An acute isosceles triangle, ABC is inscribed in a circle. Through B

angle ACB = 2(angle D), so if angle D = y, then angle ABC = ACB = 2y.

If O denotes the center of the circle, then following must be true about the interior angles of deltoid OBDC:

angle COB + angle OBD + angle D + angle OCD = 360 [Because the sum of the interior angles of any quadrilateral is always equal to 360 degrees.]

angle COB + 90 + y + 90 = 360 [Two angles are 90 degrees because each of them is defined by a tangent and a radius.]

angle COB = 180 – y

Since angle A is an inscribed angle to minor arc CB and angle COB is a central angle to minor arc CB, we have:

angle A = angle COB/2

angle A = (180 – y)/2

For triangle ABC, we have:

(180 – y)/2 + 2y + 2y = 180

180 – y + 8y = 360

y = 180/7 degrees

Therefore,

angle A = (180 – 180/7)/2 degrees

Since 180 degrees is equal to pi radian, we have:

x = angle A = (pi – pi/7)/2 radian = 3pi/7 radian

Answer: A