In the figure o is the center of a semi-circle pqr and r is the radius.
pq is a diameter of semi-circle pto. q^ = x.
determine:
1.) rq in terms of r and x , and then simplify the expression. (3 marks)
2.)the ratio area of triangle orq/area of triangle pot if x = 75 degrees. (leave your answer simplified in surd form) (5 marks)

1.) We can draw the height of triangle ORQ that intersects the center and we form right triangles. Using the definition of cosine, we get
cos x = (1/2) RQ / r
RQ = 2r cos x

2.) We simply use the expression for RQ we got from 1 to solve for the area of triangle ORQ. Similar to 1), the expression for the height of triangle ORQ is: r sin x.
For triangle POT, if we draw the height of the triangle that hits OP, we get special right triangles with angles 30-60-90. The height is simply twice that of the radius divided by the square root of 3.

So,
ratio of areas = (1/2)bh of triangle ORQ / (1/2)bh of triangle POT

b is the base  and h is the height

ratio of areas = (1/2)(2r cos 75)(r sin 75) / (1/2)(r)(2/sqrt(3))(r)
ratio of areas = sqrt(3) / 8

Answer d; a supplementary angle = 180°, so if that angle = 150°, y would = 30°.

The answer would be a