Find Probability that length of Randomly chosen chord of a circle lies between $\frac{2}{3}$ and $\frac{5}{6}$ of its diameter.
My try: I assumed unit circle with center origin. Let two randomly chosen distinct points be $A(\cos \alpha, \sin \alpha)$ and $B(\cos \beta, \sin \beta)$
Length of the chord is $$p=2\sin \left(\frac{\alpha-\beta}{2}\right)$$
Now we have to find Probability that
$$\frac{4}{3} \le 2\sin \left(\frac{\alpha-\beta}{2}\right) \le \frac{5}{3}$$
can i have any clue here?
Best Answer
HINT:-
Take a circle, let its radius be $r$.
Take any point in the circle. It doesn't matter what point you take for the first point.
Now as soon as you choose the second point, you have a chord. Let the two points of the chord make angle $2\theta$ at centre. So the length of the chord is $2\cdot r\cdot \sin\theta$.
Now the given constraint is $\frac{2}{3}\le2\sin\theta \le \frac{5}{6}$.
Now this can be solved easily.