I assume you are in calculus because calculus is required for this question. It is a related rates problem using a right triangle. The x component is the distance to the helipad which remains constant -> 50, and the y component is the height of the helicopter which is changing as the helicopter lands -> 100. We are also given the rate of change of the angle 1/15, (I am using @ in place of theta), d@/dt. We are looking for the rate of change of the y component, dy/dt.
We can set up an equation using SOH CAH TOA that uses x, y, and @ so we can calculate the desired rate. We have x and y so we will use tangent.
tan@ = y/x
Lets figure out what the initial angle is using this equation:
tan@ = 100/50=2
tan^-1 (2) = @
@ = 1.11 rads
Now let’s take the derivative of our equation with respect to time so we can find the rates. Remember that y and @ are a function of time but x is constant.
(derivative of tangent is secant*tangent)
sec@tan@ * d@/dt = 1/x * dy/dt
let’s plug in @= 1.11 d@/dt = 1/15 and x = 50 and we will find dy/dt
Sec(1.11)tan(1.11)(1/15) = 1/50 dy/dt
plug in
(2.236)(2)(1/15)=1/50 dy/dt
0.298 = 1/50 dy/dt
multiply by 50
dy/dt = 14.9 ft per second rate of descent of helicopter
Since the height is decreasing it would be appropriate to give this answer as a negative number
dy/dt = -14.9 ft/sec
