Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 8 - Right Triangles and Trigonometry - 8-5 Law of Sines - Lesson Check - Page 524: 1


$m \angle C = 43.6^{\circ}$

Work Step by Step

In this exercise, we are given the measure of an angle and two sides. We can use the law of sines to find the measure of another angle in the triangle. The law of sines states the following: $\frac{sin A}{a}$ = $\frac{sin C}{c}$, where $A$ and $C$ are the measures of two angles in the triangle and $a$ and $c$ are the measures of the sides opposite those two angles, respectively. The side opposite to $\angle A$ is $\overline{BC}$, and the side opposite to $\angle C$ is $\overline{AB}$. We want to find $m \angle C$. Let's plug in what we know into the formula: $\frac{sin 80}{10}$ = $\frac{sin m \angle C}{7}$ Multiply each side by $7$: $7\frac{sin 80}{10}$ = sin $m \angle C$ Evaluate the expression on the left: sin $m \angle C \approx 0.6894$ Take $sin^{-1}$ to solve for $C$: $m \angle C = 43.6^{\circ}$
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