Angle of elevation -- flagpole question-- Show that the angle of elevation from E to B is $\mathrm{arccot}(\sqrt{\cot^2\alpha + \cot^2 \beta})$

Question

A flagpole is erected at point A. The top of the flagpole is point B. At point C, due west of A, the angle of elevation to B is $\alpha$. At point D, due south of A, the angle of elevation to B is $\beta$. Point E is due south of C and due west of D. Show that the angle of elevation from E to B is $\mathrm{arccot}(\sqrt{\cot^2\alpha + \cot^2 \beta})$
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I am a bit stuck with the question, as there are so many variables involved. What I tried doing was to draw a cube and label A, B, C, D, E on the cube, but I wasn't sure if the question has implied that C and A are on the same line, and that A and D are on another same line; and that E is right below C and directly on the left to D.

Also, as the answer that I am trying to prove involves trig functions, I tried rewriting the variables in terms of the angle.

So far, What I have rearranged is that AB= AC $\tan \alpha$ AC= BC $\cos \alpha$ which is = DE

I tried to use pythagoras, in that AE = $\sqrt{AD^2+ DE^2}$ However, not only does the method I am trying to use involve a lot of variables, but it also doesn't seem to be arriving at the co-functions cot and inverse arccot as required in the question.

Sorry for not delineating my question properly, as I understand my phrasing of the question is a bit confusing.

Any advice would be greatly appreciated.

Answer 1

In such questions, it's always good to draw diagram.
Also, try to write the lengths in terms of one quantity.
I have tried to write everything in terms of $AE$.

Let the sought angle be $\gamma$

$$AC = AB \cot \alpha$$ $$AD = AB \cot\beta$$ By Pythagoras, $$\begin{align}AE&=\sqrt{AD^2 + AC^2}\\ AE^2 &= AB^2 \cot^2 \beta + AB^2 \ cot^2 \alpha \ \ \ \ \ \ ...... (1) \end{align}$$

Also, by Pythagoras $$EB^2 = AE^2 + AB^2$$

$$AE^2 = EB^2 - AB^2$$ But $$BE = AB\csc \gamma$$

Therefore $$\begin{align}AE^2 &= AB^2 (\csc^2\gamma - 1) \\ &= AB^2 \cot^2\gamma \ \ \ \ \ \ \ ...... (2)\end{align} $$

From 1 and 2 $$\cot^2 \gamma = \cot^2 \alpha + \cot^2\beta$$ $$\therefore \gamma = \mathrm{arccot}\left(\sqrt{ \cot^2 \alpha + \cot^2\beta}\right)$$