Let ABC be a triangle such that AB=15 and AC= | vedclass

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(C) Let $\angle ABC = 	heta$,then $\angle ACB = 2	heta$. Let $\angle BAC = 180^{\circ} - 3	heta$. Let $BC = x$.By the sine rule in $	riangle ABC$:

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$10$

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(C) Let $\angle ABC = 	heta$,then $\angle ACB = 2	heta$. Let $\angle BAC = 180^{\circ} - 3	heta$. Let $BC = x$.By the sine rule in $	riangle ABC$:$\frac{9}{\sin 	heta} = \frac{15}{\sin 2	heta} = \frac{x}{\sin 3	heta}$From $\frac{9}{\sin 	heta} = \frac{15}{2 \sin 	heta \cos 	heta}$,we get $\cos 	heta = \frac{15}{18} = \frac{5}{6}$.Using $\frac{x}{\sin 3	heta} = \frac{9}{\sin 	heta}$,we have $x = 9 \cdot \frac{\sin 3	heta}{\sin 	heta} = 9(3 - 4 \sin^2 	heta)$.Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{25}{36} = \frac{11}{36}$,we get $x = 9(3 - 4 \cdot \frac{11}{36}) = 9(3 - \frac{11}{9}) = 27 - 11 = 16$.By the Angle Bisector Theorem,$D$ divides $BC$ in the ratio $AB:AC = 15:9 = 5:3$.Thus,$BD = \frac{5}{5+3} \cdot BC = \frac{5}{8} \cdot 16 = 10$.

Let $ABC$ be a triangle such that $AB=15$ and $AC=9$. The bisector of $\angle BAC$ meets $BC$ in $D$. If $\angle ACB=2\angle ABC$,then $BD$ is

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