Good question!
Here is How I solved:
Used distance formula for number line and Pythagorean theorem
OP^2=(3-0)^2+ (1-0)^2=3+1=4 -----(1)
OQ^2=(s-0)^2 + (t-0)^2= s^2+t^2 ----(2)
Since op=oq=radius of the circle
Thus, 4=s^2+t^2 -------(3)
Now apply Pythagorean theorem,
OP^2+OQ^2=PQ^2 -------(4)
Now apply distance formula,
PQ^2=[s-(-rt(3))]^2 + (t-1)^2= [s+rt(3)]^2 + (t-1)^2 ----(5)
Equate (4) & (5), you will can cancel elements on both sides and will finally be left with
t=s*rt(3). Subsitute value of 't' in equation (3)
It will yield 4=4s^2 => s^2=1 => s=1
Ans is B WHAT IS OA?
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another tough PS problem
Source: Beat The GMAT — Problem Solving |
stop@800
can u pl explain further on ur comment "use the logic that product of slope of two perpendicular lines is -1. "
can u pl explain further on ur comment "use the logic that product of slope of two perpendicular lines is -1. "
Here you go....aditi_bc wrote:stop@800
can u pl explain further on ur comment "use the logic that product of slope of two perpendicular lines is -1. "
Product of slope of two perpendicular line is -1
so
you prd of
-1/sqrt(3) and t/s shall be -1
hence we get
t = sqrt(3) * s
now po = qo
as po = 2 (distance formula)
so
qo = 2
we know
s^2 + t^2 = qo^2 (distance formula)
s^2 + t^2 = 4
hence we get
4 s^2 = 4
s = 1 will be your answer [we can ignore -ve value here]
Hope it helps!!
The other easier way to solve this will be to use trignometery, where 30:60:90 triangle are in the ratio of 1:2:3 sq.rt
If you use this you will get the angle with given co-ordinate as 30
==> angle QO as 60 hence the value is "1"
If you use this you will get the angle with given co-ordinate as 30
==> angle QO as 60 hence the value is "1"
