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(B) Given sides of the triangle are $a=12, b=16, c=20$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{12+16+20}{2} = \frac{48}{2} = 24$. Area of the tri

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$6: 3: 2$

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(B) Given sides of the triangle are $a=12, b=16, c=20$. Semi-perimeter $s = \frac{a+b+c}{2} = \frac{12+16+20}{2} = \frac{48}{2} = 24$. Area of the triangle $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{24(24-12)(24-16)(24-20)} = \sqrt{24 	imes 12 	imes 8 	imes 4} = \sqrt{9216} = 96$. The exradii are given by $r_A = \frac{\Delta}{s-a}, r_B = \frac{\Delta}{s-b}, r_C = \frac{\Delta}{s-c}$. $r_C = \frac{96}{24-20} = \frac{96}{4} = 24$. $r_B = \frac{96}{24-16} = \frac{96}{8} = 12$. $r_A = \frac{96}{24-12} = \frac{96}{12} = 8$. The ratio $r_C : r_B : r_A = 24 : 12 : 8 = 6 : 3 : 2$.

In a triangle $ABC$, with usual notation, if $a=12, b=16, c=20$, then the ratio of the exradii of the triangle opposite to the angles in the order $\angle C, \angle B, \angle A$ is

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