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(B) Let the sides of the triangle be $a = 5k, b = 5k, c = 6k$. First, we find the value of $k$ using the inradius formula $r = \frac{\Delta}{s}$. The

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$	an^{-1}\left(\frac{24}{7}ight)$

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(B) Let the sides of the triangle be $a = 5k, b = 5k, c = 6k$. First, we find the value of $k$ using the inradius formula $r = \frac{\Delta}{s}$. The semi-perimeter $s = \frac{5k + 5k + 6k}{2} = 8k$. The area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{8k(3k)(3k)(2k)} = \sqrt{144k^4} = 12k^2$. Given $r = 6$, so $6 = \frac{12k^2}{8k}$ $\Rightarrow 6 = \frac{3k}{2}$ $\Rightarrow k = 4$. The sides are $20, 20, 24$. In the isosceles triangle, the altitude to the base $6k$ divides it into two triangles with base $3k$ and hypotenuse $5k$. The height is $\sqrt{(5k)^2 - (3k)^2} = 4k$. Let the vertex angle be $2	heta$. Then $	an 	heta = \frac{3k}{4k} = \frac{3}{4}$. The vertex angle is $2	heta$, where $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta} = \frac{2(3/4)}{1 - (9/16)} = \frac{3/2}{7/16} = \frac{24}{7}$. Thus, the largest angle is $	an^{-1}\left(\frac{24}{7}ight)$.

If the radius of the incircle of a triangle with sides $5k, 6k$ and $5k$ is $6$, then the largest angle of that triangle is

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