If $P$ is a point on the equation $\mathbf{A D}$ of the $ riangle A B C$, and $\angle A B P=\frac{2 B}{3}$, then $A P$ is equal to

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If $P$ is a point on the equation $\mathbf{A D}$ of the $	riangle A B C$, and $\angle A B P=\frac{2 B}{3}$, then $A P$ is equal to

Aicharya · Accepted Answer

$\because A D$ is altitude of $	riangle A B C$. $	herefore \angle A D B=90^{\circ} \Rightarrow \angle A P B=90^{\circ}+\frac{B}{3}$ In $	riangle A P B$, Using sine rule $ \begin{aligned} & \frac{A P}{\sin \frac{2 B}{3}}=\frac{A B}{\sin \left(90+\frac{B}{3}ight)} \ A P & =C \frac{\sin \frac{2 B}{3}}{\cos \frac{B}{3}} \ & =C \frac{2 \sin \frac{B}{3} \cos \frac{B}{3}}{\cos \frac{B}{3}} \quad[\because \sin 2 A=2 \sin A \cos A] \ & =2 C \sin \frac{B}{3} \end{aligned} $

If $P$ is a point on the equation $\mathbf{A D}$ of the $\triangle A B C$, and $\angle A B P=\frac{2 B}{3}$,…

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