What is the value of the angle $x$ of the given triangle?

Question

I have found $\angle BDE+x=130^\circ$. But I can solve from this. How can I find $x$?

Answer 1

Try expressing all angles in terms of $x$. You should get the following:

Now use Law of sines to get the following relations:

From Triangle $ABE$,
$$\ \frac{BE}{\sin(70)} = \frac{AB}{\sin(30)} \quad \text{ and } \quad \frac{BE}{\sin(70)} = \frac{AE}{\sin(80)} $$ From Triangle $ADB$, $$\ \frac{AD}{\sin(60)} = \frac{AB}{\sin(40)} $$ From Triangle $ADE$, $$\ \frac{AD}{\sin(x)} = \frac{AE}{\sin(170-x)} $$ Solving the above equations we get, $$\ \frac{\sin(60)}{\sin(40)\sin(x)} = \frac{\sin(80)}{\sin(30)\sin(170-x)} $$ Finally, $$\ \implies x=20 \deg $$ Note: All angles are in degree.
Link: Wolframalpha Solution to Trignomteric Equation