Fermats last cocktail isaac physicsWatch
this is the question:
Vervet monkeys of the West Indies have recently developed a taste for sugary cocktails. Despite this, Pierre has decided to leave his half full beverage at point A on the beach while going for a quick swim. The shoreline is very long and straight and the water is perfectly still. Pierre is at point B in the water when he spots a monkey on land looking suspiciously interested in his tasty drink. He might just have enough time to save the cocktail from the mischievous animal, but only if he takes the optimal path – that is, the path of least time – through water and land all the way back to A.
The optimal path will consist of two straight line segments
AC and CB, with C a point on the shoreline. Because this path minimizes the journey time, if C’ is another point on the shoreline which is very close to C, then the alternative path consisting of straight line segments AC’ and CB’ must give approximately the same journey time.
The ratio of Pierre's running speed over his swimming speed is r. By comparing the length of AC with the length of AC’ and similarly for BC and BC’, derive an equation for r relating the acute angles ϕ1 and ϕ2 which AC and CB make with the shoreline, respectively. (Hint: Because C and C’ are very close, the segments AC’ and C’B can be treated as being approximately parallel to AC and CB respectively.
any help would be greatly appreciated
so you might think about r being sin (90°- ϕ1) / sin (90°- ϕ2)
but isaac will not accept that... so you need to think about a different way of writing sin (90°-x)