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 .

Remember that the angle for a sector or arc is found as a percentage of the total degrees of the circle. The proportion of to is the same as to the total circumference of the circle.

The circumference of a circle is found by:

For our data, this means:

Now we can solve for using the proportions:

Cross multiply:

Divide both sides by :

Therefore, is ˚.

Inscribed , which measures , intercepts an arc with twice its measure. That arc is , which consequently has measure

.

This makes an arc which comprises

of the circle.

The circumference of a circle is multiplied by its radius, so

.

The length of is of this, or .

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

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, , which intercepts the semicircle , is such an angle. Consequently,

Inscribed intercepts an arc with twice its angle measure; this arc is , so

.

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,

.

The circumference of the path traveled by the tip of the minute hand over the course of one hour is:

feet.

Since the tip of the minute hand has traveled feet since noon, the minute hand has made

revolutions. Therefore, hours have elapsed since noon, making the time 1:15 PM.

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, , which intercepts the semicircle , is such an angle. Consequently, is a right triangle, and and are complementary angles. Therefore,

Inscribed intercepts an arc with twice its angle measure; this arc is , so

.

The major arc corresponding to this minor arc, , has measure

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

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 .