension y T R. Then T L sin 4. + T R sin T L os 4. T R os 5.3 Solving the seond eqution for T L : T L T os5.3 R os 4. TR (os5.3 )(sin 4. ) Sustituting: + T os 4. R sin x(os5.3 )(sin 4. ) Grph y 1 + x sin 5.3 nd y 11. os 4. T R 77 l, T L 7 l. 84. Let the left tension e represented y T L nd the right tension y T R. Then T L sin 45 + T R sin T L os 45 T R os 0 Solving the seond eqution for T : T L L TR os 0 os 45 TR (os 0 )(sin 45 ) Sustituting: + T os 45 R sin x(os 0 )(sin 45 ) Grph y 1 + x sin 0 nd y 500. os 45 T R 390 l, T L 518 l

12 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 8. Angle ABC 30 BC sin BC 000 kg, tension BC os 30 AB AB 000 os kg, ompression 88. Angle ABC: os(abc) 5 ABC 33. (Answers my vry due to rounding.) AB sin AB 9040 kg, ompression BC AB os os kg, tension Setion 8 4. If r is positive, θ represents the ngle etween the polr xis nd the ry onneting the point to the pole; θ my e ny of the oterminl suh ngles. If r is negtive θ represents the ngle etween the polr xis nd the ry opposite to the ry onneting the point to the pole; gin, θ my e ny of the oterminl suh ngles. 4. One set of polr oordintes is given y r x y, tn θ y x. If x 0, then θ n e given s if y is positive nd y. Answers will vry. or 3 if y is negtive

13 SECTION (, 10 ): The polr xis is rotted 10 lokwise (negtive diretion) nd the point is loted units from the pole long the negtive polr xis. (, 150 ): The polr xis is rotted 150 ounterlokwise (positive diretion) nd the point is loted units from the pole long the negtive polr xis. (, 330 ): The polr xis is rotted 330 ounterlokwise (positive diretion) nd the point is loted units long the positive polr xis os r 5. θ 4.. ( 3.500,.0) (0.8,.95)

14 4 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY ( 7.115, 5.557) (8.3, 34 ) (31, 59 ) (9.37, 15.8 ) 3. os 4os vries from vries from vries from 0 to 1 to 0 4 to 0 to 0 to 1 0 to 4 to3 1 to 0 4 to 0 3 to 0 to 1 0 to os 8os vries from vries from vries from vries from 0 to 4 0 to 1 to 0 8 to 0 4 to to 0 to 1 0 to 8 to3 4 to 3 1 to 0 8 to to 3 to 0 to 1 0 to 8 to 5 4 to 5 1 to 0 8 to to3 5 to 3 0 to 1 0 to 8 3 to7 4 3 to 7 1 to 0 8 to to 7 to 4 0 to 1 0 to sin3 sin3 vries from vries from vries from vries from 0 to 0 to 0 to 1 0 to to 3 to 1 to 0 to 0 3 to to 3 0 to 1 0 to to 3 3 to 1 to 0 to 0 3 to 5 to 5 0 to 1 0 to 5 to 5 to 3 1 to 0 to 0 to7 3 to 7 0 to 1 0 to 7 to to 5 1 to 0 to 0 M M M M

15 SECTION os 3os 3 3os vries from vries from vries from vries from 0 to 1 to 0 3 to 0 to 3 to 0 to 1 0 to 3 3 to 0 to3 1 to 0 3 to 0 0 to 3 3 to 0 to 1 0 to 3 3 to 44. os 4os 4os vries from vries from vries from vries from 0 to 1 to 0 4 to 0 to to 0 to 1 0 to 4 to to3 1 to 0 4 to 0 to 3 to 0 to 1 0 to 4 to 4. r + os θ r 4 + os θ r + 4 os θ (A) r 4 os θ r 4 os 3θ r 4 os 5θ (B) 7 leves in r 4 os 7θ (C) n leves in r os(nθ) ( > 0 nd n odd) -

