has angles of degree measure 30 and 60; the third angle must measure 90 degrees, making  a right triangle.

For the sake of simplicity, let ; the reasoning is independent of the actual length. The smaller leg of a 30-60-90 triangle has length equal to  times that of the longer leg; this is about

 The area of a right triangle is half the product of its legs, so 

Also, if , then the orange semicircle has diameter 1 and radius . Its area can be found by substituting  in the formula:

The orange semicircle has a greater area than 

That  is a diameter of the circle is actually irrelevant to the problem. Two inscribed angles of a circle that both intercept the same arc, as  and  both do here, have the same measure.

The figure referenced is below:

 is a semicircle, so  is one as well; as a semicircle, its measure is . The inscribed angle that intercepts this semicircle, , is a right angle, of measure . , and the sum of the measures of the interior angles of a triangle is , so 

 has greater measure than , so the minor arc intercepted by  , which is , has greater measure than that intercepted by , which is . It follows that the major arc corresponding to the latter, which is , has greater measure than that  corresponding to the former, which is .

Construct . The new figure is below:

, so . It follows that their respective central angles have measures

and

.

Also, since  and  -  being a semicircle - by the Arc Addition Principle, . , an inscribed angle which intercepts this arc, has half this measure, which is . The other angle of , which is , also measures , so   is equilateral.

, since all radii are congruent;

 by reflexivity;

By the Side-Angle-Side Inequality Theorem (or Hinge Theorem), it follows that . Since  is equilateral, , and since all radii are congruent, . Substituting, it follows that .

Below is the inscribed trapezoid referenced, along with its diagonals.

, so, by the Alternate Interior Angles Theorem, 

, and their intercepted angles are also congruent - that is,

By the Arc Addition Principle, 

.

Both  and  are inscribed angles of the same circle which intercept the same arc; they are therefore of the same measure. The fact that  is a diameter of the circle is actually irrelevant to the problem.

 is the arithmetic mean of  and , so

By arc addition, this becomes

Also, , or, equivalently,

, so

Solving for :

Also,

If two tangents are drawn to a circle, the measure of the angle they form is half the difference of the measures of the arcs they intercept, so

 is a right triangle whose hypotenuse  has length  times that of leg . This is characteristic of a triangle whose acute angles both have measure  -and consequently, whose acute angles are congruent. Therefore,

These inscribed angles being congruent, the arcs they intercept,  and , are also congruent.

 is an inscribed angle, so its degree measure is half that of the arc it intercepts, :

.

 and  are acute angles of right triangle . They are therefore complimentary - that is, their degree measures total . Consequently,

.