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If $P$ is a point on the altitude $A D$ of the triangle $A B C$ such that $∠ C B P = B / 3,$ then $A P$ is equal to

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$2 a \sin (C / 3)$

$2 b \sin (A / 3)$

$2 c \sin (B / 3)$

$2 c \sin (C / 3)$

answer is C.

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Detailed Solution

From $△ A B P$

$\frac{A P}{\sin (2 B / 3)} = \frac{A B}{\sin (π / 2 + B / 3)}$

$\begin{matrix} \Rightarrow A P \\ = \frac{2 c \sin (B / 3) \cos (B / 3)}{\cos (B / 3)} \\ = 2 c \sin (B / 3) \end{matrix}$

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