16 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 50. (A) r 4 os θ r 4 os 4θ r 4 os θ (B) 1 leves in r 4 os 8θ (C) n leves in r os nθ ( > 0 nd n even) x x y x x + y r os θ r r os θ 54. x + y 9 r 9 or r ±3 5. xy 1 (r os θ r sin θ) 1 r ( sin θ os θ) 1 r sin θ 1 r 1 sin r s θ 58. r( os θ + sin θ) 4 (r os θ) + r sin θ 4 x + y 4 0. r 8 os θ r 8r os θ x + y 8x. r 4 r 1 x + y 1 4. n r 1 + os (nθ) 1 1 smll petl inside 1 lrge petl smll petls etween lrge petls 3 3 smll petls inside 3 lrge petls 4 4 smll petls etween 4 lrge petls r 1 + os(nθ) will hve n lrge nd n smll petls. For n odd the smll petls re within the lrge petls. For n even the smll petls re etween the lrge petls.. r os θ (1) r sin θ () 0 θ π Divide () y (1): 1 sin os tn θ θ r, 4 4 [Note: (0, 0) is not solution to this system even though the grphs ross t the origin.]

17 SECTION (1) r sin θ () r os θ 0 θ 30 sin θ os θ 1 sin θ sin θ + sin θ 1 0 ( sin θ 1)(sin θ + 1) 0 sin θ 1 0 sin θ sin θ 1 sin θ 1 θ 30,150 θ 70 r 4, 4 r 8 (4, 30 ), (4, 150 ), ( 8, 70 ) [Note: (0, 0) is not solution to this system even though the grphs ross t the origin.] 70. P 1 (, 30 ) nd P (3, 0 ) d ( r) ( r ) rr os( ) 3 ()(3) os(030 ) 13 1os t 45 : 9k, t 90 : 14k, t 10 : 13k, t 150 : 11k 74. (A) e 0.: 11.5 (B) e 1: ellipse prol (C) e : hyperol Setion If z re iθ then the ngle θ is lled the rgument of z, usully hosen so tht π < θ π or 180 < θ 180. i 1 4. If z 1 re i nd z re z1 i( 1) re the two numers, then their quotient is e whih lies on the unit z irle with rgument θ 1 θ.

18 8 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY. If z e iθ i 3 is the numer, then w1 e i( ) 3, w e i( 4 ) 3, nd w3 e re the ue roots. They lie on the 4 unit irle with rguments,, nd, tht is, 10 prt (A) 1 + i 3 A sketh shows tht 1 + i 3 is ssoited with θ 10, r 1 + i 3 (os 10 + i sin 10 ) e 10 i (B) 3i A sketh shows tht θ 90, r 3 3i 3(os( 90 ) + i sin( 90 )) 3e ( 90 )i (C) 7 4i γ ( 7) ( 4) θ tn i 8.0(os( 150. ) + i sin( 150. )) 8.0e ( 150. )i 18. (A) 3 i A sketh shows tht θ, r 3 i os + i sin e (π/)i (B) + i A sketh shows tht θ 3, r i 3 3 os i sin 4 4 e(3π/4)i (C) 5i r ( 5) θ tn 1 5 5i 7.81(os( 0.9) + i sin( 0.9)) 7.81e 0.9i 0. (A) e 30 i (os 30 + i sin 30 ) i 3 + i

19 SECTION (B) e ( 3π/4)i 3 3 os i sin 4 4 i 1 i (C) 5.71e ( 0.48)i 5.71(os( 0.48) + i sin( 0.48)) 5.0.4i. (A) 3 e ( π/)i 3 os i sin 3 (0 i) i 3 (B) e 135 i (os i sin 135 ) i 1 + i (C).83e ( )i.83(os( ) + i sin( )).83( i( )).0.4i z 1 z r 1 e iθ 1 r e iθ r 1 r e i(θ 1 + θ ) 1 z1 re 1 r1 i z re z 1 z e 13 i 3e 93 i 18e 5 i 13i z1 e e 39 i 93 i z 3e. z 1 z r 1 e i θ1 r e i θ r 1 r e i(θ 1 + θ ) i1 z1 re 1 r1 i z re z 1 z 3e 7 i e 97 i e 14 i 1 7i z 3e 97 i z e 1.5e ( 30 )i 8. z 1 z r 1 e i θ1 r e i θ r 1 r e i(θ 1 + θ ) z1 z z 1 z 7.11e 0.79i.e 1.07i 18.91e 1.8i 18.91e 1.8i z 1 z i i1 re 1 r1 i re 7.11e.e 0.79i 1.07i r ei(θ 1 θ ) r ei(θ 1 θ ) r ei(θ 1 θ ).7e ( 0.8)i 30. z n r n e nθi (5e 15 i ) e (3 15 )i 15e 45 i 3. z n r n e nθi ( e 15 i ) 8 ( ) 8 e (8 15 )i 1e 10 i

