Related Rate Problem involving Theta

[titusrevised]

Homework Statement

A balloon rises at the rate of 8 ft/s from a point on the ground 60 feet from an observer. Find the rate of change of the angle of elevation when the balloon is 25 feet above the ground.

Homework Equations

n/a

The Attempt at a Solution

i tried several times and erased...
but...
I tried d/dt 'ing tanX = h/60 and solving for dx/dt but that gave me a number way off from the answer.
I get (secX)^2 * dx/dt = 8/60 (because dh/dt = 8)

The answer is supposed to be about .114 radians/second.

EDIT: I solved it myself, finally...

 

[titusrevised]

I got it myself. Finally...

 

[stunner5000pt]

For these related rate questions you shouldn't sub in the numbers given to you until after you differentiate

suppose the distance between observer and the point on the ground is x and the height of hte balloon and is y, then

$$\tan\theta=\frac{y}{x}$$

now differentiate with respect to time
$$\frac{d}{dt}(\tan\theta)=\frac{d}{dt}\left(\frac{y}{x}\right)$$

when you differentiate theta and y keep in mind that you have differentiate implicitly, so you will end up with terms like dy/dt or $d\theta/dt$
also x is a constant so you need not both about differentiating x with respect to t.

Hope this helps