If `P` is a point on the altitude AD of the triangle ABC such the `/_C B P=B/3,` then AP is equal to `2asinC/3` (b) `2bsinC/3` (c) `2csinB/3` (d) `2csinC/3`

Options

A. `2a sin. (C)/(3)`
B. `2b sin.(C)/(3)`
C. `2c sin.(B)/(3)`
D. `2c sin.(C)/(3)`

Answer

C

Explanation

`angle BPA = 90^(@) + ((B)/(3)), angle ABP = (2B)/(3)`
In `Delta ABP, (AP)/(sin (2B//3)) = (c)/(sin[90^(@) + (B//3)]) = (c)/(cos(B//3))` [by the sine rule]
or `AP = (c sin (2B//3))/(cos (B//3)) = (2c sin (B//3) cos (B//3))/(cos (B//3))`
`= 2 c sin (B//3)`