Chapter 6, Problem 16T

To determine

To find: The value of side labeled $x$.

Answer to Problem 16T

The value of $x$ is $19.5$.

Explanation of Solution

Given:

The value of angle is $108°$, length of side opposite to angle is $28$, and length of side adjacent to the angle is $15$.

Formula used:

Law of Sines

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Calculation:

Label the triangle.

The given triangle has two sides and one angle adjacent to one side.

Then for SSA, use law of sine.

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

First find the angle opposite to the second side which is given.

Use the law of Sines in given triangle.

$\frac{\sin A}{a}=\frac{\sin B}{b}$

Substitute $108°$ for $\angle A$, $28$ for $a$ and $15$ for $b$.

$\begin{array}{c}\frac{\sin 108°}{28}=\frac{\sin B}{15}\\ \sin B=\frac{15\times \sin 108°}{28}\\ \sin B=0.51\\ B={\sin }^{-1}\left(0.51\right)\end{array}$

Use scientific calculator to find the value of ${\sin }^{-1}\left(0.51\right)$.

Thus, the value for $\angle B$ is,

$\begin{array}{c}\angle B={\sin }^{-1}\left(0.51\right)\\ \approx 30.6\end{array}$

To find the third angle $\angle C$ opposite to the side labeled $x$, use the angle sum property of triangle.

$\angle A+\angle B+\angle C=180°$

Substitute $108°$ for $\angle A$ and $30.6°$ for $\angle B$

$\begin{array}{c}108°+30.6°+\angle C=180°\\ \angle C=180°-138.6°\\ =41.4°\end{array}$

Thus, the third angle $\angle C$ is $41.4°$.

Use law of sine to find third side $c$

$\frac{\sin A}{a}=\frac{\sin C}{c}$

Substitute $28$ for $a$, $108°$ for $\angle A$ and $41.4°$ for $\angle C$.

$\begin{array}{c}\frac{\sin 108°}{28}=\frac{\sin 41.4°}{c}\\ c=\frac{\sin 41.4°\times 28}{\sin 108°}\\ c=19.47\\ c\approx 19.5\end{array}$

The value of side labeled $x$ opposite to the sides given is $19.5$.