20 30 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 34. ( 3 + i): r, θ 30 e 30 i z n r n e nθi ( 3 + i) 8 (e 30 i ) 8 8 e (8 30 )i 5e 40 i 3. ( 1 + i): r, θ 135 e 135 i z n r n e nθi ( e 135 i ) 4 ( ) 4 e (4 135 )i 4e 540 i 4(os i sin 180 ) 4( 1 + 0i) ( 3 + i): r, θ 150 e 150 i z n r n e nθi (e 150 i ) 5 5 e (5 150 )i 3e 750 i 3e 30 i 3(os 30 + i sin 30 ) i i : r 1, θ 40 1e 40 i z n r n e nθi (1e 40 i ) e (3 40 )i 1e 70 i 1e 0 i os 0 + i sin z 1/n r 1/n e [(θ/n) + ((30 k)/n) ]i (8e 45 i ) 1/3 8 1/3 e [(45 /3) + ((30 k)/3)]i e w 1 e 15 i w e ( )i e 135 i w 3 e ( )i e 55 i ( k)i 4. ( 1 + i): r, θ 135 e 135 i z 1/n r 1/n e [(θ/n) + ((30 k)/n) ]i ( e 135 i ) 1/3 ( 1/ ) 1/3 e [(135 /3) + ((30 k)/3)]i 1/ e w 1 1/ e 45 i w 1/ e 15 i w 3 1/ e 85 i 48. z 1 1e 0 i z 1/n r 1/n e [(θ/n) + ((30 k/n)]i (1e 0 i ) 1/4 1 1/4 e [0 + ((30 k)/4)]i 1e 90 ki w 1 1e 0 i 1 + 0i w 1e 90 i 0 + i w 3 1e 180 i 1 + 0i w 4 1e 70 i 0 i 50. z 8 8e 180 i z 1/n r 1/n e [(θ/n) + ((30 k)/n]i (8e 180 i ) 1/3 8 1/3 e [(180 /3) + ((30 k)/3)]i e ( k)i w 1 e 0 i (os 0 + i sin 0 ) i w e 180 i (os i sin 180 ) w 3 e 300 i (os i sin 300 ) 1 3 i ( k)i 44. z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (1e 90 i ) 1/4 1 1/4 e [(90 /4) + ((30 k)/4)]i e w 1 e.5 i w e 11.5 i w 3 e 0.5 i w 4 e 9.5 i w 3-1 w y y w w 4 w 1 w 3 w 1 1 x x ( k)i 5. z i e 90 i z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (1e 90 ) 1/5 1 1/5 e [( 90 /5) + ((30 k)/5]i 1e w 1 1e ( 18 )i w 1e 54 i w 3 1e 1 i w 4 1e 198 i w 5 1e 70 i ( k)i -1 w 4 w 3 y 1-1 w 5 w 1 w 1 x

21 SECTION (A) x ( ) , is root of x x is degree 3 so there re two more roots. (B) y is the sping etween roots 3 w 1 From prolem 50, w i, w 1 3 i - w x w 3 (C) (1 + 3 i) [1 + 3( 3 i) + 3( 3 i) + ( 3 i) 3 ] + 8 [ i i] (D) In the sme mnner, (1 3 i) x x 3 4 x 4 1/3 (4e 0 i ) 1/3 z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (4e 0 i ) 1/3 4 1/3 e [(0 /3) + ((30 k)/3)]i 4e 10 ki x 1 4e 0 i 4 x 4e 10 i 1 3 4(os 10 + i sin 10 ) 4 i + 3 i x 3 4e 40 i 1 3 4(os 40 + i sin 40 ) 4 i 3 i 58. x x 3 7 x ( 7) 1/3 7e 180 i z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (7e 180 i ) 1/3 7 1/3 e [(180 /3) + ((30 k)/n)]i 3e ( k)i x 1 3e 0 i 3(os 0 + i sin 0 ) i i x 3e 180 i 3(os i sin 180 ) 3( 1 + 0i) 3 x 3 3e 300 i 3(os i sin 300 ) i i 0. Flse. If z 1 r 1 e i90 nd z r e i90, z1 r1 then z r ei0.. True. The squre roots of re i0 re r e i0 nd r e i180, whih re the rel numers r nd r. 4. Flse. e i0 is sixth root of 1, ut (e i0 ) e i10 is not 1. z1 r. 1 (os 1 isin 1 ) r 1 z r(os isin ) r os 1isin 1 os isin os isin os isin r 1 r os1os ios 1sin ios sin 1i sin 1sin os i sin r1 r (os1os sin 1sin ) i(ossin 1os 1sin ) r1 1 r [os(θ 1 θ ) + i sin(θ 1 θ )] r 1 r ei(θ 1 θ )

