Knowing this theorem can help a lot:

The opposite angles of quadrilaterals in circles are equal to two right angles.

If you complete the quadrilateral YCAX, then according to above theorem, ang YXC + ang YCA = 180

so,ang yca = 180-105=75

Another theorem to remember:

Angle at the centre is double the angle at the circumference if the angles have the same circumference as base.

Based on the above thm, ang YBA is double of the ang YCA.

So, ang YBA = 150

So ang YBC=180-150=30

Area of arc ybc = 0.5*r^2*pi/6

since 30 deg = pi/6

Area below = 0.5*r^2*pi/6 + pi*r^2/8

Area above = 0.5*r^2*5*pi/6 - pi*r^2/8

so ration above/below=7/5

Praetorian wrote:

This is one of Akamai's .. nice one...

http://www.manhattangmat.comPraetorian