I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door [tex]d[/tex])...
The door forms a right triangles that satisfies
[tex]\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}[/tex]
We also have
[tex]\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd[/tex]
so if you happen to know the height of the door, you can solve for [tex]b[/tex] and [tex]a[/tex].
[tex]d[/tex] is fixed, so
[tex]a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}[/tex]
We can solve for the angular velocity [tex]\dfrac{\mathrm d\theta}{\mathrm dt}[/tex]:
[tex]\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}[/tex]
At the point when [tex]\theta=40^\circ[/tex] and [tex]\dfrac{\mathrm db}{\mathrm dt}=1.8[/tex] ft/s, we get
[tex]\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}[/tex]
