If $ \triangle A B C $ is a right triangle such that $ \angle C=90^{\circ}, \angle A=45^{\circ} $ and $ B C=7 $ units. Find $ \angle B, A B $ and $ A C $.

Given:

In a right triangle \( A B C \), right angled at \( C \), \( \angle A=45^{\circ} \) and \( B C=7 \) units.

To do:

We have to find \( \angle B, A B \) and \( A C \).

Solution:  

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

Therefore,

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

$\angle B+45^o+90^o=180^o$

$\angle B=180^o-135^o$

$\angle B=45^o$     

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

$\cos\ 45^o=\frac{BC}{AB}$

$\frac{1}{\sqrt2}$=\frac{7}{AB}$          (Since $\cos 45^{\circ}=\frac{1}{\sqrt2}$)       

$AB=7\sqrt2$

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

$\sin\ 45^o=\frac{AC}{AB}$

$\frac{1}{\sqrt2}$=\frac{AC}{7\sqrt2}$          (Since $\sin 45^{\circ}=\frac{1}{\sqrt2}$)      

$AC=7$ 

The value of angle $B$ is $45^{\circ}$, the sides $AB$ and $AC$ are $7\sqrt2$units and $7$ units respectively. 

Tutorialspoint

Updated on: 10-Oct-2022

59 Views

Kickstart Your Career

Get certified by completing the course

Get Started