22 3 CHAPTER 8 ADDITIONAL TOPICS IN TRIGONOMETRY 8. For k 0, r 1/n e (θ/n + (k 30 )/n)i r 1/n e (θ/n)i For k n, r 1/n e (θ/n + (k 30 )/n)i r 1/n e (θ/n + 30 )i r 1/n e (θ/n)i 70. x x 1 x ( 1) 1/ 1e 180 i z 1/n r 1/n e [(θ/n) + ((30 k)/n)]i (1e 180 i ) 1/ 1 1/ e [(180 /) + ((30 k)/)]i 1e x 1 1e 30 i x 1e 90 i x 3 1e 150 i x 4 1e 10 i x 5 1e 70 i x 1e 330 i ( k)i 7. x 3 i 0 x 3 i x i 1/3 1e 90 i z 1/n r 1/n e [(θ /n) + ((30 k)/n)]i (1e 90 i ) 1/3 1 1/3 e [(90 /3) + ((30 k)/3)]i 1e x 1 1e 30 i x 1e 150 i x 3 1e 70 i 74. P(x) x 1; find x 1 1/ (1e 0 i ) 1/ nd write s ftors. z 1/n r 1/n e [(θ/n) + ((30 k)/n) ]i (1e 0 i ) 1/ 1 1/ e [(0 /) + ((30 k)/)]i 1e 0 ki x 1 e 0 i 1 x e 0 i i x 3 e 10 i i x 4 e 180 i 1 x 5 e 40 i 1 3 i x e 300 i 1 3 i P(x) x (x 1)(x + 1) x i x i x i x i Chpter 8 Group Ativity ( k)i Prolem 1: In the figure, the distne from P to F is r. Sine P hs retngulr oordintes (r os θ, r sin θ) nd Q hs retngulr oordintes ( p, r sin θ), the distne from P to d is r os θ ( p) r os θ + p. Then using PF e PD s stted: r e(r os θ + p) r er os θ + ep r(1 e os θ) ep ep r 1 e os Prolems 4 re left to the student. If 0 < e < 1, the oni setion is n ellipse, If e 1, the oni setion is prol. If e > 1, the oni setion is hyperol.

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

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

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

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

( ) { } [ ] { } [ ) { } ( ] { }

( ) { } [ ] { } [ ) { } ( ] { } Mth 65 Prelulus Review Properties of Inequlities 1. > nd > >. > + > +. > nd > 0 > 4. > nd < 0 < Asolute Vlue, if 0, if < 0 Properties of Asolute Vlue > 0 1. < < > or

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

H (2a, a) (u 2a) 2 (E) Show that u v 4a. Explain why this implies that u v 4a, with equality if and only u a if u v 2a.

H (2a, a) (u 2a) 2 (E) Show that u v 4a. Explain why this implies that u v 4a, with equality if and only u a if u v 2a. Chpter Review 89 IGURE ol hord GH of the prol 4. G u v H (, ) (A) Use the distne formul to show tht u. (B) Show tht G nd H lie on the line m, where m ( )/( ). (C) Solve m for nd sustitute in 4, otining

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

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

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

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

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

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

Standard Trigonometric Functions

Standard Trigonometric Functions CRASH KINEMATICS For ngle A: opposite sine A = = hypotenuse djent osine A = = hypotenuse opposite tngent A = = djent For ngle B: opposite sine B = = hypotenuse djent osine B = = hypotenuse opposite tngent

