** Q=AB is the diameter of a circle,center O .C is a point on the circumference such that /-COB=a.The area of minor segment cut off by AC is equal to twice the area of the sector BOC. Prove that **

**sin a/2 cos a/2 =pie(1/2- a/120)**

Let the radius of the circle be *r*.

Area of the sector BOC =

∠AOC = (180° – *a*) ... (Linear pair)

Area of the minor segment cut off by the chord AC

= Area of the sector AOC – Area of ΔAOC

Given, Area of the minor segment cut off the chord AC = 2 × Area of the sector BOC

These formulae are taught in Grade XI

**
**