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Triangle in a semi circle - URGENTLY need help!

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rohit_gmat: Master | Next Rank: 500 Posts; Posts: 158; Joined: Wed Jun 16, 2010 8:21 amFollowed by: 1 member; Thanked: 13 times

by rohit_gmat » Wed Nov 17, 2010 4:13 am

Hi All,

Whats the right way to resolve this question?
Need steps along with answer please

Thanks!

PS OA will be released later

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Source: Beat The GMAT — Problem Solving | Wed Nov 17, 2010 4:13 am

kevincanspain: GMAT Instructor; Posts: 613; Joined: Thu Mar 22, 2007 6:17 amFollowed by: 64 members; Location: madrid; Thanked: 171 times; GMAT Score:790

by kevincanspain » Wed Nov 17, 2010 4:36 am

First of all, do you know some often tested right triangles?

Kevin Armstrong
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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Wed Nov 17, 2010 4:40 am

Radius of the circle = âˆš(sÂ² + tÂ²) = âˆš((-âˆš3)Â² + (1)Â²) = âˆš4 = 2

Straight lines OP and OQ are perpendicular.
Thus, (Slope of OP)*(Slope of OQ) = -1

Slope of OP = 1/(-âˆš3) = -(1/âˆš3)
Slope of OQ = t/s

Therefore, (t/s)*(-1/âˆš3) = -1 => (t/s) = âˆš3 => t = (âˆš3)*s

As âˆš(sÂ² + tÂ²) = 2
=> âˆš(sÂ² + ((âˆš3)*s)Â²) = 2
=> (sÂ² + 3sÂ²) = 4
=> 4sÂ² = 4
=> sÂ² = 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: Master | Next Rank: 500 Posts; Posts: 158; Joined: Wed Jun 16, 2010 8:21 amFollowed by: 1 member; Thanked: 13 times

by rohit_gmat » Wed Nov 17, 2010 4:43 am

kevincanspain wrote:First of all, do you know some often tested right triangles?

Oh !!!!!!!! Is it a 30:60:90 => 1 : 3(1/2) : 2 .... that im supposed to use here !??!?!?!

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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beat_gmat_09: Master | Next Rank: 500 Posts; Posts: 437; Joined: Sat Nov 22, 2008 5:06 amFollowed by: 1 member; Location: India; Thanked: 50 times; GMAT Score:580

by beat_gmat_09 » Wed Nov 17, 2010 4:45 am

rohit_gmat wrote:Hi All,

Whats the right way to resolve this question?
Need steps along with answer please

Thanks!

PS OA will be released later

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: Master | Next Rank: 500 Posts; Posts: 158; Joined: Wed Jun 16, 2010 8:21 amFollowed by: 1 member; Thanked: 13 times

by rohit_gmat » Wed Nov 17, 2010 4:52 am

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]

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Tani: Legendary Member; Posts: 1255; Joined: Fri Nov 07, 2008 2:08 pmFollowed by: 90 members; Location: St. Louis; Thanked: 312 times

by Tani » Wed Nov 17, 2010 9:08 am

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

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