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

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP QUADRATIC EQUATION EXERCISE - 0 CHECK YOUR GRASP. Sine sum of oeffiients 0. Hint : It's one root is nd other root is 8 nd 5 5. tn other root 9. q 4p 0 q p q p, q 4 p,,, 4 Hene 7 vlues of (p, q) 7 equtions

More information

m m m m m m m m P m P m ( ) m m P( ) ( ). The o-ordinte of the point P( ) dividing the line segment joining the two points ( ) nd ( ) eternll in the r

m m m m m m m m P m P m ( ) m m P( ) ( ). The o-ordinte of the point P( ) dividing the line segment joining the two points ( ) nd ( ) eternll in the r CO-ORDINTE GEOMETR II I Qudrnt Qudrnt (-.+) (++) X X - - - 0 - III IV Qudrnt - Qudrnt (--) - (+-) Region CRTESIN CO-ORDINTE SSTEM : Retngulr Co-ordinte Sstem : Let X' OX nd 'O e two mutull perpendiulr

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

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

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

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

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

= 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

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

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

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

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

THREE DIMENSIONAL GEOMETRY

THREE DIMENSIONAL GEOMETRY MD THREE DIMENSIONAL GEOMETRY CA CB C Coordintes of point in spe There re infinite numer of points in spe We wnt to identif eh nd ever point of spe with the help of three mutull perpendiulr oordintes es

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

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES PAIR OF LINEAR EQUATIONS IN TWO VARIABLES. Two liner equtions in the sme two vriles re lled pir of liner equtions in two vriles. The most generl form of pir of liner equtions is x + y + 0 x + y + 0 where,,,,,,

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

PROPERTIES OF TRIANGLES

PROPERTIES OF TRIANGLES PROPERTIES OF TRINGLES. RELTION RETWEEN SIDES ND NGLES OF TRINGLE:. tringle onsists of three sides nd three ngles lled elements of the tringle. In ny tringle,,, denotes the ngles of the tringle t the verties.

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

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

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

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

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

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point

Numbers and indices. 1.1 Fractions. GCSE C Example 1. Handy hint. Key point GCSE C Emple 7 Work out 9 Give your nswer in its simplest form Numers n inies Reiprote mens invert or turn upsie own The reiprol of is 9 9 Mke sure you only invert the frtion you re iviing y 7 You multiply

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

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

Calculus Cheat Sheet. Integrals Definitions. where F( x ) is an anti-derivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx

Calculus Cheat Sheet. Integrals Definitions. where F( x ) is an anti-derivative of f ( x ). Fundamental Theorem of Calculus. dx = f x dx g x dx Clulus Chet Sheet Integrls Definitions Definite Integrl: Suppose f ( ) is ontinuous Anti-Derivtive : An nti-derivtive of f ( ) on [, ]. Divide [, ] into n suintervls of is funtion, F( ), suh tht F = f.

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

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

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

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

/ 3, then (A) 3(a 2 m 2 + b 2 ) = 4c 2 (B) 3(a 2 + b 2 m 2 ) = 4c 2 (C) a 2 m 2 + b 2 = 4c 2 (D) a 2 + b 2 m 2 = 4c 2

/ 3, then (A) 3(a 2 m 2 + b 2 ) = 4c 2 (B) 3(a 2 + b 2 m 2 ) = 4c 2 (C) a 2 m 2 + b 2 = 4c 2 (D) a 2 + b 2 m 2 = 4c 2 SET I. If the locus of the point of intersection of perpendiculr tngents to the ellipse x circle with centre t (0, 0), then the rdius of the circle would e + / ( ) is. There re exctl two points on the

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 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

Formula for Trapezoid estimate using Left and Right estimates: Trap( n) If the graph of f is decreasing on [a, b], then f ( x ) dx

Formula for Trapezoid estimate using Left and Right estimates: Trap( n) If the graph of f is decreasing on [a, b], then f ( x ) dx Fill in the Blnks for the Big Topis in Chpter 5: The Definite Integrl Estimting n integrl using Riemnn sum:. The Left rule uses the left endpoint of eh suintervl.. The Right rule uses the right endpoint

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

Exploring parametric representation with the TI-84 Plus CE graphing calculator

Exploring parametric representation with the TI-84 Plus CE graphing calculator Exploring prmetric representtion with the TI-84 Plus CE grphing clcultor Richrd Prr Executive Director Rice University School Mthemtics Project rprr@rice.edu Alice Fisher Director of Director of Technology

