In the figure above, points P and Q lie on the circle with center O. What is the value of s?

A. \frac{1}{2}

B. 1

C. \sqrt{2}

D. \sqrt{3}

E. \frac{\sqrt{2}}{2}
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Not sure if this is simpler or even different approach from the others(I maybe just formulizing qpoo's approach!)u.....

Lets say slope of OP =m1 and OQ =m2
Since OP perpendicular to OQ => m1m2=-1
Since m1= -1/sqrt(3)
=>m2=sqrt3
=>s=1,t=sqrt(3)

I know I write like this [the colored] will raise the disscussion.
In general
slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1
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The answer is B.

Lets see how:
First draw a perpendicular from the x-axis to the point P. Lets call the point on the x-axis N. Now we have a right angle triangle \triangle NOP. We are given the co-ordinates of P as (-\sqrt{3}, 1). i.e. ON=\sqrt{3} and PN=1. Also we know angle PNO=90\textdegree. If you notice this is a 30\textdegree-60\textdegree-90\textdegreetriangle with sides 1-sqrt3-2 and angle NOP=30\textdegree.

Similarly draw another perpendicular from the x-axis to point Q. Lets call the point on the x-axis R. Now we have a right angle triangle \triangle QOR. We know angle QRO=90\textdegree.

Now
angle NOP + angle POQ (= 90 -- given) + angle QOR = 180

\Rightarrow 30\textdegree+90\textdegree+angleQOR=180

\Rightarrow angleQOR=60

If you notice \triangle QOR is also a 30\textdegree-60\textdegree-90\textdegreetriangle with sides 1-sqrt3-2 and angle OQR=30\textdegree

The side opposite to this angle is OR = 1.

Hence answer is B.

There's a simple reason and misconception as to why people find this question confusing:
--> the 90 degree angle between P and Q seems like it is cut perfectly right down the centre >> to make two 45 degree angles.
--> that's why your brain instinctively thinks the answer is r = sq(3), since the distance seems the same from the left and to the right of the origin!

However, that's the trick to this question and that's why it is CRUCIAL that you train yourself to be highly critical and weary of accepting the simple answer!

Upon closer inspection of the question:

P(-sq(3),1)
Which means the triangle beneath point P has:
height = 1
base = sq(3)
meaning >> radius = 2 (30-60-90 triangle)

This follows the ideal case of the 30-60-90 triangle, which means that opposite to the height of 1, there is an angle of 30 degrees from the x-axis to point P. This means that from the y-axis to point P, there is an angle of 60 degrees (to make a 90 degrees total from x-axis to y-axis). Therefore on the right side, since the figure shows there is a 90 degree angle between point P and Q, 60 degrees have been taken up from the left of the y-axis, which leaves 30 degrees on the right of the y-axis. Furthermore, this leaves 60 degrees from Point Q to the positive x-axis. Therefore, making the exact same 30-60-90 triangle once again, since the radius is the same (the two points lie on the semi-circle). The 60 degree angle is now where the 30 degree angle was on the other triangle, which makes the height now sq(3) and base now 1 >> therefore, base r = 1!

**Have a look at the attachment and then it should make perfect sense!

The method used to find the length of OP or OQ from origin i.e. the distance formula has been used here as well to find the length of PQ with P as (-[square_root]3,1) and Q as (s,t) which gives us
{s - (-[square_root]3)}^2 + {t-1}^2 = PQ^2 = 8 ->equation 2

from Right angle Triangle POQ we had got the length of PQ^2 as 8

Correct me if I'm wrong, (and forgive my english)
the coordinates of a point p(x,y) that lays on a circle can be expressed as p(sen,cos).
we know that sen π/6 = cos π/3,
and that sen π/3 = cos π/6

in this case I just have to invert x with y and y with x -> indeed we have p(y,x)

As we know x and y have to be positive we have to change the sign for x, the new coordinates are p (y,-x).

in this way to solve the problem doesn't take more than 20 secs.
hope I was enough clear.

I solved it in the foll manner:

X is point 1 on the y axis

POQ = 90 deg - given
PX = rt 3 - given
XO = 1 - given

Hence PO = 2 (30, 60, 90 triangle)
PO is radius. So PO = OQ = RADIUS = 2

POQ is 90 deg
90 deg angle rule - a, a, rt2 a
PO = 2
QO = 2
Hence PQ = 2 rt2

XQ = 2 rt2 - rt3
approx - 2.8 - 1.7 = 1

I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.

I understood bunuel's solution ok but i have a problem with the other one.

I got every step of maths but how did you get (t-1) too. However How did you decide that "t" is bigger than 1. Because it can be less then one then it should became (1-t). Can anyone explain it plz just that part??