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Consider a triangle $A B C$ and let $a, b$ and $c$ denote the lengths of the sides opposite to vertices $A, B$ and $C$ respectively. Suppose $a=6, b=10$ and the area of the triangle is $15 \sqrt{3}$. If $\angle A C B$ is obtuse and if $r$ denotes the radius of the incircle of the triangle, then $r^2$ is equal to
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The correct answer is:
3
$\sin C=\frac{\sqrt{3}}{2}$ and $C$ is given to be obtuse
$$
\begin{aligned}
& \Rightarrow \quad C=\frac{2 \pi}{3}=\sqrt{a^2+b^2-2 a b \cos C} \\
& =\sqrt{6^2+10^2-2 \times 6 \times 10 \times \cos \frac{2 \pi}{3}}=14 \\
& \therefore \quad r=\frac{\Delta}{s} \Rightarrow r^2=\frac{225 \times 3}{\left(\frac{6+10+14}{2}\right)^2}=3
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad C=\frac{2 \pi}{3}=\sqrt{a^2+b^2-2 a b \cos C} \\
& =\sqrt{6^2+10^2-2 \times 6 \times 10 \times \cos \frac{2 \pi}{3}}=14 \\
& \therefore \quad r=\frac{\Delta}{s} \Rightarrow r^2=\frac{225 \times 3}{\left(\frac{6+10+14}{2}\right)^2}=3
\end{aligned}
$$
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