Hi All,
Whats the right way to resolve this question?
Need steps along with answer please
Thanks!
PS OA will be released later
Triangle in a semi circle - URGENTLY need help!
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- rohit_gmat
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- kevincanspain
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First of all, do you know some often tested right triangles?
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- Rahul@gurome
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Radius of the circle = √(s² + t²) = √((-√3)² + (1)²) = √4 = 2
Straight lines OP and OQ are perpendicular.
Thus, (Slope of OP)*(Slope of OQ) = -1
Slope of OP = 1/(-√3) = -(1/√3)
Slope of OQ = t/s
Therefore, (t/s)*(-1/√3) = -1 => (t/s) = √3 => t = (√3)*s
As √(s² + t²) = 2
=> √(s² + ((√3)*s)²) = 2
=> (s² + 3s²) = 4
=> 4s² = 4
=> s² = 1
=> s = 1
The correct answer is B.
Straight lines OP and OQ are perpendicular.
Thus, (Slope of OP)*(Slope of OQ) = -1
Slope of OP = 1/(-√3) = -(1/√3)
Slope of OQ = t/s
Therefore, (t/s)*(-1/√3) = -1 => (t/s) = √3 => t = (√3)*s
As √(s² + t²) = 2
=> √(s² + ((√3)*s)²) = 2
=> (s² + 3s²) = 4
=> 4s² = 4
=> s² = 1
=> s = 1
The correct answer is B.
Last edited by Rahul@gurome on Wed Nov 17, 2010 4:46 am, edited 1 time in total.
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- rohit_gmat
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Oh !!!!!!!! Is it a 30:60:90 => 1 : 3(1/2) : 2 .... that im supposed to use here !??!?!?!kevincanspain wrote:First of all, do you know some often tested right triangles?
I thought it was some other weird triangle in a circle rule that dictates that value of x in one vertice becomes value of y in the other...
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PO and OQ are perpendicular, get their equations, PO's slope can be found out, OQ's slope will be -1/PO's slope.
get equation of line OQ.
PO and OQ both are radius and are equal.
1) Get slope of PO diff y/diff x = -1/sqrt(3).
OQ's slope = sqrt(3)
2) Get equation of line OQ.
y-t/t = x-s/t
y = (t/s) x
t/s = sqrt(3)
t = sqrt(3)s
3) Radius = sqrt(sqrt(3)^2 + 1^2) = sqrt(4)
sqrt(s^2 + t^2) = sqrt(4)
substitute t = sqrt(s)
s^2 + (sqrt(3)s)^2 = 4
s^2 + 3*s^2 = 4
4 * s^2 = 4
s = +/-1
S is in I quadrant hence 1.
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- rohit_gmat
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Dear All,
Thanks for the efforts... OA is B
I guess the best way to do this would be to :
[spoiler] use the 30:60:90 rule of a triangle.... (where sides are 1:sqrt3:2) ... we can make a 90 deg triangle for P (base = sqrt 3 & height = 1) ... so we know that the radius (hyp) = 2...
Since the angle made by OP and x axis is 30 deg, & the arc PQ makes a 90 deg, we can figure out that OQ makes a 60 deg with the x-axis...
again using the same triangle rule... we know that the three angles for the triangle made on the right (with point Q) gives us 30:60:90... and the hyp (radius) is 2... so we know that the base must be 1 ( following 1:sqrt3:2) ... which means the value of x at point Q is also 1 .. i.e. s = 1 ..... choice B *phew*[/spoiler]
Thanks for the efforts... OA is B
I guess the best way to do this would be to :
[spoiler] use the 30:60:90 rule of a triangle.... (where sides are 1:sqrt3:2) ... we can make a 90 deg triangle for P (base = sqrt 3 & height = 1) ... so we know that the radius (hyp) = 2...
Since the angle made by OP and x axis is 30 deg, & the arc PQ makes a 90 deg, we can figure out that OQ makes a 60 deg with the x-axis...
again using the same triangle rule... we know that the three angles for the triangle made on the right (with point Q) gives us 30:60:90... and the hyp (radius) is 2... so we know that the base must be 1 ( following 1:sqrt3:2) ... which means the value of x at point Q is also 1 .. i.e. s = 1 ..... choice B *phew*[/spoiler]
- Tani
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The triangle in a circle rule you are thinking of is the one that says if you have a triangle inscribed in a circle and one leg of the triangle is the diameter of the circle then the angle opposite the diameter is 90 degrees.
Tani Wolff