In a triangle ABC,if angle A = 3angle B,CA = | vedclass

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(D) Let $\angle B = 	heta$. Then $\angle A = 3	heta$. Since the sum of angles in a triangle is $180^{\circ}$,$\angle C = 180^{\circ} - (A + B) = 180

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$\frac{35}{3}$

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(D) Let $\angle B = 	heta$. Then $\angle A = 3	heta$. Since the sum of angles in a triangle is $180^{\circ}$,$\angle C = 180^{\circ} - (A + B) = 180^{\circ} - 4	heta$.Using the sine rule,$\frac{AB}{\sin C} = \frac{BC}{\sin A} = \frac{AC}{\sin B}$.Substituting the given values: $\frac{AB}{\sin(180^{\circ} - 4	heta)} = \frac{16}{\sin 3	heta} = \frac{9}{\sin 	heta}$.From $\frac{16}{\sin 3	heta} = \frac{9}{\sin 	heta}$,we have $16 \sin 	heta = 9 \sin 3	heta = 9(3 \sin 	heta - 4 \sin^3 	heta)$.Dividing by $\sin 	heta$ (since $\sin 	heta 
eq 0$),$16 = 27 - 36 \sin^2 	heta$,which gives $36 \sin^2 	heta = 11$,so $\sin^2 	heta = \frac{11}{36}$.Then $\cos^2 	heta = 1 - \frac{11}{36} = \frac{25}{36}$,so $\cos 	heta = \frac{5}{6}$.Now,$\sin 4	heta = 2 \sin 2	heta \cos 2	heta = 4 \sin 	heta \cos 	heta (1 - 2 \sin^2 	heta) = 4 	imes \frac{\sqrt{11}}{6} 	imes \frac{5}{6} 	imes (1 - 2 	imes \frac{11}{36}) = \frac{20\sqrt{11}}{36} 	imes (1 - \frac{11}{18}) = \frac{5\sqrt{11}}{9} 	imes \frac{7}{18} = \frac{35\sqrt{11}}{162}$.Finally,$AB = \frac{9 \sin 4	heta}{\sin 	heta} = 9 	imes \frac{35\sqrt{11}}{162} 	imes \frac{6}{\sqrt{11}} = \frac{35}{3}$.

In a $\triangle ABC$,if $\angle A = 3\angle B$,$CA = 9$,and $BC = 16$,then the length of $AB$ is

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