Question

Question 17
Tick the correct answer and justify:
In $\Delta ABC,AB=6\sqrt{3}$ cm, AC = 12 cm and BC = 6 cm.
The angle B is:

(A) ${120}^{\circ }$
(B) ${60}^{\circ }$
(C) ${90}^{\circ }$
(D) ${45}^{\circ }$

Answer 1

(C) ${90}^{\circ }$


Given that, AB = $6\sqrt{3}cm,AC=12cm$, and BC = 6 cm
We can observe that
$A{B}^2=108$
$A{C}^2=144$
And, $B{C}^2=36$
$A{B}^2+B{C}^2=A{C}^2$
The given triangle, $\Delta ABC$, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
$\therefore \angle B={90}^{\circ }$.