It seems impossible to calculate $\alpha$ and $\beta$.
Point $A$ divides $c$ line in half, $a+b=80=d$, $h=20$, $c=50$
I tried to work from the right triangle to the left one like this:
$$\cos\alpha=\frac{x}{a}$$ $$\tan\beta=\frac{c}{2(h+x)}$$ I really don't want to put $\cos\alpha$ in the $\tan\beta$ equation, simply because I'm looking for $\alpha$ and $\beta$ and If I put them together, What do I do? Nothing.
I'm missing something with $a+b=d$. Any tips?
Show your approach to this problem.
Start by writing Pythagoras' theorem in the triangle on the right. $$a^2=\left(\frac c2\right)^2+x^2$$ $$x^2=a^2-625$$ $$x=\sqrt{a^2-625}$$ Then use this $x$ for the triangle on the left: $$(h+x)^2+\left(\frac c2\right)^2=b^2$$ $$h^2+x^2+2hx+\left(\frac c2\right)^2=b^2$$ $$400+a^2-625+40\sqrt{a^2-625}+625=(80-a)^2$$ $$400+40\sqrt{a^2-625}=6400-160a$$ $$\sqrt{a^2-625}=15-4a$$ You can square it, find $a$, then $b$, and then $\alpha$ and $\beta$. If you want, you can get $x$ as well.
Note $a = \frac{c}{2\sin\alpha}$ and $b = \frac{c}{2\sin\beta}$. Plug them into $a+b=80$ and $b\cos\beta - a\cos\alpha = h$ to get the system of equations for the angles
$$\cot\beta=\frac45+\cot\alpha, \>\>\>\>\>\csc\beta=\frac{16}5-\csc\alpha$$
Eliminate $\beta$ to get $4\csc\alpha + \cot\alpha=6$, or
$$3\tan^2\frac{\alpha}2 -12\tan\frac{\alpha}2+5=0$$ Thus, the angles are $\alpha = 2\tan^{-1}\frac{6-\sqrt{21}}3$ and $\beta= 2\tan^{-1}\frac{6-\sqrt{21}}5$.
