circle o is shown below the diagram is not drawn to scale what is the measure of angle bac

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

Explanation:

12 compartments further ways 240 more than degrees. 240 is the answer.

360/12 = 240 degrees

Circle2

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

Correct answer:

30^o

Explanation:

The mensurate of bending CEF is going to exist equal to half of the difference betwixt the measures two arcs that information technology intercepts, namely arcs AD and CD.

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

Thus, we need to detect the measure of arcs Advert and CD. Let's look at the information given and make up one's mind how it can assistance us figure out the measures of arcs Advertising and CD.

Angle BAC is an inscribed bending, which means that its meausre is 1-one-half of the measure out 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 exist 70 degrees.

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

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

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

Nosotros have the measures of arcs BC and CD. Just we still need the measure of arc AD. We can apply the last piece of information given, along with our cognition near 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 as well know that the sum of the measures of arcs AD, AB, CD, and BC must exist 360 degrees, because there are 360 degrees in a full circle.

AD+AB+CD+BC=360^o

AD+AB+80^o+70^o=360^o

AD+AB=210^o

Considering Advertizing = 2AB, nosotros can substitute 2AB for AD.

2AB+AB=210^o

3AB=210^o

AB=70^o

This ways the measure of arc AB is 70 degrees, and the measure of arc Advertising is two(70) = 140 degrees.

At present, we accept all the information we need to find the mensurate of bending CEF, which is equal to one-half the difference between the measure of arcs Advertizement and CD.

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

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

$\angle CEF=30^o$

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

Right answer:

90^o

Caption:

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 90^o.

Note: This makes sense. Since the arc S was one-4th the circumference of the circumvolve, the fundamental angle formed by arc S should be ane-fourth the total degrees of a circle.

Circle

In the circumvolve to a higher place, the length of arc BC is 100 degrees, and the segment Air-conditioning is a diameter. What is the mensurate of angle ADB in degrees?

Possible Answers:

forty

cannot be determined

eighty

100

Explanation:

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 bending ADB is an inscribed angle, its measure is equal to one-half of the measure of the angle of the arc that it intercepts. This ways that the measure of the angle is one-half of lxxx degrees, or 40 degrees.

What is the angle of a sector of area in^2 on a circle having a radius of $15\:in$ ?

Right answer:

$22.92^{\circ}$

Caption:

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

$A=\pi r^2$

For your data, this is:

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

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

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

Now, multiply this past the total 360 degrees in a circle:

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

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

What is the angle of a sector that has an arc length of 13.5 on a circle of bore ?

Right respond:

$128.92^{\circ}$

Explanation:

The commencement thing to exercise for this problem is to compute the total circumference of the circle. Detect 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, annotation that yous 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 $128.92^{\circ}$ .

Inscribed angle

Figure Not drawn to scale.

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

Possible Answers:

$66 ^{\circ }$

$77 ^{\circ }$

$88^{\circ }$

$55^{\circ }$

Insufficient information is given to reply the question.

Correct respond:

$66 ^{\circ }$

Caption:

An inscribed angle of a circle that intercepts a semicircle is a right bending; 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, then

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

(7t+1 )+ (4t+1) = 90

7t+4t+1+1 = 90

11t+2 = 90

11t+2 -2 = 90 - 2

11t = 88

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

t = 8

The measure out 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 out, so

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

Secant 2 Effigy NOT fatigued to scale

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

Correct answer:

t = 56

Explanation:

$\overline{AB}$ is a diameter, and 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 out of the angle that the segments form is equal to ane-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 $m \angle CNB = 34^{\circ }$ :

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

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

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

90 - t= 34

90-t-90 = 34 -90

-t =- 56

t = 56

Secant

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

Correct answer:

$45 ^{\circ }$

Caption:

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 out:

t+t+3t+3t = 360

8t = 360

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

t = 45

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 one-half the departure of the measures of the arcs they intercept, so

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

Substituting:

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

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

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

Inscribed quad

Figure Not drawn to calibration.

The higher up effigy shows a quadrilateral inscribed in a circumvolve. Evaluate.

Possible Answers:

u= 88

The question cannot be answered from the information given.

u= 102

u= 96

u= 92

Correct answer:

u= 92

Explanation:

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

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

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

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

88+ u = 180

88+ u - 88 = 180 - 88

u= 92 ,

the correct response.

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