Question

In $△ABC,AB=6\sqrt{3}cm,AC=12cm$ and $BC=6cm$. The $\angle B$ is :

• A. ${120}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. ${45}^o$

Answer 1

The correct option is C ${90}^o$

Given:
In $△ABC$
$AB=6\sqrt{3}cm$
$AC=12cm$
$BC=6cm$
Solution:
$A{C}^2=144$
$A{B}^2=108$
$B{C}^2=36$
Since,
$A{C}^2=A{B}^2+B{C}^2$
$\therefore$ By Converse of Pythagoras Theorem,
$△ABC$ is an Right Angle Triangle at B.
$\therefore \angle B={90}^o$