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Q30.

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Found in: Page 707

### Algebra 2

Book edition Middle English Edition
Author(s) Carter
Pages 804 pages
ISBN 9780079039903

# Solve triangle ABC using the given measurements if $B=27{}^{\circ },C=90{}^{\circ }\text{and}\phantom{\rule{-0.2em}{0ex}}\phantom{\rule{-0.2em}{0ex}}\phantom{\rule{-0.2em}{0ex}}\phantom{\rule{-0.2em}{0ex}}\text{}b=7$. Round measure of sides to the nearest tenth and angles to nearest degree.

The solution for triangle ABC is $a=13.7,c=15.4\text{and}\angle A=63°$.

See the step by step solution

## Step 1. Write down the given parameters of triangle.

The given triangle ABC is shown below whose parameters are represented as $B=27{}^{\circ },C=90{}^{\circ }\text{and}\phantom{\rule{-0.2em}{0ex}}\phantom{\rule{-0.2em}{0ex}}\phantom{\rule{-0.2em}{0ex}}\phantom{\rule{-0.2em}{0ex}}\text{}b=7$.

## Step 2. Calculation.

In triangle ABC, $B=27{}^{\circ },C=90{}^{\circ }\text{and}\phantom{\rule{-0.2em}{0ex}}\phantom{\rule{-0.2em}{0ex}}\phantom{\rule{-0.2em}{0ex}}\phantom{\rule{-0.2em}{0ex}}\text{}b=7$. Therefore, $\angle A=63°$ by angle sum property.

Now, trigonometric ratios,

$\begin{array}{c}\mathrm{sin}B=\frac{b}{c}\\ \mathrm{sin}27°=\frac{7}{c}\\ c=7\mathrm{csc}\left(27°\right)\\ c=15.4\end{array}$

Now, in right triangle ABC by Pythagoras Theorem.

$\begin{array}{c}{c}^{2}={a}^{2}+{b}^{2}\text{\hspace{0.17em}gives,}\\ {15.4}^{2}={a}^{2}+{7}^{2}\\ {a}^{2}={15.4}^{2}-{7}^{2}\\ a=\sqrt{{15.4}^{2}-{7}^{2}}\\ a=13.7\end{array}$

## Step 3. Conclusion.

The solution for triangle ABC, is $a=13.7,c=15.4\text{and}\angle A=63°$.