SAT Math - Sectors | Practice Hub

Secant

Refer to the above diagram. Evaluate the measure of $\angle BND$ .

$45 ^{\circ }$

$30^{\circ }$

$60^{\circ }$

$37 \frac{1}{2}^{\circ }$

$52 \frac{1}{2}^{\circ }$

Explanation

The total measure of the arcs that comprise a circle is $360 ^{\circ }$ , so from the above diagram,

$m \overarc { AC}+m \overarc { CD}+ m \overarc { DB} + m \overarc {BA } = 360 ^{\circ }$

Substituting the appropriate expression for each arc measure:

$8t \div 8 = 360 \div 8$

Therefore,

$m \overarc { AC}= 45^{\circ }$

and

$m \overarc { DB} = 3 t^{\circ } = 3 \cdot 45^{\circ } = 135^{\circ }$

The measure of the angle formed by the tangent segments $\overline{NB}$ and $\overline{ND}$ , which is $\angle BND$ , is half the difference of the measures of the arcs they intercept, so

$m\angle BND = \frac{1}{2} (m \overarc {DB} - m \overarc {AC})$

Substituting:

$m\angle BND = \frac{1}{2} (135 ^{\circ }- 45^{\circ } )$

$m\angle BND = \frac{1}{2} (90^{\circ } )$

$m\angle BND = 45 ^{\circ }$

Secant

Refer to the above diagram. Evaluate the measure of $\angle BND$ .

$45 ^{\circ }$

$30^{\circ }$

$60^{\circ }$

$37 \frac{1}{2}^{\circ }$

$52 \frac{1}{2}^{\circ }$

Explanation

The total measure of the arcs that comprise a circle is $360 ^{\circ }$ , so from the above diagram,

$m \overarc { AC}+m \overarc { CD}+ m \overarc { DB} + m \overarc {BA } = 360 ^{\circ }$

Substituting the appropriate expression for each arc measure:

$8t \div 8 = 360 \div 8$

Therefore,

$m \overarc { AC}= 45^{\circ }$

and

$m \overarc { DB} = 3 t^{\circ } = 3 \cdot 45^{\circ } = 135^{\circ }$

The measure of the angle formed by the tangent segments $\overline{NB}$ and $\overline{ND}$ , which is $\angle BND$ , is half the difference of the measures of the arcs they intercept, so

$m\angle BND = \frac{1}{2} (m \overarc {DB} - m \overarc {AC})$

Substituting:

$m\angle BND = \frac{1}{2} (135 ^{\circ }- 45^{\circ } )$

$m\angle BND = \frac{1}{2} (90^{\circ } )$

$m\angle BND = 45 ^{\circ }$

Secant

Figure NOT drawn to scale

Refer to the above figure. Evaluate .

Explanation

$\overline{NB}$ and $\overline{ND }$ , being secant segments of a circle, intercept two arcs such that the measure of the angle that the segments form is equal to one-half the difference of the measures of the intercepted arcs - that is,

$\frac{1}{2} (m \overarc{BD} - m \overarc {AC }) = m \angle BND$

Setting $m \overarc{BD} = 3t , m \overarc {AC } = t , m \angle BND = 28^{\circ }$ and solving for :

$\frac{1}{2} (3t - t ) = 28$

$\frac{1}{2} (2t ) = 28$

What is the angle of a sector of area on a circle having a radius of ?

$22.92^{\circ}$

$0.06^{\circ}$

$15.22^{\circ}$

$3.00^{\circ}$

$72.00^{\circ}$

Explanation

To begin, you should compute the complete area of the circle:

$A=\pi r^2$

For your data, this is:

$A=15^2\pi=225\pi$

Now, to find the angle measure of a sector, you find what portion of the circle the sector is. Here, it is:

$\frac{45}{225\pi}$

Now, multiply this by the total degrees in a circle:

$\frac{45}{225\pi}*360=22.918311805212$

Rounded, this is $22.92^{\circ}$ .

Secant 2

Figure NOT drawn to scale.

Refer to the above diagram. $\overline{AB}$ is a diameter. Evaluate $m \angle CBA$

$31^{\circ }$

$62^{\circ }$

$56^{\circ }$

$28^{\circ }$

$45^{\circ }$

Explanation

$\overline{AB}$ is a diameter, so $\overarc{ACB}$ is a semicircle - therefore, $m \overarc{ACB} = 180 ^{\circ }$ . By the Arc Addition Principle,

$m \overarc{AC}+ m \overarc{CB} = 180 ^{\circ }$

If we let $t ^{\circ }= m \overarc{AC}$ , then

$t ^{\circ }+ m \overarc{CB} = 180 ^{\circ }$ ,

and

$m \overarc{CB} = (180 -t )^{\circ }$

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since $\overline{NC}$ and $\overline{NB}$ are such segments intercepting $\overarc{AC}$ and $\overarc{CB}$ , it holds that

$\frac{1}{2}\left ( m \overarc{CB}- m \overarc{AC} \right ) = m \angle CNB$

Setting $m \overarc{AC} = t ^{\circ }$ , $m \overarc{CB} = (180 -t )^{\circ }$ , and $m \angle CNB = 28 ^{\circ }$ :

$\frac{1}{2} [ \left ( 180-t \right ) - t ] = 28$

$\frac{1}{2} \left ( 180-2 t \right ) = 28$

$m \overarc{AC} =62^{\circ }$

The inscribed angle that intercepts this arc, $\angle CBA$ , has half this measure:

$m \angle CBA = \frac{1}{2} \cdot m \overarc{AC} = \frac{1}{2} \cdot 62^{\circ } = 31^{\circ }$ .

