someone has done it in the following way (i had also posted on yahoo)

 

Related rates problem. This is going to be tricky without using diagrams, but I'll give it a shot, and hope it is clear to you. Draw a (right) triangle; Horiz leg is 'x' Vert leg is 'y' Hypotenuse is 'r' The angle is on the left, the vertical side is on the right. Let the angle be '?'. As you know: x² + y² = r² Will come back to that. You need a relation involving the angle. Let's try: x = rCos(?) Take the derivative with respect to time: dx/dt = (dr/dt)Cos(?) - rSin(?)(d?/dt) ~~~~~ Eq'n (1) From the problem we're given info....let's identify it. Horizontal rate.....(dx/dt) = 5 m/s Horizontal distance 'x' = 120 m Vertical distance 'y' = 60 m We can figure out the hypotenuse by Pythagorean Thrm. I'll let you actually do it, but 'r' = 300 m So...now what? Well, we don't know what dr/dt is in eq'n (1). Well, we know that dx/dt = 5 m/s, and we can use the following: In the same proportions: (dy/dt)/(dx/dt) = y/x ==> dy/dt = (y/x)(dx/dt) = (60 m/120 m)(5 m/s) = 5/2 m/s Now we can use is the fact that since x² + y² = r² Then (dx/dt)² + (dy/dt)² = (dr/dt)² The velocities must also be held to this relationship because it involves the same variables (in essence). So, dr/dt = ?[(dx/dt)² + (dy/dt)²] = (5/2)?5 m/s Now, we can plug this stuff into our eq'n (1) ==> dx/dt = (dr/dt)Cos(?) - rSin(?)(d?/dt) (you want the rate of change of the angle with respect to time) ==> (d?/dt) = [(dr/dt)Cos(?) - dx/dt]/[rCos(?)] Plug your numbers in and you should see that: (d?/dt) ? 0.118°/s