Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - Chapter Review - Page 656: 58

Answer

$AB=26.9$. $AC=20.0$.

Work Step by Step

The given values are $BC=18$ and $\angle A=42^{\circ}$. The triangle is shown in the image file. Use trigonometric ratios. $\Rightarrow \sin A =\frac{BC}{AB}$ Multiply both sides by $\frac{AB}{\sin A}$. $\Rightarrow \frac{AB}{\sin A}\cdot \sin A =\frac{AB}{\sin A}\cdot\frac{BC}{AB}$ Simplify. $\Rightarrow AB=\frac{BC}{\sin A}$ Substitute all values. $\Rightarrow AB=\frac{18}{\sin 42^{\circ}}$ Simplify. $AB=26.9$. Now use $\Rightarrow \tan A =\frac{BC}{AC}$ Multiply both sides by $\frac{AC}{\tan A}$. $\Rightarrow \frac{AC}{\tan A} \cdot \tan A =\frac{AC}{\tan A} \cdot \frac{BC}{AC}$ Simplify. $\Rightarrow AC =\frac{BC}{\tan A}$ Substitute all values. $\Rightarrow AC =\frac{18}{\tan 42^{\circ}}$ Simplify. $AC=20.0$.
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