A fish is reeled in at a foot per second from a point 10ft. above the water. What rate is the angle between the line and the water changing when there is 25ft. of fishing line out?

The accompanying image shows a triangle with a height of 10ft and hypotenuse marked as a variable. The sine of the angle is 10ft/hypotenuse.

How would you approach the problem or did I do it correctly?

Since the fish is being reeled in at an angle. I figured, where L is the hypotenuse, dL/dt is 1 ft/s and height is constant.
sine of theta= y/L
sine of theta=10ft/ L
(dtheta/dt) (cosine of theta)= 10ft (-dL/dt divided by L^2)
I saw theta as the inverse sine of 10/25.
dtheta/dt= sec (inverse sin of 2/5)(-10)/25^2/second
I got d(theta)/dt as about -0.017 radians per second.