The rocket's altitude [tex]y[/tex] at time [tex]t[/tex] is
[tex]y(t)=\left(800\dfrac{\rm km}{\rm h}\right)t[/tex]
so that after [tex]t=3\,\mathrm{min}[/tex], it will have traveled
[tex]y=\left(800\dfrac{\rm km}{\rm h}\right)(3\,\mathrm{min})=40\,\mathrm{km}[/tex]
The angle of elevation [tex]\theta[/tex] at time [tex]t[/tex] is such that
[tex]\tan\theta=\dfrac{y(t)}{13\,\rm km}[/tex]
At the moment when [tex]y(t)=30\,\mathrm{km}[/tex], this angle is
[tex]\tan\theta=\dfrac{30\,\rm km}{13\,\rm km}\implies\theta\approx66.57^\circ[/tex]
Differentiating both sides of the equation above gives
[tex]\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac1{13\,\rm km}\dfrac{\mathrm dy(t)}{\mathrm dt}[/tex]
and substituting the angle [tex]\theta[/tex] found above, and [tex]\dfrac{\mathrm dy(t)}{\mathrm dt}=800\dfrac{\rm km}{\rm h}[/tex], we get
[tex]\sec^2\theta66.57^\circ\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac1{13\,\mathrm km}\left(800\dfrac{\rm km}{\rm h}\right)[/tex]
[tex]\implies\boxed{\dfrac{\mathrm d\theta}{\mathrm dt}\approx9.73\dfrac{\rm rad}{\rm h}}[/tex]
