Last visit was: 04 Nov 2024, 13:35 It is currently 04 Nov 2024, 13:35

Close

GRE Prep 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 GRE 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.

Close

Request Expert Reply

Confirm Cancel
Manager
Manager
Joined: 09 Nov 2018
Posts: 88
Own Kudos [?]: 95 [0]
Given Kudos: 0
Send PM
User avatar
Senior Manager
Senior Manager
Joined: 10 Feb 2020
Posts: 496
Own Kudos [?]: 346 [0]
Given Kudos: 299
Send PM
User avatar
Senior Manager
Senior Manager
Joined: 10 Feb 2020
Posts: 496
Own Kudos [?]: 346 [0]
Given Kudos: 299
Send PM
Manager
Manager
Joined: 09 Nov 2018
Posts: 88
Own Kudos [?]: 95 [0]
Given Kudos: 0
Send PM
Re: A chord AB of a circle subtends an angle of 120° at the cent [#permalink]
1
Farina wrote:
I solved it by drawing perpendicular in between and it became 30-60-90 triangle.
x : x* sqrt3: 2x

we know that x=2
if i compare 2x=2, i get x=1 and I get correct answer of area of triangle
if i compare x* sqrt3 = 2, i get x=2/sqrt3, which is incorrect

Can anyone tell me, do we always have to compare value of any side with hypotenuse? i.e. 2x here


Relation between angle and side length is always fixed ie. 30-60-90 has the ratio of the side as x : \(\sqrt{3}\) * x: 2x

That also means, opposite to 30 the side length is x;
Opposite to 60 side length is \(\sqrt{3}\) * x
And opposite to 90, it is 2x

This is always fixed!
User avatar
Senior Manager
Senior Manager
Joined: 10 Feb 2020
Posts: 496
Own Kudos [?]: 346 [0]
Given Kudos: 299
Send PM
Re: A chord AB of a circle subtends an angle of 120° at the cent [#permalink]
kapil1 wrote:
Farina wrote:
I solved it by drawing perpendicular in between and it became 30-60-90 triangle.
x : x* sqrt3: 2x

we know that x=2
if i compare 2x=2, i get x=1 and I get correct answer of area of triangle
if i compare x* sqrt3 = 2, i get x=2/sqrt3, which is incorrect

Can anyone tell me, do we always have to compare value of any side with hypotenuse? i.e. 2x here


Relation between angle and side length is always fixed ie. 30-60-90 has the ratio of the side as x : \(\sqrt{3}\) * x: 2x

That also means, opposite to 30 the side length is x;
Opposite to 60 side length is \(\sqrt{3}\) * x
And opposite to 90, it is 2x

This is always fixed!


Ok so that is why we are comparing 2x=2 because side 2 is opposite to 90
if it were opposite to 60 then we would have compared with x*sqrt3
Verbal Expert
Joined: 18 Apr 2015
Posts: 29887
Own Kudos [?]: 36105 [0]
Given Kudos: 25917
Send PM
Re: A chord AB of a circle subtends an angle of 120° at the cent [#permalink]
1
Expert Reply
Prep Club for GRE Bot
Re: A chord AB of a circle subtends an angle of 120° at the cent [#permalink]
Moderators:
GRE Instructor
77 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
228 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne