How to calculate a solid angle (in Steradians) given only Horizontal Beam angle and Vertical Beam angle data. I would like to convert a rectangular beam shape given in Horizontal and Vertical beam angle, into solid angle representing the surface area in steradians of projected light.  For example a light projection that has a 120° H beam by 5° Vertical beam pattern.
It would be helpful if this could be represented in a geometric form that can be used in an Excel Spreadsheet that allows one to input a horizontal beam angle and a vertical beam angle and then output the area in steradians. Given that a sphere has $4\pi$ sr and a hemisphere is $2\pi$ sr I was approaching this for example above that 120°=3.1414 sr or 1/3 of a sphere, and 5°= 0.00597 sr, I thought if there are 120 sections of 5° vertical sections this becomes a total area of 0.7164 sr.  Then second guess I thought that there are only 24 vertical sections that are 5 degrees wide in the total horizontal area of 120° area.  That yields only 0.1435 sr which seems unbelievably small.  I came to the conclusion I need help and found this forum trying to find the solution.
I reviewed many questions that are similar. This is my first post, and I'd appreciate any help and guidance trying to figure where I have gone wrong in the application.  Thank you in advance
 A: Consider a beam with a rectangular profile, having horizontal & vertical beam angles $\theta_{H}$ & $\theta_{V}$, emitted by a uniform point source S (as shown in the figure below). Then consider an imaginary rectangular plane with center O, length $l$ & width $b$ at a normal distance $h$ from the point-source S, then the solid angle ($\omega$) by the rectangular plane at the point-source S is given by $$\omega=4\sin^{-1}\left(\frac{lb}{\sqrt{(l^2+4h^2)(b^2+4h^2)}}\right)$$ (Note: For derivation of the above formula, kindly go through HCR's Theory of Polygon)
Using geometry in right $\Delta SOP$ from the figure, we get $$l=2h\tan\frac{\theta_{H}}{2}$$ Similarly, in right $\Delta SOM$, we get $$b=2h\tan\frac{\theta_{V}}{2}$$  

Now, substituting these values in the formula, we get
$$\omega=4\sin^{-1}\left(\frac{2h\tan\frac{\theta_{H}}{2}2h\tan\frac{\theta_{V}}{2}}{\sqrt{\left(\left(2h\tan\frac{\theta_{H}}{2}\right)^2+4h^2\right)\left(\left(2h\tan\frac{\theta_{V}}{2}\right)^2+4h^2\right)}}\right)$$ $$=4\sin^{-1}\left(\frac{\tan\frac{\theta_{H}}{2}\tan\frac{\theta_{V}}{2}}{\sec\frac{\theta_{H}}{2}\sec\frac{\theta_{V}}{2}}\right)=4\sin^{-1}\left(\sin\frac{\theta_{H}}{2}\sin\frac{\theta_{V}}{2}\right)$$ Where, $0\leq\theta_{H}, \theta_{V}\leq\pi$
If $\theta_{H}, \theta_{V}>\pi$ apply the following formula to calculate solid angle $$\omega=4\pi-4\sin^{-1}\left(\sin\frac{\theta_{H}}{2}\sin\frac{\theta_{V}}{2}\right)$$
(Note: For more detailed explanation, kindly go through Radiometry & Photometry by H.C. Rajpoot) 
As per numerical values provided in your question, the solid angle $\omega$ (i.e. area covered\intercepted by rectangular beam shape on a unit spherical surface), subtended by the rectangular beam shape having horizontal & vertical beam angles $\theta_{H}=120^o=\frac{2\pi}{3}$ & $\theta_{V}=5^o=\frac{\pi}{36}$ at the point-source S, is given as follows $$\omega=4\sin^{-1}\left(\sin\frac{\frac{2\pi}{3}}{2}\sin\frac{\frac{\pi}{36}}{2}\right)=4\sin^{-1}\left(\sin\frac{\pi}{3}\sin\frac{\pi}{72}\right)\approx 0.15113795 \space sr$$
