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 (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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