You were on the correct track with your setup on number 8, but when you took the derivative of the #y/z# term on the right side with respect to time #t#, you did not use the Quotient Rule to perform the derivative, which rendered the rest of the problem incorrect.
The proper setup is #sin theta = y/z# as you noted. From there:
#(cos theta) (d theta)/(dt) = (z*(dy)/(dt) - y*(dz)/(dt))/(z^2)#
It is difficult to read the image included, but I believe the setup involves #y = 8#, #z = 10#, #(dy)/(dt) = 10#, #(dz)/(dt) = -6#, and #cos theta = 8/10#, which means you now have:
#(cos theta) (d theta)/(dt) = (z*(dy)/(dt) - y*(dz)/(dt))/(z^2)#
#(8/10) (d theta)/(dt) = (10(10) - 8(-6))/(10^2)#
#(4/5) (d theta)/(dt) = (100+48)/100#
#(4/5) (d theta)/(dt) = 148/100#
#(d theta)/(dt) = 37/25 * 5/4 = 37/20#
A quick sanity check on the sign of the result can help catch errors as they are in progress. As the ships continue moving, ship A will get closer to the lighthouse, while ship B will move farther away.
This action will serve to make the angle #theta# become larger and larger, much like how the angle made by a ladder to the ground increases as you push a ladder up the side of a house. (By the way, this is a common alternative word problem that tests the exact same math in many Calculus classes.)
Your original attempt ended with a negative answer, which would indicate the angle is decreasing - a result that contradicts the nature of the situation. It might not help you find the issue, but it can at least give you a hint that you're on the wrong track somewhere.
