Trigonometry (11th Edition) Clone

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

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.2 The Ambiguous Case of the Law of Sines - 7.2 Exercises - Page 311: 37

Answer

The height of the building is 218.1 ft

Work Step by Step

Let $d$ be the horizontal distance from the building to the point of observation on the ground. Let $x$ be the height of the building. We can use the law of sines and the angle to the bottom of the flagpole to find an expression for $d$: $\frac{d}{sin(90^{\circ}-26.0^{\circ})} = \frac{x}{sin(26.0^{\circ})}$ $d = \frac{x~sin(64.0^{\circ})}{sin(26.0^{\circ})}$ We can use the law of sines and the angle to the top of the flagpole to find an expression for $d$: $\frac{d}{sin(90^{\circ}-35.0^{\circ})} = \frac{x+95.0}{sin(35.0^{\circ})}$ $d = \frac{(x+95.0)~sin(55.0^{\circ})}{sin(35.0^{\circ})}$ We can equate the two expressions for $d$ and solve for $x$: $\frac{x~sin(64.0^{\circ})}{sin(26.0^{\circ})} = \frac{(x+95.0)~sin(55.0^{\circ})}{sin(35.0^{\circ})}$ $x~sin(64.0^{\circ})~{sin(35.0^{\circ}) = (x+95.0)~sin(55.0^{\circ})~sin(26.0^{\circ})}$ $0.515527~x = 0.359093~(x+95.0)$ $0.1564~x = 34.114$ $x = \frac{34.114}{0.1564}$ $x = 218.1~ft$ The height of the building is 218.1 ft
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