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

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...

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.

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]

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.