In ` A B C ,` the median `A D` divides `/_B A C` such that `/_B A D :/_C A D=2:1` . Then `cos(A/3)` is equal to `(sinB)/(2sinC)` (b) `(sinC)/(2sinB)` `(2sinB)/(sinC)` (d) `non eoft h e s e`

Options

A. `(sin B)/(2 sin C)`
B. `(sin C)/(2 sin B)`
C. `(2 sin B)/(sin C)`
D. none of these

Answer

A

Explanation

Let `(A)/(3) = angleCAD = theta`
Now, by m -n theorem,
`(1+1) cot alpha = 1 cot 2 theta - 1 cot theta`
or `2cot (B + 2 theta) = cot 2 theta - cot theta`
or `cot (B + 2 theta) + cot theta = cot 2 theta - cot (B + 2theta)`
or `(sin (B + 3 theta))/(sin (B + 2 theta) sin theta) = (sin B)/(sin (B + 2 theta) sin 2 theta)`
or `(sin (B+A))/(sin theta) = (sin B)/(sin 2 theta)`
`rArr sin C = (sin B)/(2 cos theta)`
`rArr cos.(A)/(3) = (sin B)/(2 sin C)`