More information

Vectors. Chapter14. Syllabus reference: 4.1, 4.2, 4.5 Contents:

Vectors. Chapter14. Syllabus reference: 4.1, 4.2, 4.5 Contents: hpter Vetors Syllus referene:.,.,.5 ontents: D E F G H I J K Vetors nd slrs Geometri opertions with vetors Vetors in the plne The mgnitude of vetor Opertions with plne vetors The vetor etween two points

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

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

8.3 THE HYPERBOLA OBJECTIVES

8.3 THE HYPERBOLA OBJECTIVES 8.3 THE HYPERBOLA OBJECTIVES 1. Define Hperol. Find the Stndrd Form of the Eqution of Hperol 3. Find the Trnsverse Ais 4. Find the Eentriit of Hperol 5. Find the Asmptotes of Hperol 6. Grph Hperol HPERBOLAS

More information

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx

for all x in [a,b], then the area of the region bounded by the graphs of f and g and the vertical lines x = a and x = b is b [ ( ) ( )] A= f x g x dx Applitions of Integrtion Are of Region Between Two Curves Ojetive: Fin the re of region etween two urves using integrtion. Fin the re of region etween interseting urves using integrtion. Desrie integrtion

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

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

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

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

4. Statements Reasons

4. Statements Reasons Chpter 9 Answers Prentie-Hll In. Alterntive Ativity 9-. Chek students work.. Opposite sides re prllel. 3. Opposite sides re ongruent. 4. Opposite ngles re ongruent. 5. Digonls iset eh other. 6. Students

More information

JEE Advnced Mths Assignment Onl One Correct Answer Tpe. The locus of the orthocenter of the tringle formed the lines (+P) P + P(+P) = 0, (+q) q+q(+q) = 0 nd = 0, where p q, is () hperol prol n ellipse

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

Learning Objectives of Module 2 (Algebra and Calculus) Notes:

Learning Objectives of Module 2 (Algebra and Calculus) Notes: 67 Lerning Ojetives of Module (Alger nd Clulus) Notes:. Lerning units re grouped under three res ( Foundtion Knowledge, Alger nd Clulus ) nd Further Lerning Unit.. Relted lerning ojetives re grouped under

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

15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions )

15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions ) - TRIGONOMETRY Pge P ( ) In tringle PQR, R =. If tn b c = 0, 0, then Q nd tn re the roots of the eqution = b c c = b b = c b = c [ AIEEE 00 ] ( ) In tringle ABC, let C =. If r is the inrdius nd R is the

More information

Reflection Property of a Hyperbola

Reflection Property of a Hyperbola Refletion Propert of Hperol Prefe The purpose of this pper is to prove nltill nd to illustrte geometrill the propert of hperol wherein r whih emntes outside the onvit of the hperol, tht is, etween the

More information

Topics Covered: Pythagoras Theorem Definition of sin, cos and tan Solving right-angle triangles Sine and cosine rule

Topics Covered: Pythagoras Theorem Definition of sin, cos and tan Solving right-angle triangles Sine and cosine rule Trigonometry Topis overed: Pythgors Theorem Definition of sin, os nd tn Solving right-ngle tringles Sine nd osine rule Lelling right-ngle tringle Opposite (Side opposite the ngle θ) Hypotenuse (Side opposite

More information

at its center, then the measure of this angle in radians (abbreviated rad) is the length of the arc that subtends the angle.

at its center, then the measure of this angle in radians (abbreviated rad) is the length of the arc that subtends the angle. Notes 6 ngle Mesure Definition of Rdin If circle of rdius is drwn with the vertex of n ngle Mesure: t its center, then the mesure of this ngle in rdins (revited rd) is the length of the rc tht sutends

More information

Pythagorean Theorem and Trigonometry

Pythagorean Theorem and Trigonometry Ptgoren Teorem nd Trigonometr Te Ptgoren Teorem is nient, well-known, nd importnt. It s lrge numer of different proofs, inluding one disovered merin President Jmes. Grfield. Te we site ttp://www.ut-te-knot.org/ptgors/inde.stml

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

Chapter Gauss Quadrature Rule of Integration

Chapter Gauss Quadrature Rule of Integration Chpter 7. Guss Qudrture Rule o Integrtion Ater reding this hpter, you should e le to:. derive the Guss qudrture method or integrtion nd e le to use it to solve prolems, nd. use Guss qudrture method to

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

Geometry. Trigonometry of Right Triangles. Slide 2 / 240. Slide 1 / 240. Slide 3 / 240. Slide 4 / 240. Slide 6 / 240.

