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# If the radius of the circle below (see attachment) is equal to the cho

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If the radius of the circle below (see attachment) is equal to the cho  [#permalink]

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Updated on: 29 Jun 2018, 05:44
7
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00:00

Difficulty:

15% (low)

Question Stats:

81% (02:12) correct 19% (01:42) wrong based on 149 sessions

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If the radius of the circle below (see attachment) is equal to the chord AC and O is the centre of this circle, then what is the degree measure of angle ABC?

(A) 25
(B) 30
(C) 40
(D) 45
(E) 50

P.S. 10 seconds problem:) Can you?

Attachment:

central angle.png [ 19.79 KiB | Viewed 3051 times ]

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Originally posted by smyarga on 04 Aug 2014, 13:29.
Last edited by Bunuel on 29 Jun 2018, 05:44, edited 3 times in total.
Edited the question
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Re: If the radius of the circle below (see attachment) is equal to the cho  [#permalink]

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04 Aug 2014, 13:38
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2
smyarga wrote:
If the radius of the circle below (see attachment) is equal to the chord AC and O is the centre of this circle, then what is the degree measure of angle ABC?

(A) 25
(B) 30
(C) 40
(D) 45
(E) 50

P.S. 10 seconds problem:) Can you?

Property : An inscribed angle is half the measure of a central angle. Here, <AOC is a central angle and <ABC is an inscribed angle.

Therefore :$$<AOC = 2*<ABC$$

We know that triangle AOC is equilateral, so all three angles of triangle AOC are 60 degrees. Hence <AOC = 60 degrees.
$$<ABC = \frac{<AOC}{2} = \frac{60}{2}=$$ 30 degrees

Edit : smyarga or whoever gave it to me : thanks for giving me my second little kudo
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Re: If the radius of the circle below (see attachment) is equal to the cho  [#permalink]

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04 Aug 2014, 23:03
2
I was unaware of the property explained in earlier post

Solved as below: Refer diagram

OA = OB = OC = AC = Radius of circle

$$\triangle OAC = Equilateral$$

$$\triangle OAB = Isosceles$$

$$\triangle OCB = Isosceles$$

Attachments

central%20angle.png [ 19.55 KiB | Viewed 2862 times ]

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Re: If the radius of the circle below (see attachment) is equal to the cho  [#permalink]

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05 Aug 2014, 00:04
PareshGmat wrote:
I was unaware of the property explained in earlier post

Solved as below: Refer diagram

OA = OB = OC = AC = Radius of circle

$$\triangle OAC = Equilateral$$

$$\triangle OAB = Isosceles$$

$$\triangle OCB = Isosceles$$

That's definitely not 10 seconds:) but still how did you find either 30+a or 30? I don't understand from your picture.
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Re: If the radius of the circle below (see attachment) is equal to the cho  [#permalink]

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05 Aug 2014, 01:05
smyarga wrote:
PareshGmat wrote:
I was unaware of the property explained in earlier post

Solved as below: Refer diagram

OA = OB = OC = AC = Radius of circle

$$\triangle OAC = Equilateral$$

$$\triangle OAB = Isosceles$$

$$\triangle OCB = Isosceles$$

That's definitely not 10 seconds:) but still how did you find either 30+a or 30? I don't understand from your picture.

Agreed.. It took around 110 seconds for me to solve this. I just used properties of equilateral / isosceles triangle to solve it.

Joined OA = OB = OC = AC = Radius of circle

$$\triangle OAC = Equilateral$$

$$\angle OAC = \angle ACO = \angle COA = 60^{\circ}$$

$$\triangle OAB = Isosceles$$

Say$$\angle OAB = \angle OBA = a$$

So,$$\angle BAC = 60-a; \angle BOC = 120 - 2a$$

$$\triangle OCB = Isosceles$$

$$So, \angle OCB = \frac{180 - (120-2a)}{2} = \frac{60 + 2a}{2} = 30 + a$$

$$\angle OCB = \angle OBC = \angle OBA + \angle ABC$$

$$30 + a = a + \angle ABC$$

$$\angle ABC = 30$$
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Re: If the radius of the circle below (see attachment) is equal to the cho  [#permalink]

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05 Aug 2014, 02:40
PareshGmat wrote:
smyarga wrote:
PareshGmat wrote:
I was unaware of the property explained in earlier post

Solved as below: Refer diagram

OA = OB = OC = AC = Radius of circle

$$\triangle OAC = Equilateral$$

$$\triangle OAB = Isosceles$$

$$\triangle OCB = Isosceles$$

That's definitely not 10 seconds:) but still how did you find either 30+a or 30? I don't understand from your picture.

Agreed.. It took around 110 seconds for me to solve this. I just used properties of equilateral / isosceles triangle to solve it.

Joined OA = OB = OC = AC = Radius of circle

$$\triangle OAC = Equilateral$$

$$\angle OAC = \angle ACO = \angle COA = 60^{\circ}$$

$$\triangle OAB = Isosceles$$

Say$$\angle OAB = \angle OBA = a$$

So,$$\angle BAC = 60-a; \angle BOC = 120 - 2a$$

$$\triangle OCB = Isosceles$$

$$So, \angle OCB = \frac{180 - (120-2a)}{2} = \frac{60 + 2a}{2} = 30 + a$$

$$\angle OCB = \angle OBC = \angle OBA + \angle ABC$$

$$30 + a = a + \angle ABC$$

$$\angle ABC = 30$$

Now, I got it! Thanks) But anyway it is much easier with central angles!
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Re: If the radius of the circle below (see attachment) is equal to the cho  [#permalink]

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05 Aug 2014, 03:12
Now, I got it! Thanks) But anyway it is much easier with central angles![/quote]

Agreed... However as I mentioned earlier, central angle concept was a learning lesson for me.

So I worked upon using the method stated
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Re: If the radius of the circle below (see attachment) is equal to the cho  [#permalink]

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07 Dec 2014, 00:36
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For quick solution we have to join OC ,OA AND OB. TRIANGLE OAC IS EQUILATERAL , SO ANGLE AOC= 60. AND ANGLE ABC WILL BE EQUAL TO HALF OF ANGLE AOC SINCE ANGLE ABC IS AT CIRCUMFERANCE AND ANGLE AOC IS AT CENTRE MADE BY SAME ARC.
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Re: If the radius of the circle below (see attachment) is equal to the cho  [#permalink]

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29 Jun 2018, 05:15
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Re: If the radius of the circle below (see attachment) is equal to the cho &nbs [#permalink] 29 Jun 2018, 05:15
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