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(A) Let the sides of the triangle be $a = \sqrt{3}k$,$b = \sqrt{5}k$,and $c = \sqrt{8+\sqrt{15}}k$. Since $c$ is the largest side,the largest angle $\

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$\frac{2 \pi}{3}$

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(A) Let the sides of the triangle be $a = \sqrt{3}k$,$b = \sqrt{5}k$,and $c = \sqrt{8+\sqrt{15}}k$. Since $c$ is the largest side,the largest angle $	heta$ is opposite to side $c$. Using the Law of Cosines: $\cos 	heta = \frac{a^2 + b^2 - c^2}{2ab}$. Substituting the values: $\cos 	heta = \frac{(\sqrt{3}k)^2 + (\sqrt{5}k)^2 - (\sqrt{8+\sqrt{15}}k)^2}{2(\sqrt{3}k)(\sqrt{5}k)}$. $\cos 	heta = \frac{3k^2 + 5k^2 - (8+\sqrt{15})k^2}{2\sqrt{15}k^2}$. $\cos 	heta = \frac{8k^2 - 8k^2 - \sqrt{15}k^2}{2\sqrt{15}k^2} = \frac{-\sqrt{15}k^2}{2\sqrt{15}k^2} = -\frac{1}{2}$. Since $\cos 	heta = -\frac{1}{2}$,the angle $	heta = \frac{2\pi}{3}$.

If the sides of a triangle are in the ratio $\sqrt{3} : \sqrt{5} : \sqrt{8+\sqrt{15}}$,then the largest angle in that triangle is

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