Missing angle problem

Question

In a isoceles triangle ABC having equal angles as B=C=80°, B is joined to a point D on AC such that AD=BC. Find angle BDC.

By construction, its coming out to be 30°, is there any way to it without direct measuring?

Answer 1

Build equilateral triangle $BEC$. Then $\angle EBA = 80^\circ - 60^\circ = 20^\circ = \angle BAD$ and $BE = BC = AD$. It follows by SAS that $\triangle EBA$ is congruent to $\triangle DAB$. Hence $\angle DBA = \angle BAE = 10^\circ$ and finally $\angle BDC = \angle DBA + \angle BAD = 10^\circ + 20^\circ = 30^\circ$.

Answer 2

Let the unknown angle be $x$

Then we use sin rule in triangle DBC to get

$$\frac{\sin x}{BC} = \frac{\sin 80}{BD}$$

Similarly in triangle ABD we have

$$\frac{\sin 20}{BD} = \frac{\sin (x-20)}{AD}$$

Using the fact that AD = BC, we have

$$\frac{\sin x}{\sin 80} = \frac{\sin(x-20)}{\sin 20}$$

$$\implies \frac{\sin x}{\sin 80} = \sin x\cot20 - \cos x$$

$$\implies (\sin x)(\cot 20 -\frac{1}{\sin 80}) = \cos x $$

$$\implies \tan x = \frac{\sin 80\sin 20}{\sin80\cos 20 - 1}$$

You can simplify this using the formula for $\tan(A+B)$