In a $\triangle A B C, \angle C=60^{\circ}$ and $\angle A=75^{\circ}$. If $D$ is a point on $A C$ such that the area of $\triangle B A D$ is $\sqrt{3}$ times the area of $\triangle B C D$, then the measure of $\angle A B D$ is

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Aicharya · Accepted Answer

In a $\triangle A B C$, it is given the angles $A=75^{\circ}$ and $C=60^{\circ}$ and $\quad \triangle B A D=\sqrt{3}(\triangle B C D) \Rightarrow(A D)=\sqrt{3}(C D)$ $\Rightarrow \quad \frac{A D}{C D}=\sqrt{3}$ As angle $B=45^{\circ}$, Now let $\angle A B D=\alpha$, then $\angle D B C=45^{\circ}-\alpha$ Now, by sine law $\begin{gathered} \frac{\sin \alpha}{A D}=\frac{\sin 75^{\circ}}{B D},(\text { in } \triangle A B D) \ \text { and } \frac{\sin \left(45^{\circ}-\alpha\right)}{C D}=\frac{\sin 60^{\circ}}{B D},(\text { in } \triangle D B C) \ \therefore \quad \frac{(C D) \sin \alpha}{(A D) \sin \left(45^{\circ}-\alpha\right)}=\frac{\sin 75^{\circ}}{\sin 60^{\circ}} \ \Rightarrow \frac{1}{\sqrt{3}} \times \frac{\sin \alpha}{\sin \left(45^{\circ}-\alpha\right)}=\frac{\frac{1}{2 \sqrt{2}}+\frac{\sqrt{3}}{2 \sqrt{2}}}{\frac{\sqrt{3}}{2}} \ \Rightarrow \quad \frac{\sin \alpha}{\sin \left(45^{\circ}-\alpha\right)}=\frac{\sqrt{3}+1}{\sqrt{2}} \Rightarrow \frac{\sin \left(45^{\circ}-\alpha\right)}{\sin \alpha}=\frac{\sqrt{3}-1}{\sqrt{2}} \ \Rightarrow \frac{1}{\sqrt{2}} \cot \alpha-\frac{1}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}} \Rightarrow \cot \alpha=\sqrt{3} \Rightarrow \alpha=30^{\circ} \end{gathered}$

In a \(\triangle A B C, \angle C=60^{\circ}\) and \(\angle A=75^{\circ}\). If \(D\) is a point on \(A C\)…

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