Chapter 5.2, Problem 20E

To determine

To find: The values of the six trigonometric ratios for $\angle R$ .

Answer to Problem 20E

  $\begin{array}{l}\sin R=0.9565,\cos R=0.2917,\tan R=3.2793\\ \sec R=3.4282,\csc R=1.0455,\cot R=0.3049\end{array}$

Explanation of Solution

Given information: Triangle:

Calculation:

First find the missing side

  $ST=\sqrt{{48}^2-{14}^2}=\sqrt{2304-196}=45.91$

  $\begin{array}{l}\sin R=\frac{45.91}{48}=0.9565,\cos R=\frac{14}{48}=0.2917\\ \tan R=\frac{45.91}{14}=3.2793\\ \sec R=\frac{1}{\cos R}=3.4282,\csc R=\frac{1}{\sin R}=1.0455\\ \cot R=\frac{1}{\tan R}=0.3049\end{array}$

Thus,

  $\begin{array}{l}\sin R=0.9565,\cos R=0.2917,\tan R=3.2793\\ \sec R=3.4282,\csc R=1.0455,\cot R=0.3049\end{array}$