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In the circle above the length of the minor arc AB is 10pi and the

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In the circle above the length of the minor arc AB is 10pi and the [#permalink]

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New post 24 Aug 2016, 02:45
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Difficulty:

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Question Stats:

60% (02:41) correct 40% (02:31) wrong based on 65 sessions

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In the circle above the length of the minor arc AB is 10pi and the length of the minor arc CD is 3pi. What is the measure of angle F, if the radius of the circle is 18?

(A) 20
(B) 25
(C) 30
(D) 35
(E) 40

Source: OptimusPrep

LE: Updated. Had some editing issues first time.
[Reveal] Spoiler: OA

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Last edited by Stn on 24 Aug 2016, 04:08, edited 1 time in total.

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Re: In the circle above the length of the minor arc AB is 10pi and the [#permalink]

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New post 24 Aug 2016, 03:51
Please share the options and there is no E but I see F..

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Re: In the circle above the length of the minor arc AB is 10pi and the [#permalink]

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New post 24 Aug 2016, 03:57
Stn wrote:
In the circle above the length of the minor arc AB is 10pi and the length of the minor arc CD is 3pi. What is the measure of angle E, if the radius of the circle is 18?

Source: OptimusPrep


Is it 70 degree?

Angle of Major Arc - Angle of Minor Arc = 100 - 30 = 70 Degree. (Not Sure about this).

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In the circle above the length of the minor arc AB is 10pi and the [#permalink]

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The only thing you need to remember while solving this question is the arc rule.

Arc rule: The angle subtended by the arc at the center is always twice the angle subtended by it at any other point on the circle.

IN this question since arc AB=10pi and arc CD=3pi and circumference=2.pi.r =2.pi.18 =36pi, we can
calculate the angle subtended by the arcs at the center as,

3pi/36pi = 1/12 =1/12th of whole angle =360 * 1/12 = 30 and thus making angle CBD as 15degrees.
(Coz the arc CD subtends 30deg at center, it will subtend 15deg at any other point on circumference of circle, in this case pt B)

Next, 10pi/36pi = 10/36th of whole angle = 360 * 10/36 = 100 and thus making angle ACB as 50 degrees.
Now In TRIANGLE CBF,
angle B(CBD) =15
angle C(BCF) =130
thus angle F= 180 - (B + C)
= 180 - 145
= 35

(D)

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Re: In the circle above the length of the minor arc AB is 10pi and the [#permalink]

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New post 24 Aug 2016, 07:56
Well explained Shashank. Missed to divide by 2 :(

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Re: In the circle above the length of the minor arc AB is 10pi and the [#permalink]

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New post 24 Aug 2016, 08:04
My Approach:

CB is diameter, A is a point on the semi circle..so Angle A will be 90 degrees.
Given radius = 18. So the length of the arc BAC is 18pi.
Given AB = 10pi So AC = 8pi

AC = 8pi, CD = 3pi, So AD = 11pi.
At centre, AD subtends and angle 110 degrees.( By comparing 18pi spans 180 degrees)
So at B, AD subtends an angle of 110/2 = 55 degrees

90 degrees + 55 degrees + Angle at F = 180 degrees.
Therefore F = 35 degrees

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Re: In the circle above the length of the minor arc AB is 10pi and the [#permalink]

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New post 28 Apr 2017, 02:59
Stn wrote:
In the circle above the length of the minor arc AB is 10pi and the length of the minor arc CD is 3pi. What is the measure of angle F, if the radius of the circle is 18?

(A) 20
(B) 25
(C) 30
(D) 35
(E) 40

Source: OptimusPrep

LE: Updated. Had some editing issues first time.


CB or CB is diameter how can you assume that its not given in the question stem?? and by any means with the help of info given in the question can you prove CB is the diameter??

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Re: In the circle above the length of the minor arc AB is 10pi and the [#permalink]

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New post 28 Apr 2017, 07:29
shashanksst94 wrote:
The only thing you need to remember while solving this question is the arc rule.

Arc rule: The angle subtended by the arc at the center is always twice the angle subtended by it at any other point on the circle.

IN this question since arc AB=10pi and arc CD=3pi and circumference=2.pi.r =2.pi.18 =36pi, we can
calculate the angle subtended by the arcs at the center as,

3pi/36pi = 1/12 =1/12th of whole angle =360 * 1/12 = 30 and thus making angle CBD as 15degrees.
(Coz the arc CD subtends 30deg at center, it will subtend 15deg at any other point on circumference of circle, in this case pt B)

Next, 10pi/36pi = 10/36th of whole angle = 360 * 10/36 = 100 and thus making angle ACB as 50 degrees.
Now In TRIANGLE CBF,
angle B(CBD) =15
angle C(BCF) =130
thus angle F= 180 - (B + C)
= 180 - 145
= 35

(D)

Can someone please explain how we arrived at angle C (BCF) = 130?

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Re: In the circle above the length of the minor arc AB is 10pi and the [#permalink]

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New post 02 Oct 2017, 02:06
shashanksst94 wrote:
The only thing you need to remember while solving this question is the arc rule.

Arc rule: The angle subtended by the arc at the center is always twice the angle subtended by it at any other point on the circle.

IN this question since arc AB=10pi and arc CD=3pi and circumference=2.pi.r =2.pi.18 =36pi, we can
calculate the angle subtended by the arcs at the center as,

3pi/36pi = 1/12 =1/12th of whole angle =360 * 1/12 = 30 and thus making angle CBD as 15degrees.
(Coz the arc CD subtends 30deg at center, it will subtend 15deg at any other point on circumference of circle, in this case pt B)

Next, 10pi/36pi = 10/36th of whole angle = 360 * 10/36 = 100 and thus making angle ACB as 50 degrees.
Now In TRIANGLE CBF,
angle B(CBD) =15
angle C(BCF) =130
thus angle F= 180 - (B + C)
= 180 - 145
= 35

(D)


Hi ,
But what if BC is not the diameter of the circle? It is not given in the question. Can we solve the question the same way without this info?

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Re: In the circle above the length of the minor arc AB is 10pi and the   [#permalink] 02 Oct 2017, 02:06
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