Chapter 12.2, Problem 12OE

To determine

If $\sin \theta =\frac{2}{3}$ for an acute angle $\theta$ , state two ways to find $\cos \theta ,\tan \theta$ .

Answer to Problem 12OE

  $\cos \theta =\frac{\sqrt{5}}{3}$

  $\tan \theta =\frac{2\sqrt{5}}{5}$

Explanation of Solution

Given:

  $\sin \theta =\frac{2}{3}$

Formula Used:

  $\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}=\frac{a}{c}$

  $\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}=\frac{b}{c}$

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }=\frac{a}{b}$

Calculation:

In the triangle ,

  $\sin \theta =\frac{2}{3}=\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}=\frac{a}{c}$

b = length of side adjacent to $\theta$

a = length of side opposite to $\theta$= 2 units

c = length of the hypotenuse = 3 units

Find b by using Pythagorean Theorem:

  $\begin{array}{l}{a}^2+b^2=c^2\\ \Rightarrow b=\sqrt{9-4}=\sqrt{5}\end{array}$

  $\begin{array}{l}\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}=\frac{b}{c}\\ \Rightarrow \cos \theta =\frac{\sqrt{5}}{3}\end{array}$

  $\begin{array}{l}\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }=\frac{a}{b}\\ \Rightarrow \tan \theta =\frac{2}{\sqrt{5}}\cdot \frac{\sqrt{5}}{\sqrt{5}}\\ \Rightarrow \tan \theta =\frac{2\sqrt{5}}{5}\end{array}$