Please help me with this question from Trials (1 Viewer)

[andy21lau]

AOB is a diameter of a circle with centre O and radius 1 metre.
AC is a chord of the circle such that

BAC = θ, where 0 < θ < π/2.
The area of that part of the circle contained between the diameter AB and the chord AC is equal to one quarter of the area of the circle.

i) Show that θ + (1/2)sin2θ - π/4 = 0

 

[Dumbledore]

draw a line OC, u will get a sector and an isoceles triangle
ang(OCA) = @ since its an isoceles triangle with 2 sidelengths being the radii
ang(BOC) = 2@ (forgot theorem name)
A of sector = 1/2 r^2 2@ = @
A of triangle = 1/2 ab sin C = 1/2*1*1sin(180-2@) = 1/2sin2@
A1 + A2 = pi/4 (pi is area of circle)
@ + 1/2sin2@ - pi/4 = 0

 

[andy21lau]

draw a line OC, u will get a sector and an isoceles triangle
ang(OCA) = @ since its an isoceles triangle with 2 sidelengths being the radii
ang(BOC) = 2@ (forgot theorem name)
A of sector = 1/2 r^2 2@ = @
A of triangle = 1/2 ab sin C = 1/2*1*1sin(180-2@) = 1/2sin2@
A1 + A2 = pi/4 (pi is area of circle)
@ + 1/2sin2@ - pi/4 = 0

thankyou