In right triangle:
Taken from: https://xaktly.com/MathNonRightTrig.html
We have side $a=2c_1$. Where $c_1$ is a perpendicular projection of c onto b. What is the angle A? I have tried solving using quadratic equation obtained from solving for height.
But, it fails, giving a bit incorrent answer. I think the reason may be that I may have solved the equation wrongly, or tried it in a wrong way.
Thank you for your help.
Hint
Assume that the perpendicular from $B$ into the side $b$ hits $b$ at point $D$. Then, $\triangle BCD$ is a right triangle (since $\angle D = 90^\circ$) and in that triangle, the leg $BD$ is exactly half of the hypotenuse $BC$. If that's the case, you can find $\sin(C)$ from the definition and thus compute $\angle C$, and imply $\angle A$ from it.
If you do not know trigonometry, there is a classic result that in a $30^\circ-60^\circ-90^\circ$ right triangle, the side opposite $30^\circ$ angle is exactly half the hypotenuse long.
UPDATE
With your understanding, denote $AD = h$ and observethat you can find the area of $\triangle ABC$ in two different ways: $$ A_{\triangle ABC} = \frac{ac}{2} = \frac{bh}{2} \iff h = \frac{ac}{b} $$ and so $\angle A = \arcsin(h/c) = \arcsin(a/b)$.