After 30 secs the balloon reaches a height of 120m. At that point the angle of elevation is arctan(120/200)=arctan(0.6) (about 31 degrees). The balloon continues to rise at the same rate. We can write tan(X)=4t/200=t/50 where X=angle of elevation and t is time in seconds. dX/dt, the rate of change of X is given by sec^2(X)dX/dt=1/50.
(1+tan^2(X))dX/dt=1/50 and dX/dt=1/(50(1+t^2/2500)=50/(2500+t^2). At t=30 this comes to 50/3400=1/68. So the rate of change of the angle of elevation at 30 seconds is 1/68 per second (0.0147/s).
Now, let's solve this without calculus. We know that at 30 seconds after launch, tan(X)=3/5 or 0.6. Consider a tiny change in time h and a corresponding change in angle w. tan(X+w)=(t+h)/50. This is: (tan(X)+tan(w))/(1-tan(X)tan(w))=(t+h)/50. Because w is very small tan(w)=w approximately. Put tan(X)=0.6 and t=30.
(0.6+w)/(1-0.6w)=(t+h)/50. 50(0.6+w)=(1-0.6w)(t+h); 30+50w=30-18w+h-0.6wh; 68w=h-0.6wh. 0.6wh is very small and can be ignored, so w=h/68. The rate of change of w is w/h=1/68, the result we got by calculus.