Projection of a right angle

Question

I'd appreciate your help in expressing $\delta$ in terms of $\phi$ and $\gamma$ in the following figure. .

Further, I'd appreciate any indication on what is the minimal and the maximal value of $\phi+\gamma$.

Thanks.

Answer 1

Let $PC=h$.

Thus, $AC=h\cot\phi$, $OC=h\cot\gamma$ and by Pythagoras theorem for $\Delta APO$ we obtain: $$AO=h\sqrt{\frac{1}{\sin^2\phi}+\frac{1}{\sin^2\gamma}}.$$

Thus, by law of cosines we obtain: $$\frac{1}{\sin^2\phi}+\frac{1}{\sin^2\gamma}=\cot^2\phi+\cot^2\gamma-2\cot\phi\cot\gamma\cos\delta$$ or $$2=-2\cot\phi\cot\gamma\cos\delta$$ or $$\cos\delta=-\tan\phi\tan\gamma.$$

I hope you mean to find a minimal value of $\phi+\gamma$, where $\delta=constant.$

Let $\phi+\gamma\geq90^{\circ}$.

Thus, the minimal value is $90^{\circ}$.

Now, let $\phi+\gamma<90^{\circ}$.

We'll prove that in this case $$\tan\phi\tan\gamma\leq\tan^2\frac{\phi+\gamma}{2}.$$ Indeed, we need to prove that $$\sin\phi\sin\gamma\cos^2\frac{\phi+\gamma}{2}\leq\cos\phi\cos\gamma\sin^2\frac{\phi+\gamma}{2}$$ or $$\sin\phi\sin\gamma(1+\cos(\phi+\gamma))\leq\cos\phi\cos\gamma(1-\cos(\phi+\gamma))$$ or $$\cos(\phi+\gamma)\geq\cos(\phi+\gamma)\cos(\phi-\gamma),$$ which is obvious.

Id est, $$0=\cos\delta+\tan\phi\tan\gamma\leq\tan^2\frac{\phi+\gamma}{2},$$ which gives $$\phi+\gamma\geq2\arctan\sqrt{-\cos\delta}.$$ Since $2\arctan\sqrt{-\cos\delta}<90^{\circ}$, we got the answer: $$2\arctan\sqrt{-\cos\delta}.$$