Geometry. Trigonometry of Right Triangles. Slide 2 / 240. Slide 1 / 240. Slide 3 / 240. Slide 4 / 240. Slide 6 / 240. Slide 1 / 240 Slide 2 / 240 New Jerse enter for Tehing nd Lerning Progressive Mthemtis Inititive This mteril is mde freel ville t www.njtl.org nd is intended for the non-ommeril use of students nd tehers.

More information

Inspiration and formalism

Inspiration and formalism Inspirtion n formlism Answers Skills hek P(, ) Q(, ) PQ + ( ) PQ A(, ) (, ) grient ( ) + Eerise A opposite sies of regulr hegon re equl n prllel A ED i FC n ED ii AD, DA, E, E n FC No, sies of pentgon

More information

Sect 10.2 Trigonometric Ratios

Sect 10.2 Trigonometric Ratios 86 Sect 0. Trigonometric Rtios Objective : Understnding djcent, Hypotenuse, nd Opposite sides of n cute ngle in right tringle. In right tringle, the otenuse is lwys the longest side; it is the side opposite

More information

Homework Assignment 6 Solution Set

Homework Assignment 6 Solution Set Homework Assignment 6 Solution Set PHYCS 440 Mrch, 004 Prolem (Griffiths 4.6 One wy to find the energy is to find the E nd D fields everywhere nd then integrte the energy density for those fields. We know

More information

KEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a

KEY CONCEPTS. satisfies the differential equation da. = 0. Note : If F (x) is any integral of f (x) then, x a KEY CONCEPTS THINGS TO REMEMBER :. The re ounded y the curve y = f(), the -is nd the ordintes t = & = is given y, A = f () d = y d.. If the re is elow the is then A is negtive. The convention is to consider

More information

50 AMC Lectures Problem Book 2 (36) Substitution Method

50 AMC Lectures Problem Book 2 (36) Substitution Method 0 AMC Letures Prolem Book Sustitution Metho PROBLEMS Prolem : Solve for rel : 9 + 99 + 9 = Prolem : Solve for rel : 0 9 8 8 Prolem : Show tht if 8 Prolem : Show tht + + if rel numers,, n stisf + + = Prolem

More information

Identifying and Classifying 2-D Shapes

Identifying and Classifying 2-D Shapes Ientifying n Clssifying -D Shpes Wht is your sign? The shpe n olour of trffi signs let motorists know importnt informtion suh s: when to stop onstrution res. Some si shpes use in trffi signs re illustrte

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

Trigonometric Functions

Trigonometric Functions Exercise. Degrees nd Rdins Chpter Trigonometric Functions EXERCISE. Degrees nd Rdins 4. Since 45 corresponds to rdin mesure of π/4 rd, we hve: 90 = 45 corresponds to π/4 or π/ rd. 5 = 7 45 corresponds

More information

Things to Memorize: A Partial List. January 27, 2017

Things to Memorize: A Partial List. January 27, 2017 Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors - Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved

More information

KENDRIYA VIDYALAY SANGATHAN: CHENNAI REGION CLASS XII PRE-BOARD EXAMINATION Q.No. Value points Marks 1 0 ={0,2,4} 1.

KENDRIYA VIDYALAY SANGATHAN: CHENNAI REGION CLASS XII PRE-BOARD EXAMINATION Q.No. Value points Marks 1 0 ={0,2,4} 1. KENDRIYA VIDYALAY SANGATHAN: CHENNAI REGION CLASS XII PRE-BOARD EXAMINATION 7-8 Answer ke SET A Q.No. Vlue points Mrks ={,,4} 4 5 6.5 tn os.5 For orret proof 5 LHS M,RHS M 4 du dv os + / os. e d d 7 8

More information

Log1 Contest Round 3 Theta Individual. 4 points each 1 What is the sum of the first 5 Fibonacci numbers if the first two are 1, 1?

