Lesson 6.5 xercises, pages 498 506 3. Which strategy would you use to determine the indicated measure in each triangle? a primary trigonometric ratio the osine Law the Sine Law a) cm ) cm 37 6 cm 38 d Since 2 sides and the contained angle are given, use the osine Law. Since it is a right triangle, use a primary trigonometric ratio. c) 15 cm J d) 20 cm G 22 cm k 113 H M 43 K Since 2 sides and a noncontained angle are given, use the Sine Law. Since 2 angles and a side are given, use the Sine Law. 38 6.5 The osine Law Solutions O OT OPY. P
4. etermine each measure in question 3. Give the angles to the nearest degree and the side lengths to the nearest tenth of a unit. a) Use: 2 a 2 c 2 2ac cos ) Use: cos 38 d Sustitute: a 6, c, 37 d cos 38 2 6 2 2 2(6)() cos 37 d 7.8801... 6 2 2 2(6)() cos 37 d 7.9 cm 6.3374... 6.3 cm sin H sin G k m c) Use: g d) Use: h sin K sin M Sustitute: G 22, Sustitute: K 43, h 15, g M 113, m 20 sin H 15 sin H H sin 1 15 sin 22 a H 34.1879 Á U 34 5. etermine the indicated measure in each triangle. Give the angles to the nearest degree and the side lengths to the nearest tenth of a unit. a) P Use: q 2 p 2 r 2 2pr cos Q Sustitute: p 17, r 23, Q 72 23 m q Q 72 sin 22 15 sin 22 17 m R k 20 sin 43 sin 113 20 sin 43 k sin 113 k 14.8179 Á k 14.8 cm q 2 17 2 23 2 2(17)(23) cos 72 q 17 2 23 2 2(17)(23)cos72 q 24.0072 Á q 24.0 m ) T 25 cm u 140 U 20 cm S Use: u 2 s 2 t 2 2st cos U Sustitute: s 25, t 20, U 140 u 2 25 2 20 2 2(25)(20) cos 140 u 25 2 20 2 2(25)(20) cos 140 u 42.3207 Á u 42.3 cm P O OT OPY. 6.5 The osine Law Solutions 39
c) U 18 cm 9 cm W V 13 cm Use: w 2 u 2 v 2 2uv cos W Sustitute: w 18, u 13, v 9 18 2 13 2 9 2 2(13)(9) cos W cos W 132 9 2 18 2 2(13)(9) W cos 1 a 132 9 2 18 2 2(13)(9) W 8.4356 Á U 8 d) 15.3 cm 16.5 cm 6.6 cm Use: c 2 2 d 2 2d cos Sustitute: c 15.3, 6.6, d 16.5 15.3 2 6.6 2 16.5 2 2(6.6)(16.5) cos cos 6.62 16.5 2 15.3 2 2(6.6)(16.5) cos 1 a 6.62 16.5 2 15.3 2 2(6.6)(16.5) 67.9629 Á U 68 6. security camera,, rotates through 80. The camera is m and 8 m from two doors, and, on the same wall of a uilding. To the nearest metre, how far apart are the doors? m 80 8 m Use: c 2 a 2 2 2a cos Sustitute: a 8,, 80 c 2 8 2 2 2(8)() cos 80 c 8 2 2 2(8)() cos 80 c 11.6711 Á The doors are aout 12 m apart. 40 6.5 The osine Law Solutions O OT OPY. P
7. Solve each triangle. Give the side lengths to the nearest tenth of a unit and the angle measures to the nearest degree. a) 20 cm Use: 2 c 2 d 2 2cd cos Sustitute:, c 18, d 20 2 18 2 20 2 2(18)(20) cos cm 18 cm cos 182 20 2 2 2(18)(20) cos 1 a 182 20 2 2 2(18)(20) 29.9264 Á 30 Use: c 2 2 d 2 2d cos Sustitute: c 18,, d 20 18 2 2 20 2 2()(20) cos cos 2 20 2 18 2 2()(20) cos 1 a 2 20 2 18 2 2()(20) 63.8961 Á 64 180 (64 30 ) 86 ) 30.0 m 3 P Use: p 2 m 2 n 2 2mn cos P Sustitute: m 30, n 14.7, P 3 p 2 30 2 14.7 2 2(30)(14.7) cos 3 p 30 2 14.7 2 2(30)(14.7) cos 3 p 36.2559 Á M 14.7 m Use: sin M m sin P p Sustitute: P 3, m 30, p 36.2559... sin M sin 3 30 36.2559 Á 30 sin 3 sin M 36.2559... Since M is acute: M sin 1 30 sin 3 a 36.2559... M 53.7303 Á 180 (3 53.7303 Á ) 23.2696 Á So, M 36.3 m, M 54, 23 P O OT OPY. 6.5 The osine Law Solutions 41
8. a) Use the osine Law to determine the length of to the nearest tenth of a centimetre. 7 cm Use: e 2 d 2 f 2 2df cos Sustitute: d 11, f 7, 90 e 2 11 2 7 2 2(11)(7) cos 90 e 11 2 7 2 e 13.0384 Á 13.0 cm 11 cm ) How is the Pythagorean Theorem a special case of the osine Law? When the angle in the statement of the osine Law is 90, since cos 90 0, the osine Law reduces to the Pythagorean Theorem. 