Question

Question #a948d

Answer 1

Let `h` be the length of the perpendicular drawn from B to AC and `theta=/_BCA`. So `h/a=sintheta`
Area of `DeltaABC=1/2*"base"*"height"`

`=1/2*AC*h`

`=1/2*b*a*h/a`

`=1/2ab sin theta`