Chapter 5, Problem 57RE

Solution

To find: The value of included angles $A$ , $B$ and $C$ .

The value of included angles $A$ is $44.4°$ , $B$ is $78.5°$ and $C$ is $57.1°$ .

Given information:

The length of the sides are, $a=5$ , $b=7$ , and $c=6$ .

Formula used:

Law of Cosine is given by,

  $b^2=a^2+c^2-2ac\cos B$

Law of Sine is given by,

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

Calculation:

Rewrite the Law of Cosine in terms of angle $B$.

  $B={\cos }^{-1}\frac{a^2+c^2-b^2}{2ac}$

Substitute $5$ for $a$ , $7$ for $b$ , and $6$ for $c$ in Law of Cosine.

  $\begin{array}{c}B={\cos }^{-1}\frac{a^2+c^2-b^2}{2ac}\\ ={\cos }^{-1}\frac{5^2+6^2-7^2}{2\times 5\times 6}\\ \approx 78.5°\end{array}$

Rewrite the Law of Sine for angle $A$ and angle $C$

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

Substitute $5$ for $a$ , $7$ for $b$ , $78.5°$ for angle $B$ in Law of Sine.

  $\begin{array}{c}\frac{\sin A}{5}=\frac{\sin \left(78.5°\right)}{7}\\ A\approx 44.4°\end{array}$

The sum of all included angles in a triangle is $180°$ . So,

  $\begin{array}{c}A+B+C=180\\ C=180-\left(44.4+78.5\right)\\ \approx 57.1°\end{array}$

Therefore, the value of included angles $A$ is $44.4°$ , $B$ is $78.5°$ and $C$ is $57.1°$ .