Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.5 - The Law of Sines - 6.5 Exercises - Page 515: 37

Answer

175 ft

Work Step by Step

Step 1. Represent the configuration with a new figure as shown, where T is the tree top, A is the end of shadow, AB=215 ft is the length of the shadow. $\angle TAC=52^\circ, \angle BAC=22^\circ,$ Step 2. In the right triangle $\Delta TAC$, we can obtain $\angle ATC=90-52=38^\circ$. We can also obtain $\angle TAB=52-22=30^\circ$ Step 3. In the triangle $\Delta TAB$, use the Since Law, we have $\frac{TB}{sin30^\circ}=\frac{215}{sin38^\circ}$ where TB is the height of the tree. Solve the equation, we obtain TB=$215\times\frac{sin30^\circ}{sin38^\circ}\approx175ft$
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