Please help - Geometry

Question

I know this one has been posted before, but I still don't 'get it. Can someone provide some step by step approach to this problem. I really really appreciate it. Thanks

maratikus · Answer

first, the radius of the circle is sqrt(3+1)= 2
let's define angle alpha as the angle QOT = the angle between QO and horizontal axis T.
Then s = r*cos(alpha) = 2*cos(alpha)= 2*cos(-alpha) = 2*sin(90%+alpha) = 2*sin(POT) = 2*(1/2)=1 -> B

is that specific enough?

iamcartic · Answer

Since the lines are perpendicular: Product of slopes =-1

viz. -1/√3 X t/s = -1 or t/s= √3

since both are radius: r2 + s2 = 3+1=4 hence by substitution s=1 and t=√3

HtH

maratikus · Answer

That's a good solution too. I was considering putting it but figured solving a quadratic equation is more difficult than playing with sin and cos.

alimad · Answer

call me stupid. but I still dont' get it.

all i'm seeing is a right triangle with two sides of 2 , 2 and third side of PQ.

using so PQ = 2 root (2).

we need to know what is oq correct?
so we subtract -root(3) from 2 root(2) to get oq.

This is the best understanding I have so far. Please be patient. Thanks

maratikus · Answer

QO = PO = radius of the circle. QO = sqrt((sqrt(3))^2 + 1^2)=sqrt(3+1)=2

alimad · Answer

maratikus, i've gotten this far that PO = QO

so we have right angle with 2 : 2 sides. correct.

using the pyth theorem c^2 = a ^2 + b ^2 = c^2 = 4 + 4 = root(8)
 c = 2root2.

am i on the right track?

jallenmorris · Answer

My approach to this question was completely different.

If you have 2 lines on an x-y plane that are perpendicular (90 degrees difference), then their slope is going to be inverted and the opposite sign.

In your diagram, you have Point P @  \{\sqrt{3},1\} . Since the center is at {0,0}, this makes finding the slope of the segment OP much easier.

The slope is change in y over change in x. from P to O, y changes -1, and x changes  +\sqrt{3} . So you have  \frac{-1}{\sqrt{3}  so to get the slope of a perpendicular line, you must change the sign and invert the fraction.

New slope is  \frac{\sqrt{3}}{1} . Since it's change in Y over change in x, you know the new coordinates should be  \{1,\sqrt{3}\}  and s = 1.

maratikus · Answer

No, you are not on the right track. You don't need to worry about PQ. You should calculate s. 
One way of doing that: s = QO*cos(alpha) and calculate alpha using point P and relationship of angles POT and QOT. 
Another way is to calculate tangent (slope) of alpha using the rule of products (= -1 for right angles) (t/s)*(-sqrt(3)/1) = -1 -> t = sqrt(3)*s and length of OQ = PO = 2 -> s^2+t^2 = 2^2 -> s = 1

Please help - Geometry

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