Question

AB is the diameter of a circle Centre o. c is a point on the circumference such that angle C is equal to theta the area of minor segment cut off by AC is equal to twice the area of the sector B O C prove that sin theta upon 2 cos theta upon 2 equal in bracket 1 upon 2 minus theta upon 1 20

Answer 1

Answer:

Step-by-step explanation:

Draw a diagram to understand easily.

From O draw OE ⊥ AC, which bisects AC.

Angle ∠ACO = ∠OAC as AO = OC = radius

For ΔAOC,  ∠COB is an exterior angle.  

       ∠COB = x = ∠CAO + ∠OCA 

      =>   ∠OCE = x/2

In ΔACE,  CE = OC cos x/2     =>  CA = 2 OC cos x/2

                OE = OC sin x/2

Area(ΔAOC) = 1/2 * CA * OE = OC² sin x/2  cos x/2

Area(sector COB) = π OC² * x°/360°

As given   Area(minor segment AC) = 2 π OC² x°/360°

Area of semicircle AOBC =

  = Area(minor segment AC) + Area(ΔAOC) + Area(sector COB)

So π OC² /2  = 3 π OC² * x°/360°  + OC² sin x°/2  cos x°/2

    sin x/2  cos x/2 = π (1/2 - x°/120°)