Geometry Problem - 3

Since $\angle BEC=180-40-30-40=70$ , $\triangle EBC$ is isosceles with $BC=BE$ .

$\angle CDB=180-30-30-40=80^\circ$

Let $BC=BE=1$

By sine law on $\triangle EBC$ , we have

$\dfrac{CE}{\sin 40}=\dfrac{BC}{\sin 70}$

$CE=\dfrac{\sin 40}{\sin 70}$

By sine law on $\triangle DBC$ , we have

$\dfrac{DC}{\sin 70}=\dfrac{BC}{\sin 80}$

$DC=\dfrac{\sin 70}{\sin 80}$

$\dfrac{BD}{\sin 30}=\dfrac{BC}{\sin 80}$

$BD=\dfrac{\sin 30}{\sin 80}$

By cosine law on $\triangle EDC$ , we have

$DE^2=\left(\dfrac{\sin 40}{\sin 70}\right)^2+\left(\dfrac{\sin 70}{\sin 80}\right)^2-2\left(\dfrac{\sin 40}{\sin 70}\right)\left(\dfrac{\sin 70}{\sin 80}\right)(\cos 40)$

$DE\approx 0.615132148$

By sine law on $\triangle DEB$ , we have

$\dfrac{\sin \angle DEB}{\dfrac{\sin 30}{\sin 80}}=\dfrac{\sin 30}{0.615132148}$

$\angle DEB \approx \color{#D61F06}\boxed{24.3737}$