Solve this


In $\triangle A B C, A B=6 \sqrt{3} \mathrm{~cm}, A C=12 \mathrm{~cm}$ and $B C=6 \mathrm{~cm}$. Then $\angle B$ is

(a) 45o
(b) 60o
(c) 90o
(d) 120o



$\mathrm{AB}=6 \sqrt{3} \mathrm{~cm}$

$\Rightarrow \mathrm{AB}^{2}=108 \mathrm{~cm}^{2}$

$\mathrm{AC}=12 \mathrm{~cm}$

$\Rightarrow \mathrm{AC}^{2}=144 \mathrm{~cm}^{2}$

$\mathrm{BC}=6 \mathrm{~cm}$

$\Rightarrow \mathrm{BC}^{2}=36 \mathrm{~cm}$

$\therefore \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$

Since, the square of the longest side is equal to the sum of the square of two sides, so △ABC is a right angled triangle.
∴The angle opposite to AC i.e. ∠B = 90o
Hence, the correct answer is option (c)


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