Chapter 4.3, Problem 16E

To determine

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

Answer to Problem 16E

The third side of the triangle is, perpendicular $=2\sqrt{10}$ and the other five trigonometric functions are, $\sin \theta =\frac{2\sqrt{10}}{7},\sec \theta =\frac{7}{3},\csc \theta =\frac{7}{2\sqrt{10}},\tan \theta =\frac{2\sqrt{10}}{3},\cot \theta =\frac{3}{2\sqrt{10}}$

Explanation of Solution

Given information: $$

  $\cos \theta =\frac{3}{7}$

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, $\cos \theta =\frac{3}{7}$

Therefore, Base = $3$ and Hypotenuse = $7$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ 7=\sqrt{{\left(\text{Perpendicular}\right)}^2+3^2}\\ 49={\left(\text{Perpendicular}\right)}^2+9\\ {\left(\text{Perpendicular}\right)}^2=49-9\\ \text{Perpendicular}=\sqrt{40}\\ =2\sqrt{10}\end{array}$

Hence, The third side of the triangle is, perpendicular $=2\sqrt{10}$

Let’s find the other five trigonometric functions,

  $\begin{array}{l}\sin \theta =\frac{2\sqrt{10}}{7}\\ \sec \theta =\frac{7}{3}\\ \csc \theta =\frac{7}{2\sqrt{10}}\\ \tan \theta =\frac{2\sqrt{10}}{3}\\ \cot \theta =\frac{3}{2\sqrt{10}}\end{array}$

Hence, The third side of the triangle is, perpendicular $=2\sqrt{10}$ and the other five trigonometric functions are, $\sin \theta =\frac{2\sqrt{10}}{7},\sec \theta =\frac{7}{3},\csc \theta =\frac{7}{2\sqrt{10}},\tan \theta =\frac{2\sqrt{10}}{3},\cot \theta =\frac{3}{2\sqrt{10}}$