Chapter 5.6, Problem 7E

Solution

To find: The measure of angles $A$ and $B$ , and side $c$ in the triangle.

The measures are $A={28.58}^{\circ }$ , $B={56.5}^{\circ }$ , and $c=25.1$ .

Given information: In the triangle, angle and two sides of triangle are $C={95}^{\circ }$ , $a=12$ , and $b=21$ .

Formula used:

The law of cosines to find the angles and sides are given by,

  $c^2=b^2+a^2-2ba\cos C$  ...... (1)

  $a^2=b^2+c^2-2bc\cos A$  ...... (2)

Calculation:

Substitute ${95}^{\circ }$ for $C$ , $12$ for $a$ , and $21$ for $b$ in equation (1).

  $\begin{array}{c}{c}^2={\left(21\right)}^2+{\left(12\right)}^2-2\cdot 21\cdot 12\cdot \cos {95}^{\circ }\\ c^2=441+144-2\cdot 21\cdot 12\left(-0.08715\right)\\ c^2=628.9236\\ c\approx 25.1\end{array}$

Substitute $12$ for $a$ , $21$ for $b$ , and $25.1$ for $c$ in equation (2).

  $\begin{array}{c}{\left(12\right)}^2={\left(21\right)}^2+\left(25\right)-2\cdot 21\cdot 25-\cos A\\ 144=441+625-1050\cos A\\ \cos A=\frac{922}{1050}\\ A={28.58}^{\circ }\end{array}$

The sum of interior angles in the triangle is ${180}^{\circ }$ . Calculate the measure of angle $C$ .

  $\begin{array}{l}B={180}^{\circ }-({95}^{\circ }+{28.58}^{\circ })\\ B={56.5}^{\circ }\end{array}$

Therefore, the measures are $A={28.58}^{\circ }$ , $B={56.5}^{\circ }$ , and $c=25.1$ .