Cambridge 4U Ex 1.10 Q7 (Graphs) (1 Viewer)

[Pyrobooby]

A chord AB of a circle makes an angle (theta) with the diameter passing through A. If the area of the minor segment is one-quarter the area of the circle, show that:
sin (2[theta]) = (pi/2) - (2[theta]).
Solve this equation graphically.

Even using the worked solutions, it is not making any sense to me.

Thank you in advance!

 

[deterministic]

Refer to diagram 1
Let the radius of the circle be r. Since the angle at A is theta, and triangle AOB is isosceles (two sides are radii of the circle and hence equal), and hence angle at B is also theta. Let us call angle at A=x, angle BAO= angle ABO=x (instead of writing theta).
then the angle AOB is pi-2x (use the fact that angle sum of triangle is pi)

so by the area of minor segment formula,
Area= 1/2*r^2*(pi-2x)-1/2*r^2*sin(pi-2x)
But the area of the quarter circle is pi/4*r^2. So hence
pi/4*r^2=1/2*r^2*(pi-2x)-1/2*r^2*sin(pi-2x)
Cancelling r^2 and multiplying everything by 2 gives
pi/2=pi-2x-sin(pi-2x)
pi/2-2x= sin(pi-2x)=sin(2x), since we note for any angle y, sin(pi-y)=sin(y)

 

[deterministic]

In order to solve it graphically, sketch y=sin(2x) and y=pi-2x and the point of intersection should be the solution for theta.

 

[Pyrobooby]

Ahh, thank you for the quick input! I get it now.