How to find the angle of a sector - SAT Math

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Question

In the circle above, the length of arc BC is 100 degrees, and the segment AC is a diameter. What is the measure of angle ADB in degrees?

Answer

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.

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Question

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 ?

Answer

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.

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Question

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.

Answer

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.

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

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

$\angle CEF=30^o$

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Question

A pie has a diameter of 12". A piece is cut out, having a surface area of 4.5π. What is the angle of the cut?

Answer

This is simply a matter of percentages. We first have to figure out what percentage of the surface area is represented by 4.5π. To do that, we must calculate the total surface area. If the diameter is 12, the radius is 6. Don't be tricked by this!

A = π * 6 * 6 = 36π

Now, 4.5π is 4.5π/36π percentage or 0.125 (= 12.5%)

To figure out the angle, we must take that percentage of 360°:

0.125 * 360 = 45°

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Question

Eric is riding a Ferris wheel. The Ferris wheel has 18 compartments, numbered in order clockwise. If compartment 1 is at 0 degrees and Eric enters compartment 13, what angle is he at?

Answer

12 compartments further means 240 more degrees. 240 is the answer.

360/12 = 240 degrees

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Question

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

Answer

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 $22.92^{\circ}$ .

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Question

What is the angle of a sector that has an arc length of on a circle of diameter ?

Answer

The first thing to do for this problem is to compute the total circumference of the circle. Notice that you were given the diameter. The proper equation is therefore:

$C=\pi d$

For your data, this means,

$C=12\pi$

Now, to compute the angle, note that you have a percentage of the total circumference, based upon your arc length:

$\frac{13.5}{12\pi}*360=128.9155039044336$

Rounded to the nearest hundredth, this is $128.92^{\circ}$ .

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Question

Figure NOT drawn to scale.

Refer to the above diagram. $\overarc {ABC}$ is a semicircle. Evaluate $m \overarc {AB}$ .

Answer

An inscribed angle of a circle that intercepts a semicircle is a right angle; therefore, $\angle B$ , which intercepts the semicircle $\overarc{ABC}$ , is such an angle. Consequently, $m \angle B = 90 ^{\circ }$ , and $\bigtriangleup ABC$ is a right triangle. The acute angles of $\bigtriangleup ABC$ are complementary, so

$m \angle A + m \angle C = 90 ^{\circ }$

$11t\div 11 = 88 \div 11$

The measure of inscribed $\angle C$ is

$m \angle C = (4t+1 )^{\circ } = (4 \cdot 8+1 )^{\circ } = 33 ^{\circ }$ .

An inscribed angle of a circle intercepts an arc of twice its degree measure, so

$m \overarc {AB} = 2 \cdot m \angle C= 2 \cdot 33^{\circ } = 66 ^{\circ }$ .

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Question

Figure NOT drawn to scale

Refer to the above figure. $\overline{AB}$ is a diameter of the circle. Evaluate .

Answer

$\overline{AB}$ is a diameter, so $\overarc{BC A }$ is a semicircle, and

$m \overarc{BC A } = 180^{\circ }$ ,

or, equivalently,

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

In terms of , since $m \overarc {AC } = t^{\circ }$ ,

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

$m \overarc{BC} + t^{\circ } - t ^{\circ }= 180 ^{\circ } - t^{\circ }$

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

$\overline{NB}$ and $\overline{NC }$ , being a secant segment and a tangent segment to a circle, respectively, 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,

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

Setting $m \overarc{BC} = \left (180 - t \right )^{\circ }$ , $m \overarc {AC } = t^{\circ }$ , and $m \angle CNB = 34^{\circ }$ :

$\frac{1}{2}\left [ (180 - t) - t \right ]= 34$

$\frac{1}{2} (180 - 2t) = 34$

$\frac{1}{2} \cdot 180 -\frac{1}{2} \cdot 2t = 34$

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Question

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

Answer

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:

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

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

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

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Question

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Answer

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle B$ and $\angle D$ are two such angles, so

$m \angle B + m \angle D = 180 ^{\circ }$

Setting $m \angle B = 88^{\circ }$ and $m \angle D =u ^{\circ }$ , and solving for :

,

the correct response.

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Question

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Answer

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle A$ and $\angle C$ are two such angles, so

$m \angle A + m \angle C= 180 ^{\circ }$

Setting $m \angle A = t^{\circ }$ and $m \angle C = (180-t)^{\circ }$ , and solving for :

,

The statement turns out to be true regardless of the value of . Therefore, without further information, the value of cannot be determined.

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Question

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Answer

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle A$ and $\angle C$ are two such angles, so

$m \angle A + m \angle C= 180 ^{\circ }$

Setting $m \angle A = t^{\circ }$ and $m \angle C = (t+12)^{\circ }$ , and solving for :

$2t \div 2 = 168 \div 2$

,

the correct response.

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Question

Figure NOT drawn to scale.

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

Answer

$\overline{AB}$ 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.

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Question

Figure NOT drawn to scale.

In the above figure, $\overline{AB}$ is a diameter. Also, the ratio of the length of $\overarc{BT}$ to that of $\overarc{AT}$ is 7 to 5. Give the measure of $\angle TNA$ .

Answer

$\overline{AB}$ is a diameter, so $\overarc {ATB}$ is a semicircle, which has measure $180 ^{\circ }$ . By the Arc Addition Principle,

$m \overarc {AT } + m \overarc {BT} = m \overarc {ATB}$

If we let $t ^{\circ }= m \overarc {AT }$ , then, substituting:

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

and

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

the ratio of the length of $\overarc{BT}$ to that of $\overarc{AT}$ is 7 to 5; this is also the ratio of their degree measures; that is,

$\frac{m \overarc{BT}}{m \overarc{AT}} = \frac{5}{3}$

Setting $m \overarc {AT } = t ^{\circ }$ and $m \overarc {BT} =( 180 - t ) ^{\circ }$ :

$\frac{180-t }{t} = \frac{7}{5}$

Cross-multiply, then solve for :

$\frac{12t }{12}= \frac{900}{12}$

$m \overarc {AT } =75^{\circ }$ , and $m \overarc {BT} =( 180 -75 ) ^{\circ } = 105 ^{\circ }$

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 $\overline{NB}$ and $\overline{NT}$ are such segments whose angle $\angle TNA$ intercepts $\overarc{AT}$ and $\overarc{BT}$ , it holds that:

$m \angle TNA = \frac{1}{2}\left ( m \overarc{BT}- m \overarc{AT} \right )$

$m \angle TNA = \frac{1}{2}\left ( 105 ^{\circ } - 75^{\circ } \right )$

$m \angle TNA = \frac{1}{2}\left ( 30 ^{\circ } \right )$

$m \angle TNA = 15^{\circ }$

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