Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Test - Page 98: 20

Answer

448 m

Work Step by Step

In triangle ABC, $\tan 52.5^{\circ}=\frac{h}{x}$ or $h=x\tan52.5^{\circ}$ In triangle BCD, $\tan41.2^{\circ}=\frac{h}{x+168}$ or $h=(x+168)\tan 41.2^{\circ}$ Equating both expressions for $h$, we get $x\tan52.5^{\circ}=(x+168)\tan41.2^{\circ}$ Using distributive property, we have $x\tan52.5^{\circ}=x\tan41.2^{\circ}+168\tan41.2^{\circ}$ $\implies x(\tan 52.5^{\circ}-\tan41.2^{\circ})=168\tan41.2^{\circ}$ Or $x=\frac{168\tan41.2^{\circ}}{\tan52.5^{\circ}-\tan41.2^{\circ}}$ We saw above that $h=x\tan52.5^{\circ}$ Substituting for $x$, we get $h=(\frac{168\tan41.2^{\circ}}{\tan52.5^{\circ}-\tan41.2^{\circ}})\tan52.5^{\circ}$ $=448\,m$
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