Log1 Contest Round 3 Theta Individual. 4 points each 1 What is the sum of the first 5 Fibonacci numbers if the first two are 1, 1? 008 009 Log1 Contest Round Thet Individul Nme: points ech 1 Wht is the sum of the first Fiboncci numbers if the first two re 1, 1? If two crds re drwn from stndrd crd deck, wht is the probbility of drwing

More information

CONIC SECTIONS. Chapter 11

CONIC SECTIONS. Chapter 11 CONIC SECTIONS Chpter. Overview.. Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig..). Fig.. Suppose we rotte the line m round

More information

Mathematics. Area under Curve.

Mathematics. Area under Curve. Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding

More information

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique? XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=

More information

On the diagram below the displacement is represented by the directed line segment OA.

On the diagram below the displacement is represented by the directed line segment OA. Vectors Sclrs nd Vectors A vector is quntity tht hs mgnitude nd direction. One exmple of vector is velocity. The velocity of n oject is determined y the mgnitude(speed) nd direction of trvel. Other exmples

More information

03 Qudrtic Functions Completing the squre: Generl Form f ( x) x + x + c f ( x) ( x + p) + q where,, nd c re constnts nd 0. (i) (ii) (iii) (iv) *Note t

03 Qudrtic Functions Completing the squre: Generl Form f ( x) x + x + c f ( x) ( x + p) + q where,, nd c re constnts nd 0. (i) (ii) (iii) (iv) *Note t A-PDF Wtermrk DEMO: Purchse from www.a-pdf.com to remove the wtermrk Add Mths Formule List: Form 4 (Updte 8/9/08) 0 Functions Asolute Vlue Function Inverse Function If f ( x ), if f ( x ) 0 f ( x) y f

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

Polyphase Systems. Objectives 23.1 INTRODUCTION

Polyphase Systems. Objectives 23.1 INTRODUCTION Polyphse Systems 23 Ojetives eome fmilir with the opertion of threephse genertor nd the mgnitude nd phse reltionship etween the three phse voltges. e le to lulte the voltges nd urrents for three-phse Y-onneted

More information

Position Analysis: Review (Chapter 2) Objective: Given the geometry of a mechanism and the input motion, find the output motion

Position Analysis: Review (Chapter 2) Objective: Given the geometry of a mechanism and the input motion, find the output motion Position Anlysis: Review (Chpter Ojetive: Given the geometry of mehnism n the input motion, fin the output motion Grphil pproh Algeri position nlysis Exmple of grphil nlysis of linges, four r linge. Given

More information

Trigonometry. Trigonometry. labelling conventions. Evaluation of areas of non-right-angled triangles using the formulas A = 1 ab sin (C )

Trigonometry. Trigonometry. labelling conventions. Evaluation of areas of non-right-angled triangles using the formulas A = 1 ab sin (C ) 8 8 Pythgors theorem 8 Pythgoren trids 8 Three-dimensionl Pythgors theorem 8D Trigonometri rtios 8E The sine rule 8F miguous se of the sine rule 8G The osine rule 8H Speil tringles 8I re of tringles res

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

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector

More information

Polyphase Systems 22.1 INTRODUCTION

Polyphase Systems 22.1 INTRODUCTION 22 Polyphse Systems 22.1 INTRODUTION n genertor designed to develop single sinusoidl voltge for eh rottion of the shft (rotor) is referred to s single-phse genertor. If the numer of oils on the rotor is

More information

2. Topic: Summation of Series (Mathematical Induction) When n = 1, L.H.S. = S 1 = u 1 = 3 R.H.S. = 1 (1)(1+1)(4+5) = 3

2. Topic: Summation of Series (Mathematical Induction) When n = 1, L.H.S. = S 1 = u 1 = 3 R.H.S. = 1 (1)(1+1)(4+5) = 3 GCE A Level Otober/November 008 Suggested Solutions Mthemtis H (970/0) version. MATHEMATICS (H) Pper Suggested Solutions. Topi: Definite Integrls From the digrm: Are A = y dx = x Are B = x dy = y dy dx

More information

P 1 (x 1, y 1 ) is given by,.

P 1 (x 1, y 1 ) is given by,. MA00 Clculus nd Bsic Liner Alger I Chpter Coordinte Geometr nd Conic Sections Review In the rectngulr/crtesin coordintes sstem, we descrie the loction of points using coordintes. P (, ) P(, ) O The distnce

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