Question

AB is the diameter of a circle, centre O. C is the point on circumference such that angle COB = x. The area of the minor segment cut off by AC is equal to twice the area of the sector BOC. Prove that:
sinx/2cosx/2 = pi(1/2 - x/120)

Answer 1

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°)

Answer 2

Attached.



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