Let a , b and c denote the length of sides opposite to vertices A,B and C respectively of a triangle ABC. Let a=6 ,b=10 and the area of triangle is 15 square root 3 If angle ACB is obtuse and if r denotes the radius of the incircle of the triangle then r^2 equals

Solution:     $\Delta =\frac{1}{2}ab \sin C=15\sqrt{3}=\frac{1}{2}\times 6\times 10\times \sin C$

$\Rightarrow$    $\sin C=\frac{\sqrt{3}}{2},$    or     $C=120^{\circ}$

Now ,     $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$

$\Rightarrow$           $-\frac{1}{2}=\frac{36+100-c^{2}}{120}\Rightarrow c=14.$

$2s=a+b+c\Rightarrow s=15.$

Now ,       $r=\frac{\Delta }{s}=\frac{15\sqrt{3}}{15}=\sqrt{3}\Rightarrow r^{2}=3.$

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