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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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!

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 \(30\textdegree-60\textdegree-90\textdegree\)triangle 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 \(30\textdegree-60\textdegree-90\textdegree\)triangle with sides \(1-sqrt3-2\) and angle \(OQR=30\textdegree\)

The side opposite to this angle is OR = 1.

Hence answer is B.

Since OP and OQ are the radii of the circle, and we know P has co-ordinates (-\sqrt{3}, 1), from the distance formula we have OP = OQ = 2. Therefore, s^2+t^2=4. -> (1)
Also from the right triangle, we know the length of PQ=2\sqrt{2}.
Again, using the distance formula, (s+\sqrt{3})^2 + (t-1)^2 = PQ^2 = 8. ->(2)

From (1) and (2) you can solve for s.
Hope this helps.

Line OP => y=mx => 1 = \(-sqrt{3} * m\)

=> slope of OP = \(\frac{-1}{sqrt{3}}\)

Since OP is perpendicular to OQ => m1*m2 = -1
=> slope of OQ = \(sqrt{3}\)

equation of line OQ => y = \(sqrt{3}x\)

since s and t lie on this line t= \(sqrt{3}s\)

also radius of circle = \(sqrt{ 1^2+ 3}\) = 2 = \(sqrt{ s^2+ t^2}\)

=> \(4 = s^2+ t^2 = s^2 + 3s^2 = = 4s^2\)
=> s=1
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Using Similar triangles.
In triangle APO , angle APO + AOP = 90 ---------------------1
In triangle BOQ , angle BOQ + BQO= 90 ---------------------2

Also SINCE angle POQ = 90 as given.

angle AOP + angle BOQ = 90 ------------------3

Using equations 1 and 3 we get Angle APO = Angle BQO -----------4

using equation 2 ,3, 4
we get Angle APO = Angle BQO and AOP = OQB

Since hypotenuse is common => both the triangle are congruent

=> base of triangle APO = height of triangle AQO and vise versa.

Thus s = y co-ordinate of P = 1
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The method I suggested above is very intuitive and useful. I have been thinking of how to present it so that it doesn't look ominous. Let me try to present it in another way:

Imagine a clock face. Think of the minute hand on 10. The length of the minute hand is 2 cm. Its distance from the vertical and horizontal axis is shown in the diagram below. Let's say the minute hand moved to 1. Can you say something about the length of the dotted black and blue lines?

Isn't it apparent that when the minute hand moved by 90 degrees, the green line became the black line and the red line became the blue line. So can I say that the black line \(= \sqrt{3}\) cm and blue line = 1 cm

So if I imagine xy axis lying at the center of the clock, isn't it exactly like your question? The x and y co-ordinates just got inter changed. Of course, according to the quadrants, the signs of x and y coordinates will vary. If you did get it, tell me what are the x and y coordinates when the minute hand goes to 4 and when it goes to 7?
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Note that this is not a reflection problem. The angle POQ is given as = 90, but angle POY is not the same as angle QOY.

To solve, set the product of the slopes of OP and OQ as -1
=> -1/sqrt(3) * t/s =-1
=> t = s*sqrt(3)

Also OQ=OP
=> s^2 + t^2 = 4
=> s^2 + 3*s^2 = 4
=> s^2 = 1
=> s = 1

Option (B)
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Can we assume that coordinates of the center O are (0,0) here, can we always assume so?
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