Question

If P is a point on the altitude AD of the triangle ABC such that ĐPBD = , then AP is equal to -

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

Answer 1

Answer: (C)

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

Answer 2

ĐBPA = 90⁰ + $\frac { B } { 3 }$ , ĐABP =

In DABP (sine Rule)

$\frac { \mathrm { AP } } { \sin \left( \frac { 2 B } { 3 } \right) }$ = = $\frac { \mathrm { c } } { \cos \left( \frac { \mathrm { B } } { 3 } \right) }$

̃ AP = $\frac { \operatorname { csin } \left( \frac { 2 B } { C } \right) } { \cos \left( \frac { B } { 3 } \right) }$ =

̃ AP = 2c sin .