Q=AB is the diameter of a circle,center O C is a point on the circumference such that - Maths - Areas Related to Circles

Question

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)

Ajit Singh · Accepted Answer

Let the radius of the circle be r Area of the sector BOC = $\frac{a}{360 °} \times {π r}^{2} = \frac{a π r^{2}}{360 °}$ âˆ AOC = (180Â° â€“ a) (Linear pair) Area of the minor segment cut off by the chord AC = Area of the sector AOC â€“ Area of Î”AOC $= (\frac{180 ° - a}{360 °}) {π r}^{2} - \frac{1}{2} r^{2} \sin (180 ° - a) = (\frac{180 ° - a}{360 °}) π r^{2} - \frac{1}{2} r^{2} \sin a$ Given, Area of the minor segment cut off the chord AC = 2 Ã— Area of the sector BOC $∴ (\frac{180 ° - a}{360 °}) {π r}^{2} - \frac{1}{2} r^{2} \sin a = \frac{2 a {π r}^{2}}{360 °} \Rightarrow \frac{1}{2} r^{2} \sin a = (\frac{180 ° - a}{360 °}) π r^{2} - \frac{2 a π r^{2}}{360 °} \Rightarrow \frac{1}{2} r^{2} 2 \sin \frac{a}{2} \cos \frac{a}{2} = (\frac{180 ° - a - 2 a}{360 °}) {π r}^{2} \Rightarrow \sin \frac{a}{2} \cos \frac{a}{2} = π (\frac{180 °}{360 °} - \frac{3 a}{360 °}) \Rightarrow \sin \frac{a}{2} \cos \frac{a}{2} = π (\frac{1}{2} - \frac{a}{120 °})^{} \sin (180 ° - θ) = \sin θ and sin θ = 2 \sin \frac{θ}{2} \cos \frac{θ}{2}$ These formulae are taught in Grade XI

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)

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)