Copyright © 2017, 2013, 2009 Pearson Education, Inc. Acute Angles and Right Triangles Copyright © 2017, 2013, 2009 Pearson Education, Inc. 1

Further Applications of Right Triangles 2.5 Further Applications of Right Triangles Historical Background ▪ Bearing ▪ Further Applications

Bearing There are two methods for expressing bearing. When a single angle is given, such as 164°, it is understood that the bearing is measured in a clockwise direction from due north.

Example 1 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. Right angles are formed at A and B, so angles CAB and CBA can be found as shown in the figure. Angle C is a right angle because angles CAB and CBA are complementary.

Caution A correctly labeled sketch is crucial when solving bearing applications. Some of the necessary information is often not directly stated in the problem and can be determined only from the sketch.

Bearing The second method for expressing bearing starts with a north-south line and uses an acute angle to show the direction, either east or west, from this line.

Example 2 SOLVING A PROBLEM INVOLVING BEARING (METHOD 2) A ship leaves port and sails on a bearing of N 47º E for 3.5 hr. It then turns and sails on a bearing of S 43º 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. Draw a sketch as shown in the figure. 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 43º E from point C. Because north-south lines are parallel, angle ACD is 47º by alternate interior angles. The measure of angle ACB is 47º + 43º = 90º, making triangle ABC a right triangle.

Now find c, the distance from port at point A to the ship at point B. Example 2 SOLVING A PROBLEM INVOLVING BEARING (METHOD 2) (cont.) Next, use the formula relating distance, rate, and time to find the distances from A to C and from C to B. Now find c, the distance from port at point A to the ship at point B.

(a) Find d with θ = 1°23′12″ and b = 2.0000 cm. Example 3 USING TRIGONOMETRY TO MEASURE A DISTANCE The subtense bar method is a method that surveyors use to determine a small distance d between two points P and Q. The subtense bar with length b is centered at Q and situated perpendicular to the line of sight between P and Q. Angle θ is measured, then the distance d can be determined. (a) Find d with θ = 1°23′12″ and b = 2.0000 cm. From the figure, we have

Let b = 2. Convert θ to decimal degrees: Example 3 USING TRIGONOMETRY TO MEASURE A DISTANCE (continued) Let b = 2. Convert θ to decimal degrees:

Since θ is 1″ larger, θ = 1°23′13″ ≈ 1.386944º. Example 3 USING TRIGONOMETRY TO MEASURE A DISTANCE (continued) (b) How much change would there be in the value of d if θ were measured 1″ larger? Since θ is 1″ larger, θ = 1°23′13″ ≈ 1.386944º. The difference is 82.634110 – 82.617558 = 0.016552 cm.

Example 4 SOLVING A PROBLEM INVOLVING ANGLES OF ELEVATION Francisco needs to know the height of a tree. From a given point on the ground, he finds that the angle of elevation to the top of the tree is 36.7°. He then moves back 50 ft. From the second point, the angle of elevation to the top of the tree is 22.2°. Find the height of the tree to the nearest foot. The figure shows two unknowns: x, the distance from the center of the trunk of the tree to the point where the first observation was made, and h, the height of the tree. Since nothing is given about the length of the hypotenuse, of either triangle ABC or triangle BCD, use a ratio that does not involve the hypotenuse—namely, the tangent.

Example 4 In triangle ABC: In triangle BCD: SOLVING A PROBLEM INVOLVING ANGLES OF ELEVATION (continued) In triangle ABC: In triangle BCD: Each expression equals h, so the expressions must be equal.

Since h = x tan 36.7°, we can substitute. Example 4 SOLVING A PROBLEM INVOLVING ANGLES OF ELEVATION (continued) Since h = x tan 36.7°, we can substitute. The height of the tree is approximately 45 ft.

Graphing Calculator Solution Example 4 SOLVING A PROBLEM INVOLVING ANGLES OF ELEVATION (continued) Graphing Calculator Solution Superimpose coordinate axes on the figure with D at the origin. The tangent of the angle between the x-axis and the graph of a line with equation y = mx + b is the slope of the line, m. For line DB, m = tan 22.2°. Since b = 0, the equation of line DB is

Example 4 The equation of line AB is SOLVING A PROBLEM INVOLVING ANGLES OF ELEVATION (continued) The equation of line AB is Since b ≠ 0 here, we use the point A(50, 0) and the point-slope form to find the equation.

Example 4 SOLVING A PROBLEM INVOLVING ANGLES OF ELEVATION (continued) Graph y1 and y2, then find the point of intersection. The y-coordinate gives the length of BC, or h. Thus, h ≈ 45.