Tick the correct answer and justify : In $∆ABC, AB = 6\sqrt3\ cm, AC = 12\ cm$ and $BC = 6\ cm$. The angle B is:
(a) $120^o$
(b) $60^o$
(c) $90^o$
(d) $45^o$
Given:
In $∆ABC, AB = 6\sqrt3\ cm, AC = 12\ cm$ and $BC = 6\ cm$.
To do:
We have to find angle B.
Solution:
$AC^{2}=(12)^2$
$=144$
$A B^{2}+B C^{2}=(6 \sqrt{3})^{2}+(6)^{2}$
$=108+36$
$=144$
$AC^2=A B^{2}+B C^{2}$
This implies,
Triangle ABC is a right angle triangle, right angled at $B$.
$\angle B =90^{\circ}$
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