Chapter 4.3, Problem 17E

To determine

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

Answer to Problem 17E

The third side of the triangle is, hypotenuse $=\sqrt{10}$ and the other five trigonometric functions are, $\sin \theta =\frac{3}{\sqrt{10}},\cos \theta =\frac{1}{\sqrt{10}},\mathrm{Sec}\theta =\sqrt{10},\csc \theta =\frac{\sqrt{10}}{3},\cot \theta =\frac{1}{3}$

Explanation of Solution

Given information: $$

  $\tan \theta =3$

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, $\tan \theta =3$

Therefore, Perpendicular = $3$ and Base = $1$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ \text{Hypotenuse}=\sqrt{3^2+1^2}\\ \text{Hypotenuse}=\sqrt{9+1}\\ =\sqrt{10}\end{array}$

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

Let’s find the other five trigonometric functions,

  $\begin{array}{l}\sin \theta =\frac{3}{\sqrt{10}}\\ \cos \theta =\frac{1}{\sqrt{10}}\\ \mathrm{Sec}\theta =\sqrt{10}\\ \csc \theta =\frac{\sqrt{10}}{3}\\ \cot \theta =\frac{1}{3}\end{array}$

Hence, The third side of the triangle is, hypotenuse $=\sqrt{10}$ and the other five trigonometric functions are, $\sin \theta =\frac{3}{\sqrt{10}},\cos \theta =\frac{1}{\sqrt{10}},\mathrm{Sec}\theta =\sqrt{10},\csc \theta =\frac{\sqrt{10}}{3},\cot \theta =\frac{1}{3}$