Math Law of Sines and Cosines 6 Solution

Law of Sines and Cosines 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? cos 40° = e / 3300 e = 2527.9 km 1) 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 = 1/2 * ab * sin C 9 = 1/2 * a² * sin 150° 36 = a² a = 6 c² = b² + b² - 2(b)(b)cos 150° c = 11.6 2) 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? tan 70° = n / 30 n = 82.4 mi 3) 82.4 miles 4. The area of ΔABC is 45 square units. If a = 10 and b = 15, find the measure of angle C to the nearest degree. Area = 1/2 * ab * sin C 45 = 1/2 * 10 * 15 * sin C C = 36.9° or 143.1° Need to find supplement too! 4) 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)cos Y Y = 78.5° (ZN)² = 2² + 5² - 2(2)(5)cos 78.5° ZN = 5.0 5) ZN = 5 6. Solve triangle XYZ if x = 52, y = 70 and z = 100. 100² = 52² + 70² - 2(52)(70)cos Z ∠Z = 109.2° sin 109.2°/100 = sin X/52 ∠X = 29.4° ∠Y = 41.4° 7. Solve triangle XYZ and find its area. x = 92.9 y = 110.8 ∠Z = 58° Area = 3489.2 units² 8. Solve triangle XYZ and find its area. x = 33.5° y = 94.5° z = 126.5° Area = 1246.5 units² 9. Eli Cooley flew his plane 800 km north, turned 135° and flew 1150 km. How far is Mr. Cooley from his starting point? x² = 800² + 1150² - 2(800)(1150)cos 135° 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. sin 35° = h / 5.74 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)cos 75° x = 373.8 sin 135°/373.8 = sin 15.2°/y sin 135°/373.8 = sin θ/250 θ = 40.2° or 139.8° y = 138.6 w = 262.7 A1 = ½(138.6)(262.7)sin 135° = 12872.9 A2 = ½(350)(250)sin 75° = 42259.25 Total Area = 55132.21 ft² Lakefront footage = 401.3 ft 12. The diagram shows the dimensions for a sail on a wooden model ship. Find the area of the sail to the nearest square inch. s = (12.5 + 26.9 + 18) / 2 = 28.7 Area I = √(28.7 * (28.7 - 26.9) * (28.7 - 18) * (28.7 - 12.5)) ≈ 94.6 in² Area II = ½ * (16.5) * (26) * sin 75° ≈ 207.2 in² Total Area = 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)cos 123° 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)cos 49° x ≈ 6042.8 m 8000² = 5000² + 6042.8² - 2(5000)(6042.8)cos θ θ = 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° Area = 5 * ½(4)(4)sin 72° = 38.04 in² 16. Observers at points A and B, 30 km apart, sight an airplane between them at angles of elevation of 40° and 75°, respectively. How far is the plane from each observer? sin 65° / 30 = sin 40° / a a = 21.3 km sin 65° / 30 = sin 75° / b b = 31.97 km 17. Two hikers follow a trail that splits into two forks. Each hiker takes a different fork. The forks diverge at an angle of 67° and both hikers walk at a speed of 3.5 mph. How far apart are the hikers after 1 hour? x² = 3.5² + 3.5² - 2(3.5)(3.5)cos 67° x = 3.86 mi 18. After leaving an airport, a plane flies for 1.5 hours at a speed of 200 km/h on a course of 200°. Then, on a course of 340°, the plane flies for 2 hours at a speed of 250 km/h. At this time, how far from the airport is the plane? 1.5 * 200 = 300 km 2 * 250 = 500 km x² = 300² + 500² - 2(300)(500)cos 40° x = 331.9 km