In the figure above, points P and Q lie on the circle with

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Re: Coordinate System and Geometry [#permalink]

20 Nov 2012, 00:55

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agarboy wrote:

Can someone explain how to do this problem.. I have no clue how to even start on this.

In the attached picture, from question OA = \sqrt{3}
PA =1
Thus radius, OP = 2

Thus OPA is a 30, 60, 90 triangle. where angle POA is 30.
Also angle POQ is 90, given in question.

therefore, angle QOB = 60
And hence, angle OQB =30

trianlge QOB is also 30, 60, 90 triangle. in which we know OQ =OP =2 (radius)
Therefore BQ = \sqrt{3}
and OB = 1

We need value of r which is BQ = \sqrt{3}

Ans D it is!

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Re: Coordinate System and Geometry [#permalink]

20 Nov 2012, 00:59

Vips0000 wrote:

agarboy wrote:

Can someone explain how to do this problem.. I have no clue how to even start on this.

In the attached picture, from question OA = \sqrt{3}
PA =1
Thus OA = 2

Thus OPA is a 30, 60, 90 triangle. where angle POA is 30.
Also angle POQ is 90, given in question.

therefore, angle QOB = 60
And hence, angle OQB =30

trianlge QOB is also 30, 60, 90 triangle. in which we know OQ =OP =2
Therefore BQ = \sqrt{3}
and OB = 1

We need value of r which is BQ = \sqrt{3}

Can you please explain the red part. It should OP i sppose.
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Re: Coordinate System and Geometry [#permalink]

20 Nov 2012, 01:02

good job vips.
Thanks for the explaination
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Re: Coordinate System and Geometry [#permalink]

20 Nov 2012, 03:25

agarboy wrote:

Can someone explain how to do this problem.. I have no clue how to even start on this.

Merging similar topics.
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Re: Coordinate System and Geometry [#permalink] 20 Nov 2012, 03:25

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In the figure above, points P and Q lie on the circle with

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In the figure above, points P and Q lie on the circle with

In the figure above, points P and Q lie on the circle with

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