Chapter 12.2, Problem 20WE

To determine

The length of one side of right angled triangle is given . Give the lengths of the other two sides and the angle.

Answer to Problem 20WE

  $b=6\sqrt{3},c=12$

  $\angle B=60°$

Explanation of Solution

Given:

  

Formula Used:

  $\begin{array}{l}\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}=\frac{a}{c}\\ \Rightarrow c\sin \theta =a\end{array}$

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

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

Calculation:

In the triangle ,

  $\angle A=30°$

b = length of side adjacent to $\theta$

a = length of side opposite to $\theta$= 6 units

c = length of the hypotenuse

  $\begin{array}{l}c\sin \angle A=a\\ \Rightarrow c\sin 30°=\mathrm{6........................}[\angle A=30°,a=6]\\ \Rightarrow c=\frac{6}{\left(\frac{1}{2}\right)}\mathrm{..............................}[\sin 30°=\frac{1}{2}]\\ \Rightarrow c=12\end{array}$

  $\begin{array}{l}c\cos \angle A=b\\ \Rightarrow 12\cos 30°=b\mathrm{.............}[\angle A=30°,c=12]\\ \Rightarrow 12\cdot \frac{\sqrt{3}}{2}=b\mathrm{....................}[\cos 30°=\frac{\sqrt{3}}{2}]\\ \Rightarrow b=6\sqrt{3}\end{array}$

So, $b=6\sqrt{3},c=12$

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

So,

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