Finding the measure of an angle inside triangle

Let $BC = 1$ and use sine rule on $\triangle DBC$ :

$\begin{aligned} \frac {DC}{\sin \angle BDC} & = \frac {BC}{\sin \angle BDC} \\ \frac {DC}{\sin 50^\circ} & = \frac 1{\sin 60^\circ} \\ \implies DC & = \frac {\sin 50^\circ}{\sin 60^\circ} \end{aligned}$

Using sine rule on $\triangle EBC$ :

$\begin{aligned} \frac {EC}{\sin \angle EBC} & = \frac {BC}{\sin \angle BEC} \\ \frac {EC}{\sin 70^\circ} & = \frac 1{\sin 50^\circ} \\ \implies EC & = \frac {\sin 70^\circ}{\sin 50^\circ} \end{aligned}$

Let $\angle DEC = \theta$ and use sine rule on $\triangle DEC$ :

$\begin{aligned} \frac {\sin \angle DEC}{DC} & = \frac {\sin \angle EDC}{EC} & \small \color{#3D99F6} \text{Note that }\angle EDC = 180^\circ - 10^\circ - \theta \\ \frac {\sin \theta}{DC} & = \frac {\sin (170^\circ - \theta)}{EC} \\ EC \sin \theta & = DC \color{#3D99F6} \sin (170^\circ - \theta) & \small \color{#3D99F6} \text{Note that }\sin (180^\circ - x) = \sin x \\ EC \sin \theta & = DC \color{#3D99F6} \sin (10^\circ + \theta) \\ \frac {\sin 70^\circ}{\sin 50^\circ} \times \sin \theta & = \frac {\sin 50^\circ}{\sin 60^\circ} \times (\sin 10^\circ \cos \theta + \cos 10^\circ \sin \theta) \\ \implies \tan \theta & = \frac {\sin 10^\circ}{\frac {\sin 70^\circ \sin 60^\circ}{\sin^2 50^\circ} - \cos 10^\circ} \\ & \approx 0.43198722 \\ \implies \theta & \approx \boxed{23}^\circ \end{aligned}$