9. fire spotter sees smoke on a earing of 060. t a point 20 km due east of the fire spotter, a ranger sees the same smoke on a earing of 320. a) How far is the smoke from each location? SRV is: 360 320 40 In SR, is: 90 60 30 R is: 90 40 50 S is: 180 (30 50 ) 0 f Use: sin s sin S Sustitute: 30, S 0, s 20 f 20 sin 30 sin 0 5 km 60 30 20 km S 0 r 20 Use: sin R sin 0 Sustitute: R 50 r 20 sin 50 sin 0 20 sin 50 r sin 0 50 320 20 sin 30 f r 15.5572 Á sin 0 f.1542... The smoke is approximately km and 16 km from each location. V R ) fire crew is 5 km due north of the fire spotter. How far is the crew from the smoke? Give the answers to the nearest kilometre. In S, Use: f 2 c 2 s 2 2cs cos Sustitute: c 15.5572..., s 5, 60 f 2 15.5572... 2 5 2 2(15.5572...)(5) cos 60 f 15.5572... 2 5 2 2(15.5572...)(5) cos 60 f 13.7565... The crew is approximately 14 km from the fire. 42 6.5 The osine Law Solutions O OT OPY. P
. In, = 2, = 120, and = 14 cm; determine the exact lengths of and. Let x centimetres Then 2x centimetres Use: 2 a 2 c 2 2ac cos Sustitute: 14, a 2x, c x, 120 14 2 (2x) 2 x 2 2(2x)(x) cos 120 196 5x 2 4x 2 (0.5) 196 7x 2 x 2 28 x 28, or 2 7 So, 2 7 cm and 4 7 cm x 120 14 cm 2x 11. Three circles have radii 3 cm, 4 cm, and 5 cm. ach circle just touches the other 2 circles externally. To the nearest tenth of a square centimetre, determine the area of the triangle formed y the centres of the circles. The centres of the circle form, with: : 3 cm 4 cm 7 cm : 4 cm 5 cm 9 cm 7 cm 8 cm : 3 cm 5 cm 8 cm raw the perpendicular from to meet at. In, use: 9 cm c 2 a 2 2 2a cos Sustitute: c 7, a 9, 8 7 2 9 2 8 2 2(9)(8) cos 144 cos 96 96 cos 144 48.1896... In, use: sin 8 sin 48.1896... 5.9628... The area of is: 0.5()() 0.5(9)(5.9628...) 26.8328... So, the area of the triangle formed y the centres of the circles is approximately 26.8 cm 2. P O OT OPY. 6.5 The osine Law Solutions 43
12. Here is a sketch of the route taken y an orienteering group on a one-day trek. nd 3 km G 80 5 km Start 15 a) To the nearest tenth of a kilometre, what is the straight-line distance from the start to the end? is: 90 80 G 15 So, is: 15 90 115 In, use: 2 a 2 c 2 2ac cos Sustitute: a 3, c 5, 115 2 3 2 5 2 2(3)(5) cos 115 3 2 5 2 2(3)(5) cos 115 6.8321... The straight-line distance is approximately 6.8 km. ) To the nearest degree, what is the earing of the end point from the start point? In, determine ; use: sin a sin Sustitute: 115, a 3, 6.8321... sin sin 115 3 6.8321... 3 sin 115 sin 6.8321... Since is acute: sin 1 3 sin 115 a 6.8321... 23.4506... 23 So, is approximately 90 (23 15 ) 52 The earing of the end point from the start point is approximately 052. 44 6.5 The osine Law Solutions O OT OPY. P
13. Oservers on two ships see an aircraft flying at 000 m. Oserver reports the earing of the aircraft as 048 and its angle of elevation as 70. Oserver reports the earing of the aircraft as 285 and its angle of elevation as 15. The earing of oserver from oserver is 096. To the nearest tenth of a kilometre, how far apart are the ships? In, 90 70 20 km etermine ; use: 48 70 15 tan 20 96 285 tan 20 In, The diagram is not drawn to scale. 90 15 75 etermine ; use: tan 75 tan 75 180 96 84 360 285 75 So, 84 75 9 In, 96 48 48 9 So, 180 (48 9 ) 123 Use: d 2 a 2 2 2a cos Sustitute: a tan 75, tan 20, 123 d 2 ( tan 75 ) 2 ( tan 20 ) 2 2( tan 75 )( tan 20 ) cos 123 d 2 2 [(tan 75 ) 2 (tan 20 ) 2 2(tan 75 )(tan 20 ) cos 123 ] d (tan 75 ) 2 (tan 20 ) 2 2(tan 75 )(tan 20 ) cos 123 d 39.4211... The ships are approximately 39.4 km apart. P O OT OPY. 6.5 The osine Law Solutions 45
14. The lengths of the sides of a parallelogram are 9 cm and cm. The shorter diagonal is 12 cm long. To the nearest tenth of a centimetre, what is the length of the longer diagonal? In PQR, determine Q. Use: q 2 p 2 r 2 2pr cos Q Sustitute: q 12, p 9, r 12 2 9 2 2 2(9)() cos Q 180 cos Q 37 cos Q 37 180 Q 78.1379... So, QPS 180 78.1379... 1.8620... In PQS, determine QS. Use: p 2 q 2 s 2 2qs cos P Sustitute: q 9, s, P 1.8620... p 2 9 2 2 2(9)() cos 1.8620... p 9 2 2 2(9)() cos 1.8620... p 14.7648... The longer diagonal is approximately 14.8 cm long. S P cm 12 cm R Q 9 cm 46 6.5 The osine Law Solutions O OT OPY. P