Chapter 4.3, Problem 18E

To determine

To find: The third side of the triangle and other five trigonometric functions.

Answer to Problem 18E

The third side of the triangle is, base $=\sqrt{273}$ and the other five trigonometric functions are, $\sin \theta =\frac{4}{17},\cos \theta =\frac{\sqrt{273}}{17},\mathrm{Sec}\theta =\frac{17}{\sqrt{273}},\tan \theta =\frac{4}{\sqrt{273}},\cot \theta =\frac{\sqrt{273}}{4}$

Explanation of Solution

Given information: $$

  $\csc \theta =\frac{17}{4}$

Calculation: From the properties of trigonometric function,

  $\begin{array}{l}\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}},\text{ csc}\theta \text{=}\frac{\text{Hypotenuse}}{\text{Perpendicular}}\\ \cos \theta =\frac{\text{Base}}{\text{Hypotenuse}},\text{ sec}\theta \text{=}\frac{\text{Hypotenuse}}{\text{Base}}\\ \tan \theta =\frac{\text{Perpendicular}}{\text{Base}},\text{ cot}\theta =\frac{\text{Base}}{\text{Perpendicular}}\end{array}$   

Given, $\csc \theta =\frac{17}{4}$

Therefore, Perpendicular = $4$ and Hypotenuse = $17$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ 17=\sqrt{4^2+{\left(\text{Base}\right)}^2}\\ 289=16+{\text{Base}}^2\\ \text{Base}=\sqrt{289-16}\\ \text{Base}=\sqrt{273}\end{array}$

Hence, The third side of the triangle is, base $=\sqrt{273}$

Let’s find the other five trigonometric functions,

  $\begin{array}{l}\sin \theta =\frac{4}{17}\\ \cos \theta =\frac{\sqrt{273}}{17}\\ \mathrm{Sec}\theta =\frac{17}{\sqrt{273}}\\ \tan \theta =\frac{4}{\sqrt{273}}\\ \cot \theta =\frac{\sqrt{273}}{4}\end{array}$

Hence, The third side of the triangle is, base $=\sqrt{273}$ and the other five trigonometric functions are, $\sin \theta =\frac{4}{17},\cos \theta =\frac{\sqrt{273}}{17},\mathrm{Sec}\theta =\frac{17}{\sqrt{273}},\tan \theta =\frac{4}{\sqrt{273}},\cot \theta =\frac{\sqrt{273}}{4}$