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555-605 (Medium)|   Geometry|      
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smyarga
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If the radius of the circle below (see attachment) is equal to the cho Updated on: 29 Jun 2018, 05:44
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.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

35% (medium)

Question Stats:

74% (01:44) correct 26% (01:39) wrong based on 205 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?

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central angle.png
central angle.png [ 19.79 KiB | Viewed 6974 times ]
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If the radius of the circle below (see attachment) is equal to the cho 04 Aug 2014, 13:38
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smyarga
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?
Show more

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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PareshGmat
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If the radius of the circle below (see attachment) is equal to the cho 04 Aug 2014, 23:03
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I was unaware of the property explained in earlier post :(

Solved as below: Refer diagram

OA = OB = OC = AC = Radius of circle

\(\triangle OAC = Equilateral\)

\(\triangle OAB = Isosceles\)

\(\triangle OCB = Isosceles\)



Answer = 30
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central%20angle.png
central%20angle.png [ 19.55 KiB | Viewed 6534 times ]

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If the radius of the circle below (see attachment) is equal to the cho 05 Aug 2014, 00:04
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PareshGmat
I was unaware of the property explained in earlier post :(

Solved as below: Refer diagram

OA = OB = OC = AC = Radius of circle

\(\triangle OAC = Equilateral\)

\(\triangle OAB = Isosceles\)

\(\triangle OCB = Isosceles\)



Answer = 30
Show more

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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If the radius of the circle below (see attachment) is equal to the cho 05 Aug 2014, 01:05
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PareshGmat
I was unaware of the property explained in earlier post :(

Solved as below: Refer diagram

OA = OB = OC = AC = Radius of circle

\(\triangle OAC = Equilateral\)

\(\triangle OAB = Isosceles\)

\(\triangle OCB = Isosceles\)



Answer = 30

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

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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If the radius of the circle below (see attachment) is equal to the cho 05 Aug 2014, 02:40
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PareshGmat
I was unaware of the property explained in earlier post :(

Solved as below: Refer diagram

OA = OB = OC = AC = Radius of circle

\(\triangle OAC = Equilateral\)

\(\triangle OAB = Isosceles\)

\(\triangle OCB = Isosceles\)



Answer = 30

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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Now, I got it! Thanks) But anyway it is much easier with central angles!
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If the radius of the circle below (see attachment) is equal to the cho 05 Aug 2014, 03:12
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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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If the radius of the circle below (see attachment) is equal to the cho 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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If the radius of the circle below (see attachment) is equal to the cho 10 May 2019, 22:37
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smyarga
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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Do u know more questions of inscribed angle? im having trouble with those
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If the radius of the circle below (see attachment) is equal to the cho 27 Dec 2022, 17:22
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