Math Law of Sines and Cosines Applications Solution

Honors Pre-Calculus Law of Sines & Cosines Applications Name: Where appropriate, give angle measures to the nearest tenth of a degree and lengths of sides in simplest radical form or to the nearest hundredth. 1. Find the area   of triangle PQR if q = 6, r = 7, and ∠P = 50°. Also find p. Area = 1/2 * 6 * 7 * sin 50° = 16.1 units² p² = 6² + 7² - 2(6)(7)cos 50° p = 5.6 2. Find the measure of the largest angle in a triangle with sides having lengths 3√6, 6√3, and 9√2. (9√2)² = (3√6)² + (6√3)² - 2(3√6)(6√3)cos x 0 = -54√2 cos x 0 = cos x x = cos⁻¹(0) x = 90° 3. In ΔRST, ∠R = 75°, ∠S = 45°, and t = 3. Find r and s. sin 60°/3 = sin 75°/r = sin 45°/s r = 3.35 s = 2.45 4. Three measurements in ΔABC are given as ∠A = 60°, a = 4, and b = 5. Show that at least one of the measurements is incorrect. sin 60°/4 = sin B/5 sin B = 5sin 60°/4 sin B = 1.0825 Sine cannot be greater than 1 5. A regular polygon with 180 sides is inscribed in a circle with radius 1. Find its area. Compare your answer with π. Central angle = 360°/180 = 2° Area of one triangle = 1/2 * 1 * 1 * sin 2° = 0.01745 Total area = 180 * 0.01745 ≈ 3.141 6. A submarine dives at an angle of 16° with the horizontal. If it takes 4 minutes to dive from the surface to a depth of 300 feet, how fast does it move along its sloping path downward? Give your   answer in feet per minute. Then convert it to nautical miles per hour. Note: 1 nautical mile per hour   ≈ 6080 feet per hour. sin 16° = 300/x x = 300/sin 16° ≈ 1088.4 ft in 4 min So 1088.4/4 = 272.1 ft/min 272.1 ft/min * 60 min/1 hr * 1 nm/6080 ft ≈ 2.69 nm/hr 7. In ΔXYZ, ∠X = 21.1°, x = 6, and y = 9. Find the measure(s) of ∠Y. sin 21.1°/6 = sin Y/9 Y = 32.7° and 147.3° 8. In parallelogram ABCD, ∠A = 60°, AB = 5, and AD = 8. a. Find the area of ABCD. Area = 2 * 1/2 * 5 * 8 * sin 60° = 34.6 units² b. Find the lengths of both diagonals. x² = 5² + 8² - 2(5)(8)cos 120° x = 11.4 y² = 5² + 8² - 2(5)(8)cos 60° y = 7 9. A triangle has an area of 21 cm² and two of its sides are 9 cm and 14 cm long. Find the possible measures of the angle formed by these sides. 21 = 1/2 * 9 * 14 * sin A sin A = 1/3 A ≈ 19.5° or 160.5° 10. In the diagram below, ΔABC is similar to ΔDEF and ∠A = 120°. a. Find the lengths a, e, and f. a = 14 e = 8.57 f = 14.3 b. Find the ratio of the areas of the triangles. ratio of sides = 14/20 = 7/10 ratio of areas = (7/10)² = 49/100 11. The diagonals of a parallelogram have lengths 8 and 14 and they meet at a 60° angle. Find the area and perimeter of the parallelogram. Area = 2 * 1/2 * 4 * 7 * sin 60° = 48.5 units² Perimeter = 2(6.1) + 2(9.6) = 31.4 units 12. An obtuse triangle with sides 12 and 10 has two sides of lengths 4 and 10. Find the length of the third side. There are two answers. x² = 4² + 10² - 2(4)(10)cos 143.1° x = 7.2 or 13.4 13. The perimeter of a regular decagon (10 sides) is 240. Find its area. Area of decagon = 10 * (area of one triangle) = 4424.38 units² 14. If fencing costs $2.50 per foot, how much will it cost to buy fencing to go around the plot of land shown below? Perimeter = 50 + 38.1 + 73.4 + 30 = 191.50 ft Cost = 191.50 * 2.50 = $478.75 15. In the township of Madison, rural undeveloped land is taxed at a rate of $115 per acre. Find the tax on the plot of land shown. Note: 1 acre = 43,560 ft² x² = 100² + 130² - 2(100)(130)cos 120° x = 199.7 Area = 1/2(125)(199.7)sin 54.3° + 1/2(100)(130)sin 120° ≈ 15764.98 ft² 15764.98 / 43560 ≈ 0.36 acres $115 * 0.36 = $41.40 16. A ship is steaming north at 6 knots (nautical miles per hour) when the captain sights a small island at an angle of 15° to the east of the ship's course. After 10 minutes, the angle is 28°. How far away is the island at this moment? 1 nm/hr * 1/6 hr = 1/15 nm sin 15°/x = sin 13°/1 x = 1.15 nm 17. In ΔABC, a = 5, b = 8, and c = 7. a. Solve ΔABC. A = 38.2° B = 81.8° C = 60° b. Find the area of ΔABC. Area = 17.32 units² c. Find the length of the altitude to AC. 4.33 18. Ship A sights ship B on a compass bearing of 080°. Make a sketch and give the compass bearing of ship A from ship B. 260° or S 80° W 19. Ship X sights ship Y on a bearing of 308°. What is the bearing of X from Y? 128° or S 38° E 20. An airplane flies on a course of 110° at a speed of 1200 km/hr. How far east of its starting point is it after 2 hours? 2255.3 km 21. A hunter walks east for 1 hour and then north for 1 1/2 hours. What course should the hunter take to return to his starting point? What assumptions do you make to answer the question? 1 Rate is constant S 33.7° W or 213.7° tan θ = 1/1.5 θ = 33.7° S 33.7° W or 213.7° 22. Point B is 10 km north of point A, and point C is 10 km north of point B on a bearing of 060° from B. Find the bearing and distance of C from A. Bearing: 30° Distance: 17.3 km x² = 10² + 10² - 2(10)(10)cos120° x = 17.3 km 23. Point S is 4 km west of point R, and point T is 4 km southwest of S. Find the bearing and distance of R from T. Bearing: 67.5° Distance: 7.4 km x² = 4² + 4² - 2(4)(4)cos135° x = 7.4 km 24. Traveling at a speed of 10 knots, a ship proceeds south from its port for 1 1/2 hours and then changes course to 130° for 1/2 hour. At this time, how far from port is the ship? 18.6 nm x² = 5² + 15² - 2(5)(15)cos130° x = 18.6 nm 25. A sailboat leaves its dock and proceeds east for 2 miles. It then changes course to 205° until it is due south of its dock. How far south is it? 4.3 miles tan 65° = x/2 x = 4.3 miles 26. Two ships, A and B, leave port at the same time. Ship A proceeds at 12 knots on a course of 040°, while ship B proceeds at 9 knots on a course of 115°. After 2 hours, ship A loses power and radios for help. How far and on what course must ship B travel to reach ship A? Bearing: 358° Distance: 26 nm x² = 18² + 24² - 2(18)(24)cos75° x = 26 nm Bearing: 270° + 25° = 295° = 358°