PLEASE HELP!! --> A balloon rises at a rate of 4 meters per second from a point on the ground 40 meters from an observer. What is the rate of change of the angle of elevation (in rad/sec) when the balloon is 40 meters above the ground?

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PLEASE HELP!! --> A balloon rises at a rate of 4 meters per second from a point on the ground 40 meters from an observer. What is the rate of change of the angle of elevation (in rad/sec) when the balloon is 40 meters above the ground?

Mathematics
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Step 1: draw a diagram. If you done that, tell us what an expression for angle of elevation, call it A in terms of 40 meters and the balloon's height, h. And what is height h as a function of time t?
I don't understand
When the balloon is h meters off the ground, what is the trigonometric expression for the angle of elevation. It involves something like sin or cos or tan or cot or sec or cosec.

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Other answers:

sin
The horizontal distance of 40 m and the vertical height of h, form two sides of right-angled triangle. The observer is sitting at the end of the horizontal 40 m looking up at the balloon. The angle at which she's looking is the angle of elevation, call it A. What is the trig relationship between that angle, h and 40?
That is, she's looking along the diagonal towards the balloon.
opp/adj
I'm still waiting for an equation in which A appears.
tan A = h/40 That's the equation.
Now, the height h = 4t, because it's rising at a constant velocity. Hence tan A = 4t/40 = t/10 or A = arctan(t/10). The question now asks you for dA/dt when the balloon is 40 meters off the ground. So you need to a) calculate dA/dt in general b) find the value of t for which h = 40 m c) evaluate dA/dt for that value of t.
ok i got .24

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