# Thread: cant understand this solution in certain points..

1. ## cant understand this solution in certain points..

on a circle with radius 1 there are two points one is static the other is
distributed evenly on the circle (i dont know the proper english term)
we sign X to be the hypotenuse on the circle

find the density function of the variable X
?

solution(i partialy understood it)the questions come after the photo

they say that we take x to be an angle from o til pi/2
we scan only the northern hemisphere because its symmetric.

what is the curv graph represents
what is t?
what is the meaning of y<=t?
why if they say that x should go from o til pi/2?
in the probability function they put x fro 0 till arcsin t/2 where t goes from
0<t<=2
?

after this step they did the derivative to find the density function
i understand that step

2. Originally Posted by transgalactic
on a circle with radius 1 there are two points one is static the other is
distributed evenly on the circle (i dont know the proper english term)
we sign X to be the hypotenuse on the circle

find the density function of the variable X
?

solution(i partialy understood it)the questions come after the photo

they say that we take x to be an angle from o til pi/2
we scan only the northern hemisphere because its symmetric.

what is the curv graph represents the length of the side "y" as the angle x varies from 0 to 90 degrees
what is t? the various values of y fom 0 to 2
what is the meaning of y<=t? y ranges from 0 to "t" for a specific "t"
why if they say that x should go from o til pi/2? x is the angle BCA
in the probability function they put x fro 0 till arcsin t/2 where t goes from
0<t<=2
? Since the triangle ABC is always right-angled at B, $Sinx=\frac{y}{2},\ y=t,\ Sinx=\frac{t}{2},\ x=Sin^{-1}\frac{t}{2}$

after this step they did the derivative to find the density function
i understand that step
Attached is a diagram to help

3. its not sin its cosine

AC*cos(X)=AB
2*cos(x)=y

4. If you place the angle x at "A", then yes, that's what it will be.
I placed it at "B" to correspond with your work,
(as x was in the wrong position according to the worked solution).

5. look at my sketch there is 90 degree angle on point B

6. Yes, the angle at the circumference for a triangle whose hypotenuse is the diameter of a circle is 90 degrees.

Do you want to work with the angle at A or C ?
and the side opposite A or adjacent A

7. i want to work with angle A
like in my sketch which is signed by x

8. Ok, well it doesn't really matter, because the situation is symmetrical.
You'd get the same answers from the other end at B, once you use Sine and Cosine correctly.

So, from your sketch $Cosx=\frac{y}{2}$

$y=t=2Cosx$

This gives rise to $Cosx=\frac{y}{2}=\frac{t}{2}$

Then $x=arcCos\frac{t}{2},\ or\ Cos^{-1}\frac{t}{2}$