Chapter 5.5, Problem 12CFU

To determine

To solve: The triangle, given the triangle at the right. Round to the nearest tenth if necessary, $a=11,b=21$

Answer to Problem 12CFU

  $\begin{array}{l}c=23.7,\\ m\angle A={27.6}^{\circ },\\ m\angle B={62.4}^{\circ },\\ m\angle C={90}^{\circ }\end{array}$

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 two sides, so we start with an arc function. We need to pick an angle and it makes no difference which one. I will pick $\angle A$ from this angle we are looking at the opposite and adjacent sides so we need

  ${\tan }^{-1}\frac{11}{21}=\theta$

You can solve for c with either the cosine (or sine now that you know b ) or the Pythagorean Theorem.

  $\begin{array}{l}{\tan }^{-1}\frac{11}{21}={27.6}^{\circ }\\ \text{This is }\angle A\text{ so }\angle B={90}^{\circ }-{27.6}^{\circ }={62.4}^{\circ }\\ c=\sqrt{\left({11}^2+{21}^2\right)}=\sqrt{121+441}=23.7\\ m\angle A={90}^{\circ }-{78}^{\circ }={12}^{\circ }\end{array}$

  $\begin{array}{l}c=23.7,\\ m\angle A={27.6}^{\circ },\\ m\angle B={62.4}^{\circ },\\ m\angle C={90}^{\circ }\end{array}$