Q:

What is the minimum time to get from X to Q on a circle by running, swimming, or both?

I am having trouble with this intro calculus question... The question is:
You are at point P on one side of a large artificial circular pool 60 m across and you wish to get to the point Q on the exact opposite side of the pool. To do this you may run around the perimeter, swim directly across, or swim at an angle A (from line PQ) to a point X on the circumference and then run the rest of the way. Among all possible choices, which will get you to Q in the shortest possible time, assuming you can run at 4 m/sec and swim at 2 m/sec? What is that minimum time?
Note that the central angle between X and Q is 2A.
What I did so far:
For the swimming + running time, I found the chord length between P and X to be 20sinA. (since chord length is 2rsin(A/2) = 60sin(2A/2) = 60sinA). But the problem doesn't even tell me whether the angle is in radians or degrees so I am unsure which chord formula I should use, so I assumed A is in radians...
For arc length (X to Q), it's rA = 60(2A) = 120A, right? I am really lost which formulas to use! And I always end up getting a weird formula with the terms sinA - A or something which I don't know how to find critical values with.

A:

Let us use the following picture to do our calculations:
Part $1$, Just Running: Note that as the radius of the circle is $30$ m, then the circumference of the semicircle equals $30\pi$, which is the distance we require. As the running speed is $4$ m/sec, the time taken is:$$T_{\text{ Just Running }}=\frac{30\pi}{4}$$
Part $2$, Running and Swimming: Note that if the swimmer swims at an angle $\theta$ to the diameter $PQ$, then she will surely land at a point away from $Q$. Name that $X$. Now as angle $XPQ$ is $\theta$, and the angle subtended by the arc $XQ$ to the centre (say, $O$) is $2\theta$, we can conclude that angle $OXP =\theta$. Then, using sine rule, and calculating, we get, $$\frac{\sin \theta}{30}=\frac{\sin(\pi-2\theta)}{PX} \implies PX=30\cos \theta$$ Also, the distance along the arc $XQ$ can be calculated  as $\frac{2\pi (30)}{2\pi}(2\theta) =60\theta$ where $\theta$ is expressed in radians. Using the relevant speeds, we get, $$T_{\text{ Swimming + Running }}=\frac{30\cos \theta}{4}+\frac{60\theta}{2}$$
Hope you can take it from here.

