How to find the angle of a sector - Math

Since

,

the triangle is a right triangle with right angle .

is an inscribed angle on the circle, so the arc it intercepts is a semicricle. Therefore, is also a semicircle, and it measures .

The measure of an inscribed angle of a circle is one-half that of the arc it intercepts. Therefore, .

The measure of an arc intercepted by an inscribed angle of a circle is twice that of the angle. Therefore,

has twice the radius of , so has four times the area of . This means that for a sector of to have the same area as a sector of , the central angle of the latter sector must be four times that of the former sector. This makes (b) greater than (a), which is only twice that of the former sector.

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 area of the circle.

The area 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 ˚.

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 area of the circle.

The area 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 ˚.

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 ˚.

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 ˚.

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 .

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.

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,

.

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 .

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,

.

Since we know that segment AC is a diameter, this means that the length of the arc ABC must be 180 degrees. This means that the length of the arc AB must be 80 degrees.

Since angle ADB is an inscribed angle, its measure is equal to half of the measure of the angle of the arc that it intercepts. This means that the measure of the angle is half of 80 degrees, or 40 degrees.

The circumference of the circle is .

The length of the arc S is .

A ratio can be established:

Solving for __yields 90o.

Note: This makes sense. Since the arc S was one-fourth the circumference of the circle, the central angle formed by arc S should be one-fourth the total degrees of a circle.