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A chord AB of a circle subtends an angle of 120° at the cent

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kapil1
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A chord AB of a circle subtends an angle of 120° at the cent [#permalink] New post   Updated on: 14 Jul 2020, 01:18
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A chord AB of a circle subtends an angle of 120° at the centre O. If the radius of the circle is 2. What is the area of the triangle AOB?​

A. 1​

B. √3/2​

C. √2​

D. √3​

E. Cannot be determined​

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Originally posted by kapil1 on 14 Jul 2020, 00:57.
Last edited by kapil1 on 14 Jul 2020, 01:18, edited 1 time in total.
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Re: A chord AB of a circle subtends an angle of 120° at the cent [#permalink] New post   14 Jul 2020, 01:14
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OA is missing
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Re: A chord AB of a circle subtends an angle of 120° at the cent [#permalink] New post   14 Jul 2020, 16:35
1
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
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Re: A chord AB of a circle subtends an angle of 120° at the cent [#permalink] New post   15 Jul 2020, 02:40
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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!
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Re: A chord AB of a circle subtends an angle of 120° at the cent [#permalink] New post   15 Jul 2020, 04:50
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
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Re: A chord AB of a circle subtends an angle of 120° at the cent [#permalink] New post   15 Jul 2020, 04:54
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Read this https://gre.myprepclub.com/forum/gre-quant ... tml#p53081

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