Chapter 5.6, Problem 5E

Solution

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

The measures are $B={89.56}^{\circ }$ , $C={35.44}^{\circ }$ , and $a=9.8$ .

Given information: In the triangle, the measures are $A={55}^{\circ }$ , $b=12$ , and $c=7$ .

Formula used:

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

  $a^2=b^2+c^2-2bc\cos A$  ...... (1)

  $b^2=a^2+c^2-2ac\cos B$  ...... (2)

Calculation:

Substitute ${55}^{\circ }$ for $A$ , $12$ for $b$ , and $7$ for $c$ in equation (1).

  $\begin{array}{c}{a}^2={\left(12\right)}^2+{\left(7\right)}^2-2\cdot 12\cdot 7\cdot \cos {55}^{\circ }\\ =144+49-168\cdot 0.6736\\ =96.6352\end{array}$

Further, simplify.

  $a=9.8$

Substitute $9.8$ for $a$ , $12$ and $7$ for $c$ in equation (2).

  $\begin{array}{c}{\left(12\right)}^2={\left(9.8\right)}^2+{\left(7\right)}^2-2\left(9.8\right)\left(7\right)\cos B\\ 144=96.04+49-137.2\cos B\\ \cos B=\frac{13}{1715}\\ B={89.56}^{\circ }\end{array}$

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

  $\begin{array}{c}A+B+C={180}^{\circ }\\ C={180}^{\circ }-\left({55}^{\circ }+{89.56}^{\circ }\right)\\ C={35.44}^{\circ }\end{array}$

Therefore, the measures are $B={89.56}^{\circ }$ , $C={35.44}^{\circ }$ , and $a=9.8$ .