571 5.4 Solutions and Applications of Right Triangles EXAMPLE 5 Solving a Problem Involving Bearing (Method 1) Radar stations A and B are on an east-west line, 3.7 km apart. Station A detects a plane at C, on a bearing of 61°. Station B simultaneously detects the same plane, on a bearing of 331°. Find the distance from A to C. SOLUTION Begin with a sketch showing the given information. See Figure 52. A line drawn due north is perpendicular to an east-west line, so right angles are formed at A and B. Angles CBA and CAB can be found as follows. ∠CBA = 331° - 270° = 61° and ∠CAB = 90° - 61° = 29° A right triangle is formed. The distance from A to C, denoted b in the figure, can be found using the cosine function for angle CAB. cos 29° = b 3.7 Cosine ratio b = 3.7 cos 29° Multiply by 3.7 and rewrite. b ≈3.2 km Two significant digits S Now Try Exercise 69. 61° N A 29° B C b N 61° 331° 3.7 km Figure 52 Expressing Bearing (Method 2) Start with a north-south line and use an acute angle to show the direction, either east or west, from this line. Figure 53 shows several sample bearings using this method. Either N or S always comes first, followed by an acute angle, and then E or W. N 428 N 428 E S 318 E 318 S S 408 W 408 S N 528 N 528 W Figure 53 EXAMPLE 6 Solving a Problem Involving Bearing (Method 2) A ship leaves port and sails on a bearing of N 47° E for 3.5 hr. It then turns and sails on a bearing of S 43° E for 4.0 hr. If the ship’s rate is 22 knots (nautical miles per hour), find the distance that the ship is from port. SOLUTION Draw and label a sketch as in Figure 54. Choose a point C on a bearing of N 47° E from port at point A. Then choose a point B on a bearing of S 43° E from point C. Because north-south lines are parallel, angle ACD measures 47° by alternate interior angles. The measure of angle ACB is 47° + 43° = 90°, making triangle ABC a right triangle. B D N A Port 438 478 S S C 478 N c b = 77 nautical mi a = 88 nautical mi Figure 54
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