# Langley’s Adventitious Angles (Part 2)

As a follow-up to yesterday’s triangle problem, here’s another one in the same equivalence class that I found at http://thinkzone.wlonk.com/MathFun/Triangle.htm (via the comments at Math With Bad Drawings). The author of this webpage tantalizingly calls this the World’s Hardest Easy Geometry Problem: solve for $x$ in the figure below.

This figure is similar to the figure in yesterday’s post, except the values of $m\angle BAE$ and $m\angle DAE$ have changed.

So as to not ruin the fun, I won’t give the answer here. Instead, I’ll leave a thought bubble so you can think about the answer. In case you’re wondering: yes, I did figure this out for myself without using the Laws of Sines and Cosines. But I needed over an hour to solve this problem , and that’s after I had time to read and reflect upon the solution to the problem I posed in yesterday’s post.

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