`C F` is the internal bisector of angle `C` of ` A B C` , then `C F` is equal to (a) `(2a b)/(a+b)cos(C/2)` (b) `(a+b)/(2a b)cosC/2` (c) `( bsinA)/(sin(B+C/2))` (d) none of these

Options

A. `(2ab)/(a + b) cos.(C)/(2)`
B. `(a +b)/(2ab) cos.(C)/(2)`
C. `(b sin A)/(sin(B + (C)/(2)))`
D. none of these

Answer

A, C

Explanation

`Delta = Delta_(1) + Delta_(2)`
`rArr (1)/(2) ab sin C = (1)/(2) b (CF) sin .(C)/(2) + (1)/(2) a (CF) sin.(C)/(2)`
or `CF = (ab sin C)/((a + b) sin.(C)/(2)) = (2ab cos.(C)/(2))/(a+b)`
Again in `DeltaCFB`, by the sine rule, we have
`(CF)/(sin B) = (a)/(sin (pi -theta)) = (a)/(sin theta) = (a)/(sin (B+(C)/(2))) " " ( :' theta + B + (C)/(2) = pi)`
or `CF = (a sin B)/(sin (B +(C)/(2))) = (b sin A)/(sin (B +(C)/(2)))` [by sine rule]