Last visit was: 21 Apr 2026, 14:25 It is currently 21 Apr 2026, 14:25
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
555-605 (Medium)|   Geometry|      
User avatar
smyarga
User avatar
Tutor
Joined: 20 Apr 2012
Last visit: 06 Aug 2020
Posts: 82
Own Kudos:
822
 [22]
Given Kudos: 39
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE:Education (Education)
Expert
Expert reply
GMAT 2: 730 Q51 V38
Posts: 82
Kudos: 822
 [22]
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.
Edited the question
8
Kudos
Add Kudos
14
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
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

History

Date
Time
Result
Not Attempted Yet

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 Spoiler
Attachment:
central angle.png
central angle.png [ 19.79 KiB | Viewed 6779 times ]
ShowHide Answer
Official Answer
B
Most Helpful Reply
User avatar
oss198
User avatar
Joined: 18 Jul 2013
Last visit: 01 Jan 2023
Posts: 68
Own Kudos:
431
 [13]
Given Kudos: 120
Location: Italy
Schools: LBS '21 Insead Sept19 Intake (A) IESE '21 (A$)
GMAT 1: 600 Q42 V31
GMAT 2: 700 Q48 V38
GPA: 3.75
Schools: LBS '21 Insead Sept19 Intake (A) IESE '21 (A$)
GMAT 2: 700 Q48 V38
Posts: 68
Kudos: 431
 [13]
If the radius of the circle below (see attachment) is equal to the cho 04 Aug 2014, 13:38
10
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
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 :)
General Discussion
avatar
PareshGmat
avatar
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,531
Own Kudos:
8,270
 [3]
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,531
Kudos: 8,270
 [3]
If the radius of the circle below (see attachment) is equal to the cho 04 Aug 2014, 23:03
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
Attachments

central%20angle.png
central%20angle.png [ 19.55 KiB | Viewed 6364 times ]

User avatar
smyarga
User avatar
Tutor
Joined: 20 Apr 2012
Last visit: 06 Aug 2020
Posts: 82
Own Kudos:
822
 [1]
Given Kudos: 39
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE:Education (Education)
Expert
Expert reply
GMAT 2: 730 Q51 V38
Posts: 82
Kudos: 822
 [1]
If the radius of the circle below (see attachment) is equal to the cho 05 Aug 2014, 00:04
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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.
avatar
PareshGmat
avatar
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,531
Own Kudos:
8,270
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,531
Kudos: 8,270
If the radius of the circle below (see attachment) is equal to the cho 05 Aug 2014, 01:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
smyarga
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,\(\angle BAC = 60-a; \angle BOC = 120 - 2a\)

\(\triangle OCB = Isosceles\)

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

\(\angle OCB = \angle OBC = \angle OBA + \angle ABC\)

\(30 + a = a + \angle ABC\)

\(\angle ABC = 30\)
User avatar
smyarga
User avatar
Tutor
Joined: 20 Apr 2012
Last visit: 06 Aug 2020
Posts: 82
Own Kudos:
822
Given Kudos: 39
Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE:Education (Education)
Expert
Expert reply
GMAT 2: 730 Q51 V38
Posts: 82
Kudos: 822
If the radius of the circle below (see attachment) is equal to the cho 05 Aug 2014, 02:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
PareshGmat
smyarga
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,\(\angle BAC = 60-a; \angle BOC = 120 - 2a\)

\(\triangle OCB = Isosceles\)

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

\(\angle OCB = \angle OBC = \angle OBA + \angle ABC\)

\(30 + a = a + \angle ABC\)

\(\angle ABC = 30\)
Show more


Now, I got it! Thanks) But anyway it is much easier with central angles!
avatar
PareshGmat
avatar
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,531
Own Kudos:
8,270
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,531
Kudos: 8,270
If the radius of the circle below (see attachment) is equal to the cho 05 Aug 2014, 03:12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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 :)
avatar
mridupawandas
avatar
Joined: 02 May 2014
Last visit: 01 Feb 2017
Posts: 93
Own Kudos:
46
 [1]
Given Kudos: 46
Status:Applied
Location: India
Concentration: Operations, General Management
Schools: Terry '18 (WL) Katz '18 (D) Tippie '18 (D) Eller FT'18 (A) Schulich Sept"18 (S) Desautels '18 (I) Sauder '18 (S) Olin '18 (S) Neeley '18 (WL) Tulane '18 (A) Moore '18 Fox(Temple)'18 UCSD '18 Weatherhead '18 (S) GWU '18 Simon '18 Madison '18
GMAT 1: 690 Q47 V38
GPA: 3.35
WE:Information Technology (Computer Software)
Schools: Terry '18 (WL) Katz '18 (D) Tippie '18 (D) Eller FT'18 (A) Schulich Sept"18 (S) Desautels '18 (I) Sauder '18 (S) Olin '18 (S) Neeley '18 (WL) Tulane '18 (A) Moore '18 Fox(Temple)'18 UCSD '18 Weatherhead '18 (S) GWU '18 Simon '18 Madison '18
GMAT 1: 690 Q47 V38
Posts: 93
Kudos: 46
 [1]
If the radius of the circle below (see attachment) is equal to the cho 07 Dec 2014, 00:36
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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.
User avatar
AlN
User avatar
Joined: 02 Jan 2017
Last visit: 19 Dec 2023
Posts: 53
Own Kudos:
7
Given Kudos: 23
Location: India
Schools: Oxford "21
Schools: Oxford "21
Posts: 53
Kudos: 7
If the radius of the circle below (see attachment) is equal to the cho 10 May 2019, 22:37
Kudos
Add Kudos
Bookmarks
Bookmark this Post
oss198
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 :)
Show more

Do u know more questions of inscribed angle? im having trouble with those
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,955
Own Kudos:
1,117
Posts: 38,955
Kudos: 1,117
If the radius of the circle below (see attachment) is equal to the cho 27 Dec 2022, 17:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109728 posts
Krunaal
Tuck School Moderator
853 posts