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Re: GRE Math Challenge #16-O is the center of the circle [#permalink]
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Vijay0808 wrote:
How can the triangle above can be concluded as a 30-60-90 triangle since the figure is not drawn to scale? Can anybody help?



Hey,

Sorry for the poor quality of the figure. Your point is very correct. OPQ is a right angle triangle. I have added that to the question description.

Thanks for pointing that out.

Regards,
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Re: GRE Math Challenge #16-O is the center of the circle [#permalink]
sandy wrote:
Vijay0808 wrote:
How can the triangle above can be concluded as a 30-60-90 triangle since the figure is not drawn to scale? Can anybody help?



Hey,

Sorry for the poor quality of the figure. Your point is very correct. OPQ is a right angle triangle. I have added that to the question description.

Thanks for pointing that out.

Regards,


Hi,

Sorry for my ignorance, but even if that is a right angled triangle how you would make out that the other two angles are 30 and 60?
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Re: GRE Math Challenge #16-O is the center of the circle [#permalink]
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We have a right triangle OPQ right angled at Q.

PQ= 6 or OP= 12

Sin (angle POQ)= \(\frac{PQ}{OP}\). or \(\frac{6}{12}\) or \(\frac{1}{2}\).


Hence angle POQ = 30 degree.


So Quantity A is 30; and Quantity B is 45. Thus Quantity B is greater.
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Re: GRE Math Challenge #16-O is the center of the circle [#permalink]
OP is radius of circle how can you use this in sine angle calculation

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GRE Math Challenge #16-O is the center of the circle [#permalink]
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MariaMughees wrote:
OP is radius of circle how can you use this in sine angle calculation

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Hi, here given, PQ = 6, \(OP=OR= 12\) (being radius of the circle), On using Pythagoras theorem, we get OQ = 10. So, side of the triangle are on \(6:10:12\) i.e \(1:\sqrt{3}:2\), which is clear indication of being 30:60:90 triangle.

ask still doubt remain!
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Re: GRE Math Challenge #16-O is the center of the circle [#permalink]
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Re: GRE Math Challenge #16-O is the center of the circle [#permalink]
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