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Problem
Problem
If
P
P
P
is a point on the altitude
A
D
AD
A
D
of the
Δ
A
B
C
\Delta ABC
Δ
A
B
C
such that
∠
C
B
P
=
B
3
\angle CBP = \frac{B}{3}
∠
C
B
P
=
3
B
, then
A
P
AP
A
P
is equal to a.
2
a
sin
C
3
2a \sin \frac{C}{3}
2
a
sin
3
C
b.
2
b
sin
A
3
2b \sin \frac{A}{3}
2
b
sin
3
A
c.
2
c
sin
B
3
2c \sin \frac{B}{3}
2
c
sin
3
B
d.
2
c
sin
C
3
2c \sin \frac{C}{3}
2
c
sin
3
C
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