This is the correct response.

What is the angle of a sector of area on a circle having a radius of ?

$22.92^{\circ}$

$0.06^{\circ}$

$15.22^{\circ}$

$3.00^{\circ}$

$72.00^{\circ}$

Explanation

To begin, you should compute the complete area of the circle:

$A=\pi r^2$

For your data, this is:

$A=15^2\pi=225\pi$

Now, to find the angle measure of a sector, you find what portion of the circle the sector is. Here, it is:

$\frac{45}{225\pi}$

Now, multiply this by the total degrees in a circle:

$\frac{45}{225\pi}*360=22.918311805212$

Rounded, this is $22.92^{\circ}$ .

Secant

Figure NOT drawn to scale

Refer to the above figure. Evaluate .

Explanation

$\frac{1}{2} (m \overarc{BD} - m \overarc {AC }) = m \angle BND$

Setting $m \overarc{BD} = 3t , m \overarc {AC } = t , m \angle BND = 28^{\circ }$ and solving for :

$\frac{1}{2} (3t - t ) = 28$

$\frac{1}{2} (2t ) = 28$

Secant 2

Figure NOT drawn to scale.

Refer to the above diagram. $\overline{AB}$ is a diameter. Evaluate $m \angle CBA$

$31^{\circ }$

$62^{\circ }$

$56^{\circ }$

$28^{\circ }$

$45^{\circ }$

Explanation

$\overline{AB}$ is a diameter, so $\overarc{ACB}$ is a semicircle - therefore, $m \overarc{ACB} = 180 ^{\circ }$ . By the Arc Addition Principle,

$m \overarc{AC}+ m \overarc{CB} = 180 ^{\circ }$

If we let $t ^{\circ }= m \overarc{AC}$ , then

$t ^{\circ }+ m \overarc{CB} = 180 ^{\circ }$ ,

and

$m \overarc{CB} = (180 -t )^{\circ }$

$\frac{1}{2}\left ( m \overarc{CB}- m \overarc{AC} \right ) = m \angle CNB$

Setting $m \overarc{AC} = t ^{\circ }$ , $m \overarc{CB} = (180 -t )^{\circ }$ , and $m \angle CNB = 28 ^{\circ }$ :

$\frac{1}{2} [ \left ( 180-t \right ) - t ] = 28$

$\frac{1}{2} \left ( 180-2 t \right ) = 28$

$m \overarc{AC} =62^{\circ }$

The inscribed angle that intercepts this arc, $\angle CBA$ , has half this measure:

$m \angle CBA = \frac{1}{2} \cdot m \overarc{AC} = \frac{1}{2} \cdot 62^{\circ } = 31^{\circ }$ .

This is the correct response.

The length of an arc, , of a circle is $8\pi$ and the radius, , of the circle is . What is the measure in degrees of the central angle, $\theta$ , formed by the arc ?

Explanation

The circumference of the circle is $2\pi r$ .

$2\pi(16)=32\pi$

The length of the arc S is $8\pi$ .

A ratio can be established:

$\frac{S}{\text{circumference}}=\frac{\theta}{360^o}$

$\frac{8\pi}{32\pi}=\frac{\theta}{360^o}$

Solving for _ $\theta$ _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.

Circle2

In the figure above that includes Circle O, the measure of angle BAC is equal to 35 degrees, the measure of angle FBD is equal to 40 degrees, and the measure of arc AD is twice the measure of arc AB. Which of the following is the measure of angle CEF? The figure is not necessarily drawn to scale, and the red numbers are used to mark the angles, not represent angle measures.

Explanation

The measure of angle CEF is going to be equal to half of the difference between the measures two arcs that it intercepts, namely arcs AD and CD.

$\angle CEF=\frac{1}{2}(AD-CD)$

Thus, we need to find the measure of arcs AD and CD. Let's look at the information given and determine how it can help us figure out the measures of arcs AD and CD.

Angle BAC is an inscribed angle, which means that its meausre is one-half of the measure of the arc that it incercepts, which is arc BC.

$\angle BAC=\frac{1}{2}BC$

$35^o=\frac{1}{2}BC\rightarrow BC=70^o$

Thus, since angle BAC is 35 degrees, the measure of arc BC must be 70 degrees.

We can use a similar strategy to find the measure of arc CD, which is the arc intercepted by the inscribed angle FBD.

$40^o=\frac{1}{2}CD\rightarrow CD=80^o$

Because angle FBD has a measure of 40 degrees, the measure of arc CD must be 80 degrees.

We have the measures of arcs BC and CD. But we still need the measure of arc AD. We can use the last piece of information given, along with our knowledge about the sum of the arcs of a circle, to determine the measure of arc AD.

We are told that the measure of arc AD is twice the measure of arc AB. We also know that the sum of the measures of arcs AD, AB, CD, and BC must be 360 degrees, because there are 360 degrees in a full circle.

Because AD = 2AB, we can substitute 2AB for AD.

This means the measure of arc AB is 70 degrees, and the measure of arc AD is 2(70) = 140 degrees.

Now, we have all the information we need to find the measure of angle CEF, which is equal to half the difference between the measure of arcs AD and CD.

$\angle CEF=\frac{1}{2}(AD-CD)$

$\angle CEF=\frac{1}{2}(140^o-80^o)=\frac{1}{2}(60^o)$

$\angle CEF=30^o$

Sectors

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