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MPM2D Grade 10 Academic SOLVING RIGHT TRIANGLES Practice 1. Solve each triangle. Round each side length to the nearest tenth of a unit, and each angle to the nearest degree. a) Solution β C = 90o β β E = 90o β 33o = 57o, πΆπ· πΆπ· sin β E = βΉ sin 33o = πΆπΈ 22 CD = 22 sin 33o β12.0 m, πΈπ· π π· cos β E = βΉ cos 33o = πΆπΈ 22 ED = 22 cos 33o β18.5 m. b) Solution β B = 90o β β G = 90o β 65o = 25o, π΅πΉ 29 sin β G = βΉ sin 65o = π΅πΊ BG = tan β G = π΅πΉ πΊπΉ 29 π΅πΊ β 32.0 cm, sin 65° 29 o βΉ tan 65 = GF = πΊπΉ 29 tan 65° β 13.5 cm. c) Solution β P = 90o β β R = 90o β 29o = 61o, π π 51 cos β R = βΉ cos 29o = π π RP = tan β R = ππ π π 51 π π β 58.3 m, cos 29° ππ o βΉ tan 29 = 51 PM = 51 tan 29o β 28.3 m. d) Solution β X = 90o β β Z = 90o β 30o = 60o, ππ 6 sin β Z = βΉ sin 30o = ππ XZ = tan β Z = ππ ππ 6 sin 30° 6 o βΉ tan 30 = YZ = ππ 6 β 12 cm, ππ tan 30° β 10.4 cm. e) Solution β S = 90o β β N = 90o β 54o = 36o, ππ ππ sin β N = βΉ sin 54o = ππ 17 OS = 17 sin 54o β13.8 km, ππ ππ cos β N = βΉ cos 54o = ππ 17 ON = 17 cos 54o β10.0 km. f) Solution β T = 90o β β M = 90o β 13o = 77o, ππ ππ sin β M = βΉ sin 13o = ππ 13 TX = 13 sin 13o β 2.9 m, ππ ππ cos β M = βΉ cos 13o = ππ 13 XM = 13 cos 13o β 12.7 m, 2. Find all the unknown angles, to the nearest degree, and all the unknown side lengths, to the nearest tenth of a unit. a) Solution π΄π΅ 13 cos β A = βΉ cos β A = π΄πΆ β1 13 15 β A = cos (15) β 30 , β C = 90o β β A β 90o β 30o = 60o. o By the Pythagorean Theorem, determine BC: BC = βπ΄πΆ 2 β π΄π΅2 = β152 β 132 = β56 β 7.5 cm. b) Solution tan β G = πΌπ» πΊπ» βΉ tan β G = 9 9 7 β G = tan β1 (7) β 52o, β I = 90o β β G β 90o β 52o = 38o. By the Pythagorean Theorem, determine GI: GI = βπΌπ» 2 + πΊπ» 2 = β92 + 72 = β130 β 11.4 cm. c) Solution πΎπΏ 5 cos β L = βΉ cos β L = π½πΏ 6 5 β L = cos β1 (6) β 34o, β J = 90o β β L β 90o β 34o = 56o. By the Pythagorean Theorem, determine JK: JK = βπ½πΏ2 β πΎπΏ2 = β62 β 52 = β11 β 3.3 cm. d) Solution ππ tan β P = βΉ tan β P = 1.1 ππ β1 β P = tan (1.1) β 48o, β R = 90o β β P β 90o β 48o = 42o. By the Pythagorean Theorem, determine PR: PR = βππ 2 + ππ2 = β112 + 102 = β221 β 14.9 cm. e) Solution ππ tan β R = βΉ tan β R = 1.125 π π β R = tan β1 (1.125) β 48o, β T = 90o β β R β 90o β 48o = 42o. By the Pythagorean Theorem, determine TR: TR = βππ 2 + π π 2 = β182 + 162 = β580 β 24.1 m. f) Solution sin β W = ππ ππ βΉ sin β W = 25 25 54 β W = sin β1 (54) β 28o, β Y = 90o β β W β 90o β 28o = 62o. By the Pythagorean Theorem, determine WX: WX = βππ 2 β ππ 2 = β542 β 252 = β56 β 47.9 cm. Application and Problem Solving 3. Peggyβs Cove Lighthouse, in Nova Scotia, is possibly the most photographed lighthouse in the world. The observation deck is about 20 m above sea level. From the observation deck, the angle of depression of a boat is 60 . How far is the boat from the lighthouse, to the nearest metre? Solution π΅πΆ 20 tan A = βΉ tan 6o = AB = π΄π΅ 20 tan 6° π΄π΅ β 190 m. The boat is about 190 m from the lighthouse. 