Geometry: Measurements of right triangle inscribed in a circle

So, I've got a triangle, ABC, inscribed in a circle--Thale's theorem states that it is therefore a right triangle. It is also given that $\overline{BA}$ is the diameter of the circle, and hence angle ACB is the right angle. That's all well and good.

My question is... What is the probabilty that the measure of angle CAB is less than/equal to 60 degrees?

My thinking: Since it's a right triangle, for CAB to be == 60 degrees, CBA would have to be == 30 degrees. This seems...perfectly plausible. But then what if CBA is 45 degrees...that's possible too... This is where I kind of ran into a dead end, because I couldn't figure out the probabilities of each scenario occurring. What am I missing?

• @amWhy I am given that AB is the diameter. – Roy Brewer Jun 15 '13 at 18:36

Let's say we have our triangle $\triangle ABC$ inscribed in a circle, with $\overline{BA}$ its diameter.

It is not too hard to show that the angle labeled $\theta$ in the diagram is equal to $2\angle CAB$. This is sometimes known as the central angle theorem.

This means that, assuming a "random" triangle is chosen by randomly choosing the point $C$ on the top half of the circle, we are randomly choosing an angle $\theta$ between $0^\circ$ and $180^\circ$, and wanting it to come up less than or equal to $120^\circ$. The probability of this would be $\frac{2}{3}$.

• cute graphic! ;-) Good ole' drawing by hand is still in style! +1 – Namaste Jun 15 '13 at 18:48
• Thanks! I am too rusty with TikZ to come up with it fast enough for math.SE answers anyway :) – Zev Chonoles Jun 15 '13 at 18:51

This is Bertrand's random chord paradox in another form. Once you throw in "probability," then some sort of randomization is going on. The question under consideration doesn't specify how such randomization is going to take place.

Consider the figure that Zev Chonoles has drawn. Let's label the origin as O. Now select a point uniformly along AB at random and draw a perpendicular from that point to intersect the semicircle. Now the probability that the angle $\theta$ is greater than $60^\circ$ is exactly $\dfrac{3}{4}$.

Further consider the +y intercept to be Y. Now select a point uniformly along OY at random and draw a perpendicular to intersect the circle in the first quadrant of the graph. Now the probability that the angle $\theta$ is greater than $60^\circ$ is: $$\text{P(}\theta\text{>60)} = 1 - \text{sin}(60) \approx 0.86603$$