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tutonaranjo wrote:No luck solving this. Please help.

Options are:
a) 1/2
b) 1
c) sqrt 2
d) sqrt 3
e) sqrt 2 / 2

B is correct answer.

(s,t) will be (1, sqrt 3), radius is 2 .
Angle O is 60 degress on left and 30 degrees on right since sin 360=sqrt 3/2

Got it.... 30:60:90 triangles... should've spotted that when seing 1, sqrt 3
thanks

In the fig. side OP = OQ, then shouldn't angle OPQ = OQP = 45 degree ?

I think another way to look at it is this.

The 2 lines are perpendicular to each other, so their slopes are the negative inverse to each other. Also since they are both the radius of the semi-circle, they are both equidistant to the origin.

Hence, s,t has the coordinate of (1,sq 3).

akram wrote:

tutonaranjo wrote:No luck solving this. Please help.

Options are:
a) 1/2
b) 1
c) sqrt 2
d) sqrt 3
e) sqrt 2 / 2

B is correct answer.

(s,t) will be (1, sqrt 3), radius is 2 .
Angle O is 60 degress on left and 30 degrees on right since sin 360=sqrt 3/2

akram wrote:

tutonaranjo wrote:No luck solving this. Please help.

Options are:
a) 1/2
b) 1
c) sqrt 2
d) sqrt 3
e) sqrt 2 / 2

B is correct answer.

(s,t) will be (1, sqrt 3), radius is 2 .
Angle O is 60 degress on left and 30 degrees on right since sin 360=sqrt 3/2

B is correct, but akram's reasoning is not. I have no idea where he comes up with sin 360 = sqrt 3/2

It is a 30-60-90 triangle, and Angle O is 30 degrees on the left and 60 on the right (not what akram said). This may clear things up:

https://www.themathpage.com/aTrig/30-60-90-triangle.htm

tutonaranjo can you pls confirm the right answer, I keep getting it as sq root of 3. I can explain how if this is correct. If this is wrong I need to understand where i went wrong.

thanks

Perpendicular Property:
If the angle b/w 2 points P (x1,y1) and Q (x2,y2) is 90 i.e right angle then
then the product of their slopes is -1 i.e y1/x1 * y2/x2 = -1 .
y2/x2 = -1 x1/y1;
= -1 -RT(3)/1
y2/x2 =RT(3)/1 =====> Q(1,RT(3))

Easy to Remember:
if Point P (a,b) then it's Perpendicular point is (-b,a) :
in this case : P (-RT(3),1 ) , Q(s,t) (1,RT(3) )

thanks a lot krishnamurthyu for your reply and explaination

The rule that krishnamurthyu mentioned is correct but the way to remember it could be misleading.

In the case of this problem if point P coordinates are (a,b), then point Q coordinates are (b,-a) NOT (-b,a).

It really depends on the quadrant where point Q lies. If it was in QIII then the coordinates would have been (-b,a).

Hi,

I found that while the diagrams are generally helpful and accurate on the GMAT, in this particular case, the diagram was misleading.

I was stuck on D (root 3) for an answer until I saw from one of the previous posts, using the 30-60-90 rule, that the "right half" of angle POQ MUST be be 30 degrees.

From this I quickly realized that while the diagram make it look like the right angle at POQ is bisected into two 45 degree angles, that is NOT the case. I think root 3 is the trap answer for those of us who mistakenly thought PQ is parallel to the x-axis, and that t equals 1. Really, the only thing to be had from P is that the radius of the circle (side OP) is 2.

Using the 30-60-90 rule, we get s = 1.

Tricky.

i think koushik is absolutely write


In the fig. side OP = OQ=2...

since 2 sides are equal it is an isoceles triangle
then angle OPQ = OQP = 45 degree ...

since we know length of 2 sides then length of PQ=?

by pythagaros theorem

PQ^2=OP^2+OQ^2
=4+4
=8
PQ=2sqrt(2)

there fore length of PQ=2sqrt(2)

Formula:
Distance between 2 points (x1,y1) and (x2,y2) is sqrt (x2-x1)^2+(y2-y1)^2....

then calculate the distance between O and Q

since O is origin theno(0,0)
Q(s,t)

after calculating the distance then
s^2+t^2=4----(1)

then the distance of P and Q

s^2+t^2+t+4+s sqrt(3)=8----(2)

substiting the 1 value in 2 we have

4+t+4+s sqrt(3)=8
t=- s sqrt(3) ---- (3)

substiting the value of 3 in 1 we have
s^2+3 s^2=4
4s^2=4
s=1

and substitute the value s=1 in eq 1 we have

t=sqrt(3)

krishnamurthyu wrote:Perpendicular Property:
If the angle b/w 2 points P (x1,y1) and Q (x2,y2) is 90 i.e right angle then
then the product of their slopes is -1 i.e y1/x1 * y2/x2 = -1 .
y2/x2 = -1 x1/y1;
= -1 -RT(3)/1
y2/x2 =RT(3)/1 =====> Q(1,RT(3))

Easy to Remember:
if Point P (a,b) then it's Perpendicular point is (-b,a) :
in this case : P (-RT(3),1 ) , Q(s,t) (1,RT(3) )

Krishanmurhyu, the slope of a line is Y2-Y1/x2-x1 , for a given line we have only one coordinates (i.e. P or Q), unless we are assuiming that O is (0,0).

Please explain?

Thanks,

Vineet