A geometry problem by abhishek anand

By approximation method. Let $AB=1$ .

Apply sine law at $\triangle ADB$ .

$\dfrac{DA}{\sin 60}=\dfrac{1}{\sin 40}$ $\implies$ $DA \approx 1.347296355$

$\dfrac{BD}{\sin 80}=\dfrac{1}{\sin 40}$ $\implies$ $BD \approx 1.532088886$

Apply sine law at $\triangle AEB$ .

$\dfrac{EB}{\sin 70}=\dfrac{1}{\sin 30}$ $\implies$ $EB \approx 1.879385242$

Apply cosine law at $\triangle EDB$ .

$(DE)^2=(1.532088886)^2+(1.879385242)^2-2(1.532088886)( 1.879385242)(\cos 20)$

$DE \approx 0.684040287$

Finally, apply sine law at $\triangle DEA$ .

$\dfrac{\sin x}{1.347296355}=\dfrac{\sin 10}{0.684040287}$

$\sin x=0.342020143$

$x=\sin^{-1}0.342020143=$ $\boxed{20}$