Search any question & find its solution
Question:
Answered & Verified by Expert
If $P$ is a point on the equation $\mathbf{A D}$ of the $\triangle A B C$, and $\angle A B P=\frac{2 B}{3}$, then $A P$ is equal to
Options:
Solution:
2101 Upvotes
Verified Answer
The correct answer is:
$2 C \sin \frac{B}{3}$
$\because A D$ is altitude of $\triangle A B C$.
$\therefore \angle A D B=90^{\circ} \Rightarrow \angle A P B=90^{\circ}+\frac{B}{3}$
In $\triangle A P B$,
Using sine rule
$$
\begin{aligned}
& \frac{A P}{\sin \frac{2 B}{3}}=\frac{A B}{\sin \left(90+\frac{B}{3}\right)} \\
A P & =C \frac{\sin \frac{2 B}{3}}{\cos \frac{B}{3}} \\
& =C \frac{2 \sin \frac{B}{3} \cos \frac{B}{3}}{\cos \frac{B}{3}} \quad[\because \sin 2 A=2 \sin A \cos A] \\
& =2 C \sin \frac{B}{3}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.