Specifically, suppose $A$ and $B$ are points on the circle with center $C$. The angle $\angle{ACB}$ is $15^o$. If a point $D$ is randomly chosen on the circle, then what is the probability that the triangle $\triangle{ABD}$ is $obtuse$?
I found this problem in an old textbook. It says the answer is 23/24, but does not provide a solution and I am lost.
My first approach was drawing lines perpendicular to line $AB$ on points $A$ and $B$, as shown below
I know that if point $D$ lies in the outer regions, the triangle must be obtuse because $\angle{ADB}$ would be greater than $90^o$ since it is past the perpendiculars. After that, I'm not really sure where I should go. I know that there is a small region in the middle that consists of $\angle{C}$ being greater than $90^o$, but I'm having trouble picturing or working out how to calculate that region. It also doesn't seem like the area would calculate to 23/24, so I feel like I might have gone somewhere wrong with my diagram.
Best Answer
You are going in the right track. However, your picture might be causing a bit of confusion. The problem states that the point $D$ is randomly chosen $on$ the circle, not $in$ the circle. I think a better picture would be this:
You had mentioned that there is a small area for $\angle{C}$. The region that you are referring to is the orange part of the circle.
As shown in the picture, if you were to place the point $D$ anywhere on the red or orange sections of the circle you will get an obtuse triangle. So we are only left with the blue section of the circle.
Due to symmetry, the orange and blue sections have the same length. So, if we can find the length of the orange section then we can find the length of the blue section.
If we think of the circle as a unit circle, $ r = 1$ then the length of the orange arc is:
$$ s = r\theta$$ $$ s = \theta $$
Since the angle between $A$ and $B$ is $15^o$, the arc length in radians is $$15^o = \frac{\pi}{12}$$ So, the only part of the circle that does not satisfy the conditions has a length of $\frac{\pi}{12}$
So the probability of not getting an obtuse triangle is $$\frac{\frac{\pi}{12}}{2\pi} = \frac{1}{24}$$
Therefore, the probability of getting an obtuse triangle is $$ 1 - \frac{1}{24} = \frac{23}{24} $$