In a triangle ABC,angle C = 60^{circ} and ang | vedclass

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(A) In $	riangle ABC$,the sum of angles is $180^{\circ}$. Thus,$\angle B = 180^{\circ} - (75^{\circ} + 60^{\circ}) = 45^{\circ}$.Since $	riangle BAD

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$30$

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(A) In $	riangle ABC$,the sum of angles is $180^{\circ}$. Thus,$\angle B = 180^{\circ} - (75^{\circ} + 60^{\circ}) = 45^{\circ}$.Since $	riangle BAD$ and $	riangle BCD$ share the same altitude from vertex $B$ to the base $AC$,the ratio of their areas is equal to the ratio of their bases: $\frac{	ext{Area}(	riangle BAD)}{	ext{Area}(	riangle BCD)} = \frac{AD}{CD} = \sqrt{3}$.Let $\angle ABD = \alpha$,then $\angle DBC = 45^{\circ} - \alpha$.Using the sine rule in $	riangle BAD$ and $	riangle BCD$:$\frac{AD}{\sin \alpha} = \frac{BD}{\sin 75^{\circ}}$ and $\frac{CD}{\sin(45^{\circ} - \alpha)} = \frac{BD}{\sin 60^{\circ}}$.Dividing these equations gives $\frac{AD}{CD} = \frac{\sin \alpha}{\sin(45^{\circ} - \alpha)} \cdot \frac{\sin 60^{\circ}}{\sin 75^{\circ}} = \sqrt{3}$.$\frac{\sin \alpha}{\sin(45^{\circ} - \alpha)} = \sqrt{3} \cdot \frac{\sin 75^{\circ}}{\sin 60^{\circ}} = \sqrt{3} \cdot \frac{(\sqrt{6} + \sqrt{2})/4}{\sqrt{3}/2} = \frac{\sqrt{6} + \sqrt{2}}{2} = \frac{\sqrt{2}(\sqrt{3} + 1)}{2} = \frac{\sqrt{3} + 1}{\sqrt{2}}$.Solving $\frac{\sin \alpha}{\sin 45^{\circ} \cos \alpha - \cos 45^{\circ} \sin \alpha} = \frac{\sqrt{3} + 1}{\sqrt{2}}$ leads to $\cot \alpha = \sqrt{3}$,hence $\alpha = 30^{\circ}$.

In a $\triangle ABC$,$\angle C = 60^{\circ}$ and $\angle A = 75^{\circ}$. If $D$ is a point on $AC$ such that the area of $\triangle BAD$ is $\sqrt{3}$ times the area of $\triangle BCD$,then the measure of $\angle ABD$ is (in $^{\circ}$)

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