Question

Tick the correct answer and justify:

In $\Delta ABC,AB=6\sqrt{3}cm$, $AC=12cm$and $BC=6cm$.

The angle $B$ is:

• A. $120$
• B. $60$
• C. $90°$
• D. $45$

Answer 1

The correct option is C

$90°$

We have to find the measure of angle B

On applying Pythagoras theorem to the $\Delta ABC$ we get:

${\left(AC\right)}^2={\left(12\right)}^2=144$

${\left(AB\right)}^2={\left(6\sqrt{3}\right)}^2=6\times 6\times \sqrt{3}\times \sqrt{3}=36\times 3=108$

${\left(BC\right)}^2={\left(6\right)}^2=36$

${\left(AB\right)}^2+{\left(BC\right)}^2=108+36=144$

Thus $\angle B=90°$