Question

AB is the diameter of a circle with centre O. C is a point on circumference such that angle COB =theta. The area of the minor segment cut off by AC is equal to twice the area of the sector BOC prove that sin theta /2*cos thet/2=π(1/2-theta/120)​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given AB is diameter of circle with centre O.

∠COB = θ

Area of segment cut off, by AC = (area of sector) - (area of ΔAOC).

∠AOC = 180 - θ [∠AOC and ∠BOC form linear pair]

Area of sector = (180 - θ)/360 * πr²

                        = πr²/2 - πθr²/360

In ΔAOC, drop a perpendicular AM, this bisects ∠AOC and side AC.

Now, In ΔAMO,

sin∠AOM = AM/DA

sin(180 - θ)/2 = AM/R

⇒ AM = R sin(90 - θ/2) = R cos θ/2

cos ∠ADM = OM/OA

⇒ cos(90 - θ/2) = OM/Y

⇒ OM = R sin θ/2

Now,

Area of segment = πr²/2 - πθr²/360 - 1/2(AC * OM)

⇒ πr²/2 - πθr²/360 - 1/2 * (2 R cos θ/2 sinθ/2)

⇒ r²[π/2 - πθ/360 - cosθ/2sinθ/2]

Area of segment by AC = 2 (Area of sector BDC)

r²[π/2 - πθ/360 - cosθ/2sinθ/2]

= 2r²[πθ/360]

Now,

cosθ/2 sinθ/2

= π/2 - πθ/360 - 2πθ/360

= π/2 - πθ/360[1 + 2]

= π(1/2 - θ/120°)

Hope it helps!