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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 $\angle ADE=180°-80°=100°$.
