A man at point A on the shore of a circular lake of radius 1mi wants to reach the opposite point C as soon as possible. He can walk 6mi/h and row his boat 3mi/h. At what angle theta to the diameter AC should he row?
The Attempt at a Solution
My diagram looks like this: I first drew the diameter of the circle AC with A being on the left. Next, I called the point to which the man should row B and I drew a line from A to B (B being somewhere between theta = 0 and theta = pi/2). Finally, I drew another line from O (the center of the lake) to B. Therefore, angle CAB = theta and angle COB = 2theta. Moreover, AC=2, so OB=1, AO=1, and OC=1.
My initial equation for the shortest total time that it will take the man to reach point C is:
T = AB/3 + 2theta/6
Now, I know that I need to calculate the derivative of T and set it equal to zero to minimize T. However, I'm not sure how to calculate AB.
By the law of cosines:
(AB)2=AO2 + OB2 -2ab(cosAOB), which equals:
(AB)2=12 + 12 - 2cos(AB) or 2-2cos(AOB)
This is where I get confused. I set cos(AOB) equal to cos(2theta - 2pi), but that isn't working out for me. The correct answer has AB = 2cos(theta). I've been working at this for a while and I don't understand how to get AB = 2cos(theta). I would greatly appreciate your help.