If P is a point on the altitude AD of the tri | vedclass

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(B) Given that $AD$ is the altitude of $	riangle ABC$,so $\angle ADB = 90^{\circ}$.In $	riangle BDP$,$\angle BPD = 180^{\circ} - 90^{\circ} - \frac{

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$2C \sin \frac{B}{3}$

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(B) Given that $AD$ is the altitude of $	riangle ABC$,so $\angle ADB = 90^{\circ}$.In $	riangle BDP$,$\angle BPD = 180^{\circ} - 90^{\circ} - \frac{B}{3} = 90^{\circ} - \frac{B}{3}$.Therefore,$\angle APB = 180^{\circ} - \angle BPD = 180^{\circ} - (90^{\circ} - \frac{B}{3}) = 90^{\circ} + \frac{B}{3}$.In $	riangle ABP$,using the sine rule:$\frac{AP}{\sin(\angle ABP)} = \frac{AB}{\sin(\angle APB)}$$\frac{AP}{\sin(\frac{2B}{3})} = \frac{c}{\sin(90^{\circ} + \frac{B}{3})}$$AP = \frac{c \sin(\frac{2B}{3})}{\cos(\frac{B}{3})}$$AP = \frac{c \cdot 2 \sin(\frac{B}{3}) \cos(\frac{B}{3})}{\cos(\frac{B}{3})}$$AP = 2c \sin \frac{B}{3}$

If $P$ is a point on the altitude $AD$ of the $\triangle ABC$,and $\angle ABP = \frac{2B}{3}$,then $AP$ is equal to

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