1. The problem statement, all variables and given/known data A weather balloon is rising vertically at a speed of 10 meters per second. An observer is 40 meters away from the launch point of the balloon. the distance between the observer and the balloon in the air is 50 meters. How fast is the angle of elevation changing? 2. Relevant equations Given the length of the hypotenuse is 50, and the bottom leg of the triangle is 40, it follows that the height of the balloon from its launch site is 30 meters. 3. The attempt at a solution tan[itex]\theta[/itex] = opposite side "a" / adjacent side "b" = 30 / 40 = 3/4 Taking the derivative of both sides, we get d[itex]\theta[/itex]/dt*sec^2([itex]\theta[/itex]) = (a'b - b'a) / b^2 We are given a' is 10, and we know b' is 0 since the observer isn't moving on the ground. Therefore for the right hand side we get [10(40)]/(40^2). We divide this by sec^(theta) and get the answer right?