573 5.4 Solutions and Applications of Right Triangles ALGEBRAIC SOLUTION Figure 56 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. See Figure 57 in the Graphing Calculator Solution. Because nothing is given about the length of the hypotenuse of either triangle ABC or triangle BCD, we use a ratio that does not involve the hypotenuse—namely, the tangent. In triangle ABC, tan 36.7° = h x or h = x tan 36.7°. In triangle BCD, tan 22.2° = h 50 + x or h = 150 + x2 tan 22.2°. Each expression equals h, so the expressions must be equal. x tan 36.7° = 150 + x2 tan 22.2° Equate expressions for h. x tan 36.7° = 50 tan 22.2° + x tan 22.2° Distributive property x tan 36.7° - x tan 22.2° = 50 tan 22.2° Write the x-terms on one side. x1tan 36.7° - tan 22.2°2 = 50 tan 22.2° Factor out x. x = 50 tan 22.2° tan 36.7° - tan 22.2° Divide by the coefficient of x. We saw above that h = x tan 36.7°. Substitute for x. h = a 50 tan 22.2° tan 36.7° - tan 22.2°b tan 36.7° Use a calculator. tan 36.7° = 0.74537703 and tan 22.2° = 0.40809244 Thus, tan 36.7° - tan 22.2° = 0.74537703 - 0.40809244 = 0.33728459 and h = a 5010.408092442 0.33728459 b 0.74537703 ≈45. To the nearest foot, the height of the tree is 45 ft. 22.28 50 ft 36.78 A C B h D x Figure 56 EXAMPLE 8 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°. See Figure 56. Find the height of the tree to the nearest foot. GRAPHING CALCULATOR SOLUTION* In Figure 57, we have superimposed Figure 56 on coordinate axes with the origin at D. By definition, 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°. Because b = 0, the equation of line DB is y1 = 1tan 22.2°2x. The equation of line AB is y2 = 1tan 36.7°2x + b. Because b≠0 here, we use the point A150, 02 and the point-slope form to find the equation. y2 - y0 = m1x - x02 Point-slope form y2 - 0 = m1x - 502 x0 = 50, y0 = 0 y2 = tan 36.7°1x - 502 Lines y1 and y2 are graphed in Figure 58. The y-coordinate of the point of intersection of the graphs gives the length of BC, or h. Thus, h ≈45 . 22.28 50 ft 36.78 A C B h D x x y Figure 57 −10 −20 75 150 Figure 58 S Now Try Exercise 77. *Source: Reprinted with permission from The Mathematics Teacher, copyright 1995 by the National Council of Teachers of Mathematics. All rights reserved.
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