4. A kite is 32 m above the ground. The angle the kite string makes with the ground is 390 . How long is the kite string, to the nearest metre? Solution π΄π΅ 32 sin β C = βΉ sin 39o = AC = π΄πΆ 32 sin 39° π΄πΆ β 51 m. The kite string ia about 51 m long. 5. The worldβs fastest roller coaster is at Six Flags Magic Mountain. At the start, riders are shot forward and then up a tower. They then begin the backward descent. From a point 100 m from the foot of the tower, the angle of elevation of the top of the tower is 520 . Find the height of the tower, to the nearest metre. Solution β tan 52o = βΉ h = 100 tan 52o β 128 m 100 The height of the tower is about 128 m. 6. Montrealβs Marathon building is 195 m tall. From a point level with, and 48 m from, the base of the building, what is the angle of elevation of the top of the building, to the nearest degree? Solution 195 195 tan x = βΉ x = tan β1 ( ) β 79o 38 38 The angle of elevation of the top of the building is about 79o. 7. Ropes are used to pull a totem pole upright. Then, the ropes are anchored in the ground to hold the pole until the hole is filled. One of the ropes holding this totem pole is 18 m long and forms an angle of 480 with the ground. Find, to the nearest metre, a) the height of the totem pole Solution β sin 48o = βΉ h = 18 sin 48o β 13 m 18 The height of the totem pole is about 13 m. b) how far the anchor point is from the base of the totem pole Solution π₯ cos 48o = βΉ x = 18 cos 48o β 12 m. 18 The anchor point is about 12 m from the base of the totem pole. 8. The two guy wires supporting a flagpole are each anchored 7 m from the flagpole and form an angle of 520 with the ground. What is the total length of guy wire, to the nearest metre, needed to support this flagpole? Solution 7 7 cos 52o = βΉ x = β 11.4 m. π₯ cos 52° l = 2x = 2(11.4) β 23 m. The total length of guy wire is about 23 m. 9. Edmontonβs CN Tower is a high-rise office building. It has 27 storeys. From a point 35 m from the base of the building and level with the base, the angle of elevation of the top is 72.5o. a) Find the height of Edmontonβs CN Tower, to the nearest metre. Solution β tan 72.5o = βΉ h = 35 tan 72.5oβ 111 m. 35 The height of Edmontonβs CN Tower is about 111 m. b) Torontoβs CN Tower is a tourist attraction, with a height of 555 m. If an office building were the height of Torontoβs CN Tower, how many stories would you expect it to have? Explain. Solution π 27 = βΉ n = 27× 5 = 135 m 555 111 I would expect it has 135 stories. 10. A coast guard patrol boat is 14.8 km east of the Brier Island lighthouse. A disabled yacht is 7.5 km south of the lighthouse. a) How far is the patrol boat from the yacht, to the nearest tenth of a kilometre? Solution By Pythagorean Theorem, determine BY: BY2 = 7.52 + 14.82 BY2 = 275.29 BY β16.6 km The patrol boat from is 16.6 km from the yacht. b) At what angle south of due west, to the nearest degree, should the patrol boat travel to reach the yacht? Solution 7.5 tan π = β 0.506757 14.8 π = tan β1 (0.506757) β 27o To reach the yacht, the patrol boat should travel in the direction W27o S. 11. A sign shows that a hill has a grade of 9%. What angle does the hill make with the horizontal, to the nearest tenth of a degree? Solution tan π = 0.09 βΉ π = tan β1 (0.09) β 5.1o. 12. Use right triangles ABC and DEF to complete the table. Leave all ratios in fraction form. Triangle ABC DEF tan x sin x cos x tan (90° β x) 5 5 12 /12 /13 12/13 /5 3 3 4 4 /4 /5 /5 /3 a) How is tan x related to tan (90° β x)? Answer tan x × tan (90° β x) = 1 b) How is sin x related to cos (90° β x)? Answer sin x = cos (90° β x) c) How is cos x related to sin (90° β x)? Answer cos x = sin (90° β x) sin (90° β x) cos (90° β x) 12 5 /13 /13 4 3 /5 /5