geometry

[vgmat_24]

In the given figure, points P and Q lies on the circle with center O. Figure attached

What is the value of S?

a)1
b)1/2
c)sqrt(2)
d)sqrt(3)
e)sqrt(2)/2

[eravi11]

B..

First determine the angle made by OP with x-axis. The angle is 30 degree.
so angle made by OQ with x-axis is 60 degree.
so s/OQ=cos60=1/2
since OQ=1
s=1/2

[touche]

Originally Posted by eravi11

B..

First determine the angle made by OP with x-axis. The angle is 30 degree.
so angle made by OQ with x-axis is 60 degree.
so s/OQ=cos60=1/2
since OQ=1
s=1/2

All fine till finding OQ

Finding OQ....
sq rt3^2+1^2=OQ^2
OR 3+1=OQ^2
OQ=2

cos 60=1/2=s/2(2 being radius)
s=1
A

[UttamA]

Good Explanation touche

Answer is A

[eravi11]

oops..OQ=2 phbbt..seems i was sleeping..i took OQ=1

[clock60]

one more way to solve
point P (3^1/2,1)
tg(P)=1/(3^1/2)=(3^1/2)/3=pi/6=30
30+90=120, 180-120=60
tg(Q)=tg(60)=tg(pi/3)=(3^1/2)/1
(3^1/2)/1=t/s
s=1

[mitzi]

IMO A.

it is a simple quadrant question. It has nothing to do with geometry.

point P(- sqrt 3, 1), point Q is (1, sqrt 3)

so answer is A.

[mitzi]

Originally Posted by eravi11

B..

First determine the angle made by OP with x-axis. The angle is 30 degree.
so angle made by OQ with x-axis is 60 degree.
so s/OQ=cos60=1/2
since OQ=1
s=1/2

eravi11,

how did you get the angle 30? it should be 45. isn't it?

[touche]

Originally Posted by mitzi

IMO A.

it is a simple quadrant question. It has nothing to do with geometry.

point P(- sqrt 3, 1), point Q is (1, sqrt 3)

so answer is A.

You are absolutely right!
No calculations needed in this case, where ever point P(X,Y) is, Q will be (y,x)!

[eravi11]

Originally Posted by mitzi

eravi11,

how did you get the angle 30? it should be 45. isn't it?

No mitzi..angle will be 30. see attched..but i liked thy ingenuity in solving the pblm..I made it unneccessarily complicated..thanks!!