Chapter 5.5, Problem 13CFU

To determine

To solve: The triangle, given the triangle at the right. Round to the nearest tenth if necessary, $A={32}^{\circ },c=13$

Answer to Problem 13CFU

  $\begin{array}{l}b=11,\\ a=6.9,\\ m\angle A={12}^{\circ },\\ m\angle B={58}^{\circ },\\ m\angle C={90}^{\circ }\end{array}$

Explanation of Solution

Given information: Triangle:

Calculation:

Lots of ways to completely solve but often you have to start with what’s given. We have a side and an angle. We start from $\angle A,\angle B={90}^{\circ }-{32}^{\circ }={58}^{\circ }$

Now let’s get the sides. For side a

  $\begin{array}{l}\sin {32}^{\circ }=\frac{a}{13}\\ a=13\times \sin {32}^{\circ }\\ a=13\times 0.5299\\ a=6.9\end{array}$

Solve for b with the cosine,

  $\begin{array}{l}\cos {32}^{\circ }=\frac{b}{13}\\ b=13\times \cos {32}^{\circ }\\ b=13\times 0.8480\\ b=11.0\\ m\angle A={90}^{\circ }-{78}^{\circ }={12}^{\circ }\end{array}$

  $\begin{array}{l}b=11,\\ a=6.9,\\ m\angle A={12}^{\circ },\\ m\angle B={58}^{\circ },\\ m\angle C={90}^{\circ }\end{array}$