In a right triangle $ A B C $, right angled at $ C $, if $ \angle B=60^{\circ} $ and $ A B=15 $ units. Find the remaining angles and sides.

Given:

In a right triangle \( A B C \), right angled at \( C \), \( \angle B=60^{\circ} \) and \( A B=15 \) units.

To do:

We have to find the remaining angles and sides.

Solution:  

We know that sum of the angles in a triangle is $180^o$.

Therefore,

$\angle A+\angle B+\angle C=180^o$

$\angle A+60^o+90^o=180^o$

$\angle A=180^o-150^o$

$\angle A=30^o$

$\sin\ B=\frac{AC}{AB}$

$\sin\ 60^o=\frac{AC}{AB}$

$\frac{\sqrt3}{2}=\frac{AC}{15}$          (Since $\sin 60^{\circ}=\frac{\sqrt3}{2}$)       

$AC=\frac{15\sqrt3}{2}$          

$\cos\ B=\frac{BC}{AB}$

$\cos\ 60^o=\frac{BC}{AB}$

$\frac{1}{2}=\frac{BC}{15}$          (Since $\cos 60^{\circ}=\frac{1}{2}$)       

$BC=\frac{15}{2}$    

The values of angles $B$ and $C$ are $60^{\circ}$ and $90^{\circ}$ respectively, the sides $BC$ and $AC$ are $\frac{15}{2}$units and $\frac{15\sqrt3}{2}$ units respectively.

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Updated on: 10-Oct-2022

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