Chapter 6, Problem 18T

To determine

To find: The value of angle labeled $\theta$.

Answer to Problem 18T

The value of $\theta$ is $40.2°$.

Explanation of Solution

Given:

The value of one side is $5$, second side is $7$ and angle between both the sides is $75°$.

Formula used:

Law of Sines

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

Law of Cosines

$a^2=b^2+c^2-2bc\cos A$

Calculation:

The below figure shows the labeling of the given triangle.

The given triangle has two sides and an angle between both the sides.

Then for SAS, use law of cosine.

$a^2=b^2+c^2-2bc\cos A$

First find the side opposite to the given angle

Substitute $5$ for $b$, $7$ for $c$ and $75°$ for $\angle A$.

$\begin{array}{c}{a}^2={\left(5\right)}^2+{\left(7\right)}^2-2\times 5\times 7\times \cos 75°\\ a^2=74-70\times \cos 75°\\ a^2=55.8\\ a\approx 7.5\end{array}$

Thus the side opposite to the given angle is $7.5$.

Now find the second angle, use Law of Sine

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

Substitute $75°$ for $\angle A$, $7.5$ for $a$ and $5$ for $b$

$\begin{array}{c}\frac{\sin 75°}{7.5}=\frac{\sin B}{5}\\ \sin B=\frac{5\times \sin 75°}{7.5}\\ \sin B=0.646\\ B={\sin }^{-1}\left(0.646\right)\end{array}$

Use scientific calculator with to find the value of ${\sin }^{-1}\left(0.646\right)$.

The value of $\angle B$ is,

$\begin{array}{c}\angle B={\sin }^{-1}\left(0.646\right)\\ \approx 40.2\end{array}$

The value of angle labeled $\theta$ for ${\sin }^{-1}\left(0.646\right)$ is $40.2°$.