Geometric Problem. Find an angle. In△ABC , AB=AC ∠ACB=72∘. D is a point in △ABC such that ∠DBC=42∘,∠DCB=54∘.Find an angle∠BAD. Is there any geometric method? 

Let, $AB=AC=a$ in isosceles $\triangle ABC$ 
Using Sine rule  in isosceles $\triangle ABC$  $$\frac{\sin \angle BAC}{BC}=\frac{\sin \angle ABC}{AC}$$ $$\frac{\sin 36^\circ}{BC}=\frac{\sin 72^\circ}{a}\implies BC=2a\sin 18^\circ$$ Similarly applying Sine rule in $\triangle BCD$
$$\frac{\sin \angle DCB}{BD}=\frac{\sin \angle BDC}{BC}$$  $$\frac{\sin 54^\circ}{BD}=\frac{\sin 84^\circ}{2a\sin 18^\circ}$$  $$\implies BD=\frac{2a\sin 18^\circ\sin 54^\circ}{\cos 6^\circ}$$
Let, $\angle BAD=\alpha$
Applying Sine rule in $\triangle ABD$
$$\frac{\sin \angle BAD}{BD}=\frac{\sin \angle ADB}{AB}$$  $$\frac{\sin \alpha}{\frac{2a\sin 18^\circ\sin 54^\circ}{\cos 6^\circ}}=\frac{\sin (180^\circ-(\alpha+30^\circ))}{a}$$ $$\cos 6^\circ\sin \alpha=2\sin 18^\circ\sin 54^\circ\sin (\alpha+30^\circ)$$  $$\frac{\sin(\alpha+30^\circ)}{\sin \alpha}=\frac{\cos 6^\circ}{2\sin 18^\circ\sin 54^\circ}$$ $$\frac{\sin \alpha\cos 30^\circ+\cos \alpha\sin 30^\circ}{\sin \alpha}=\frac{\cos 6^\circ}{2\sin 18^\circ\sin 54^\circ}$$
$$\frac{\sqrt 3}{2}+\frac{1}{2}\cot \alpha=\frac{\cos 6^\circ}{2\sin 18^\circ\sin 54^\circ}$$
$$\frac{1}{2}\cot \alpha=\frac{\cos 6^\circ}{2\sin 18^\circ\sin 54^\circ}-\frac{\sqrt 3}{2}$$
$$\cot \alpha=\frac{\cos 6^\circ-\sqrt 3\sin 18^\circ\sin 54^\circ}{\sin 18^\circ\sin 54^\circ}$$
$$\tan \alpha=\frac{\sin 18^\circ\sin 54^\circ}{\cos 6^\circ-\sqrt 3\sin 18^\circ\sin 54^\circ}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\alpha=\tan^{-1}\left(\frac{\sin 18^\circ\sin 54^\circ}{\cos 6^\circ-\sqrt 3\sin 18^\circ\sin 54^\circ}\right)=24^\circ}}$$
