## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Acute Angles and Right Triangles - Section 2.4 Solving Right Triangles - 2.4 Exercises - Page 81: 61b

#### Answer

Accounting for the curvature of the earth, Mt. Everest would appear shorter than it actually is being over 100 miles away. 22,357 ft $\lt$ 29,000 ft

#### Work Step by Step

Total Mt. Everest height = T $\approx$ 29,000 ft Distance away = 100 miles = 528,000 ft TOA = $\tan\theta$ = $\frac{Opposite}{Adjacent}$ $\tan\theta$ = $\frac{29,000}{528,000}$ $\theta$ = $\tan^{-1}(\frac{29,000}{528,000})$ $\theta$ = 3.144 $^{\circ}$ Plugging this value, $\theta$, into part (a), you see that the height appears smaller than it normally would. $\sin\theta$ = $\frac{P}{h}$ $\sin3.14^{\circ}$ = $\frac{P}{142,631}$ P = $142,631\times\sin3.14^{\circ}$ P = 7,812.7 ft T = 7,812.7 + 14,545.0 T = 22,357.7 ft 22,357 ft $\lt$ 29,000 ft

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