Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 2 - Section 2.3 - Solving Right Triangles - 2.3 Problem Set: 53

Answer

Chapter 2 - Section 2.3 Problem Set: 53 (Answer) Refer to Figure 1 $x = 17$ (To 2 significant digits)
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Work Step by Step

Chapter 2 - Section 2.3 Problem Set: 53 (Solution) Refer to Figure 1 In $\triangle$BAC, $\tan A$ = $\frac{h}{(x+y)}$ $\tan 43^{\circ}$ = $\frac{h}{(x+11)}$ $h = (x+11) \cdot \tan 43^{\circ}$ $\longrightarrow (1)$ In $\triangle$BDC, $\tan \angle BDC$ = $\frac{h}{x}$ $\tan 57^{\circ}$ = $\frac{h}{x}$ $h = x \cdot \tan 57^{\circ}$ $\longrightarrow (2)$ Equating $(1) to (2)$ $(x+11) \cdot \tan 43^{\circ}$ = $x \cdot \tan 57^{\circ}$ $x = \frac{11\tan 43^{\circ}}{\tan 57^{\circ} - \tan 43^{\circ}}$ $x = 17$ (To 2 significant digits)
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