Chapter 12.2, Problem 24WE

To determine

Find the lengths of the sides and the measures of the angles that are not given.

Answer to Problem 24WE

  $b=3\sqrt{3}$

  $\angle A=30°,\angle B=60°$

Explanation of Solution

Given:

  

a = 3 , c = 6

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 ,

b = length of side adjacent to $\angle A$ = $\sqrt{c^2-a^2}=\sqrt{36-9}=\sqrt{27}=3\sqrt{3}$

a = length of side opposite to $\angle A$= 3 units

c = length of the hypotenuse = 6 units

Find b by using Pythagorean Theorem:

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

  $\begin{array}{l}\sin \angle A=\frac{\text{length of side opposite to }\angle \text{A}}{\text{length of the hypotenuse}}=\frac{a}{c}=\frac{3}{6}=\frac{1}{2}\\ \Rightarrow \sin \angle A=\sin 30°\\ \Rightarrow \angle A=30°\end{array}$

Since the sum of the angles of a triangle is $180°$

So,

  $\begin{array}{l}\angle B=180°-\angle A-\angle C=180°-30°-90°=60°\\ \Rightarrow \angle B=60°\end{array}$

Hence, the solution is :

  $b=3\sqrt{3}$

  $\angle A=30°,\angle B=60°$