# How do we evaluate the degree of $x$ using sine law?

Given that

ABC is a triangle, $$|EC| = |BC| = |BD|$$, $$\angle CBA= 80^\circ,\angle ACB= 60^\circ, \angle EDA= x^\circ$$

Evaluate $$x$$

I want to solve this for $$x$$ using law of sines if possible.

My attempt:

From the property of triangle, the sum of the angles will be equal to $$180$$.

$$\angle BAC = 180 - 80 - 60 = 40^\circ$$

In $$\triangle ABC$$,

$$\frac{\sin 40}{|BC|} = \frac{\sin 80}{|AC|} \implies \frac{|AC|}{|BC|} = \frac{\sin 80}{\sin40}$$

Could you help me take it from there?

Regards

You don't need trigonometry to solve for $$x$$. Observe that $$BCE$$ is an equilateral triangle, so $$BE=BC$$, so $$BDE$$ is an isosceles triangle. From this, it's easy to see that $$x=100°$$.
Since $$|BC|=|EC|$$ and $$\angle BCE=60$$ then $$\triangle BCE$$ is equilateral.
Hence $$|BE|=|BC|=|BD|$$, so that $$\triangle BDE$$ is isosceles with $$\angle DBE=80°-60°=20°$$.
Hence, $$\angle BDE=(180°-20°)/2=80°$$ so that $$\angle ADE=180°-80°=100°$$.