Question

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

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

Answer 1

Answer: (C)

${90}^o$

In $△ABC$, $AB=6\sqrt{3}$cm,$AC=12$cm and $BC=6$ cm
we know,
$\Rightarrow b^2=a^2+c^2-2ac\cos B$
$\Rightarrow A{C}^2=A{B}^2+B{C}^2-2.AB.BC\cos B$
$\Rightarrow {12}^2=(6\sqrt{3}{)}^2+6^2-2\times 6\sqrt{3}\times 6\times \cos B$
$\Rightarrow 144=108+36-72\sqrt{3}\times \cos B$
$\Rightarrow \cos B=0$
$\Rightarrow \cos B=\cos 90$
$\therefore B={90}^{\circ }$