University:
Morehouse College
Course:
HMTH 116 | Trigonometry
Academic year:

2022
Views:

213

Pages:

4
Author:

5CelestialChronicle

Law of Sines and Cosines Name: Key 1. An airplane flies on a course of 130° at a speed of 1100 km/h. How far east of its starting point is it after 3 hours? cos40° = e/3300 e = 2527.9 km 2. One angle of an isosceles triangle has a measure of 150°. If the area of the triangle is 9 cm², what is the perimeter of the triangle? Area = 1/2absinC 9 = 1/2(a)(a)sin15° 9 = 1/2(a²) 36 = a² a = 6 c² = b² + b² - 2(6)(6)cos150° c = 11.6 Perimeter = 6 + 6 + 11.6 = 23.6 cm 3. A ship leaves port and proceeds west 30 miles. It then changes course to 020° until it is due north of its origin. How far north of its origin is it? tan70° = n/30 n = 82.4 miles 4. The area of triangle ABC is 45 square units. If a = 10 and b = 15, find the measure of angle C to the nearest degree. Area = 1/2absinC 45 = 1/2(10)(15)sinC C = 36.9° or 143.1° 5. Given the diagram below, find ZN to the nearest whole unit. Y is the largest angle. Find Y first! Then find ZN. 7² = 5² + 6² - 2(5)(6)cosY Y = 78.5° (ZN)² = 2² + 5² - 2(2)(5)cos78.5° ZN = 5.0 6. Solve △XYZ if x = 52, y = 70 and z = 100. 100² = 52² + 70² - 2(52)(70)cos∠Y ∠Y = 41.4° ∠Z = 109.2° ∠X = 29.4° 7. Solve △XYZ and find its area. Given: x = 52, ∠Y = 70°, and ∠X = 52° sin58°/100 = sin52°/x x = 92.9 sin58°/100 = sin70°/y y = 110.8 ∠Z = 58° Area = 1/2(52)(100)sin58° = 2196.5 units² 8. Solve △XYZ and find its area. Given: x = 52°, y = 70°, and z = 100 100² = 52² + 70² - 2(52)(70)cos∠X ∠X = 33.5° ∠Y = 146.5° ∠Z = 126.5° Area = 1/2(70)(100)sin126.5° = 3489.2 units² 9. Eli Cooley flew his plane 900 km north, turned 135°, and flew 1150 km. How far is Mr. Cooley from his starting point? x² = 900² + 1150² - 2(900)(1150)cos135° x ≈ 2032.7 km 10. The angle of elevation to the peak of a mountain is 30°. A kilometer closer, the angle of elevation is 35°. Find the height of the mountain. sin35°/h = sin5°/1 h ≈ 3.3 km 11. A lakefront plot of land is shown below. What is its area and lakefront footage (bolded sides are lakefront footage)? x² = 250² + 350² - 2(250)(350)cos75° x ≈ 373.8 sin75°/373.8 = sinθ/250 θ = 40.2° or 139.8° y = 138.6 w = 262.7 A1 = 1/2(138.6)(262.7)sin135° = 12872.9 A2 = 1/2(350)(250)sin75° = 42259.25 Area = 55132.21 ft² Lakefront footage = 401.3 x² = 26² + 16.5² - 2(26)(16.5)cos75° x ≈ 26.9 S = (26.9 + 26 + 16.5)/2 = 34.7 Area I = √(34.7(34.7-26.9)(34.7-26)(34.7-16.5)) ≈ 94.6 in² Area II = 1/2(16.5)(26)sin75° ≈ 207.2 in² Area I + Area II = 94.6 + 207.2 ≈ 301.8 in² 13. A ship leaves port and travels 36 miles west, then 24 miles on a course bearing 213°. How far is it from its starting point? x² = 24² + 36² - 2(24)(36)cos123° x ≈ 53.04 miles 14. A hiker walks 8000 m on a course of S 81° E. She then changes direction and hikes 5000 m on a course of N 32° W. How far is she from her starting point, and on what course must she travel to return to the starting point? x² = 5000² + 8000² - 2(5000)(8000)cos49° x ≈ 6042.8 m 8000² = 5000² + 6042.8² - 2(5000)(6042.8)cosα α ≈ 92.4° θ = 92.4° - 32° = 60.4° S 60.4° W 15. A regular pentagon is inscribed in a circle of radius 4 in. Find the area of the pentagon. 360°/5 = 72° A = 1/2(4)(4)sin72° = 7.61 7.61 * 5 = 38.04 in²

Math Law of Sines and Cosines Solution Key

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