.We have $\frac{\sin \alpha}{\sin (180-\beta)} = \frac{\sin 20}{\sin 80}$.
The first thing we use is that $\alpha + \beta = 160$ from the triangle $ABD$. From here, $180 - \beta = 180 - (160-\alpha) = 20 + \alpha$.
Next, we note that:
$$
\frac{\sin 20}{\sin 80} = \frac{\sin 20}{\cos(90-80)} = \frac{\sin 20}{\cos 10} = \frac{2 \sin 10 \cos 10} {\cos 10} = 2 \sin 10
$$
So, we have the equation :
$$
\frac{\sin \alpha}{\sin (\alpha + 20)} = 2 \sin 10\\ \implies \sin \alpha = 2 \sin 10 \sin (20+\alpha) = 2 \sin 10 \sin 20 \cos \alpha + 2 \sin 10 \cos 20 \sin \alpha
$$
Now, collecting terms of $\sin \alpha$ on one side, and facctorizing it out,
$$
\sin \alpha(1 - 2 \sin 10 \cos 20) = 2 \sin 10 \sin 20 \cos \alpha \\ \implies
\tan \alpha = \frac{2 \sin 10 \sin 20}{1 - 2 \sin 10 \cos 20}
$$
The right hand side is some fixed number which we have to find.
To do this, we first simplify the denominator, using the formulas : $$2 \sin A\cos B = \sin(A+B) + \sin(A-B) \quad ; \quad \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
We will also use the fact that $\sin 30 = \frac 12$.
In our case,
$$
1 - 2 \sin 10 \cos 20 = 1- (\sin 30 + \sin (-10)) = 2 \sin 30 - (\sin 30 - \sin 10) \\ = \sin 30 +\sin 10 = 2 \sin 20 \cos 10
$$
Therefore,
$$
\tan \alpha = \frac{2 \sin 10 \sin 20}{1 - 2 \sin 10 \cos 20} = \frac{2\sin 10 \sin 20}{2 \cos 10 \sin 20} = \tan 10
$$
Now, since $0 < \alpha < 180$, we get that $\alpha = 10$. From here, $80-\alpha = 70$ is the desired angle.
In the standard notation we need to prove that
$$\frac{a^2+b^2-c^2}{2ab}<\cos60^{\circ}$$ or
$$c^2>a^2-ab+b^2$$ or
$$\sqrt[3]{(a^3+b^3)^2}>a^2-ab+b^2$$ or
$$(a+b)^2(a^2-ab+b^2)^2>(a^2-ab+b^2)^3$$ or
$$ab>0,$$ which is true.
Id est, $$\measuredangle ACB>60^{\circ}$$ and we are done!
Best Answer
Another way.
Let $BC=x$.
Thus, by law of cosines for $\Delta ADB$ we obtain: $$x^2+100-2x\cdot10\cdot\frac{1}{2}=16.$$ Can you end it now?
I also got that it's impossible.