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Q30.

Expert-verifiedFound in: Page 707

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** $a=13.7,c=15.4\text{and}\angle A=63\xb0$.**

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

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, $\angle A=63\xb0$ by angle sum property.

Now, trigonometric ratios,

$\begin{array}{c}\mathrm{sin}B=\frac{b}{c}\\ \mathrm{sin}27\xb0=\frac{7}{c}\\ c=7\mathrm{csc}\left(27\xb0\right)\\ c=15.4\end{array}$

Now, in right triangle ABC by Pythagoras Theorem.

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

The** **solution for triangle *ABC*, is $a=13.7,c=15.4\text{and}\angle A=63\xb0$.

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