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

BSFU
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  • Inspect each triangle and write down all the angle measures. – Andrew Chin Apr 29 '20 at 06:56
  • @AndrewChin, maybe I'm wrong, but I don't want to confuse the OP. $\measuredangle=40^{\circ}$ and $\measuredangle BCA=20^{\circ}$, but this isn't a right-angle triangle... and $\Delta ABD$ isn't isosceles.... so it is useless, isn't it? – Matcha Latte Apr 29 '20 at 08:02
  • See https://math.stackexchange.com/questions/3620825/langleys-adventitious-angles/3621250#3621250 – Quanto Apr 29 '20 at 12:57

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

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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

Shiv Tavker
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