Question

If P is a point on the altitude AD of the triangle ABC such that ∠CBP = B/3, then AP is equal to

• A. 2a sin
• B. 2b sin
• C. 2c sin$\frac { B } { 3 }$
• D. 2c sin

Answer 1

Answer: (C)

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

Answer 2

Here ∠BPA = 90⁰ + B/3, ∠ABP = 2B/3.

In ∆ABP (by sine rule)

$\frac { \mathrm { AP } } { \sin ( 2 \mathrm {~B} / 3 ) } = \frac { \mathrm { C } } { \sin \left( 90 ^ { \circ } + \mathrm { B } / 3 \right) } = \frac { \mathrm { c } } { \cos ( \mathrm { B } / 3 ) }$ ⇒AP ⇒ AP = 2 c sin(B/3)