# Geometry/trig help - having trouble finding an angle [duplicate]

I've been trying stuff with angle sums, Sine law and Thales theorem, but either I'm making very bad mistakes or I'm just tired - either way I would love some outside input. Thank you.

• This is essentially a duplicate of "Finding angle in isosceles triangle", although a different angle is sought. The sole answer uses no trig. (I seem to recall another duplicate, but I haven't been able to locate it.) – Blue Nov 8 '18 at 10:23
• Ah, here's another: "Find an angle of an isosceles triangle ", with a variety of answers. There could be more. – Blue Nov 8 '18 at 10:26
• Thanks a lot for the help, I appreciate it. Btw, when you searched for these questions did you just used keywords from my question or is there some cleverer way of searching on here? – Rock Dekasba Nov 8 '18 at 11:48
• Searching Math.SE can be tricky. The task is made that much harder when question titles are really generic, instead of describing what the question is actually about. (hint, hint) ;) In this case, I just searched for something like "isosceles angle 20". (Thinking I might've answered the question myself at some point, I tried a search with "user:409" as well.) Since I was pretty sure I'd seen the question before, I probably sifted-through a few more generically-titled questions than a casual searcher might have done. – Blue Nov 8 '18 at 12:02

Let'd denote $$\gamma$$ as the angle BDA. Then the angle BDC $$=180 - \gamma$$.
Let's also denote $$\beta$$ as the angle DBC. From symmetry, we get that the angle ACB $$=\alpha + \beta$$. Now, looking at the lower triangle, we obtain $$(\beta) + (\alpha + \beta) + (180-\gamma) = 180 \qquad \Rightarrow \qquad \alpha + 2\beta - \gamma = 0$$ From the upper triangle we get $$20 + \alpha + \gamma = 180 \qquad \Rightarrow \qquad \alpha + \gamma = 160$$ Maybe you can continue from here?
• You can additionally use the en.wikipedia.org/wiki/Law_of_sines for $\alpha$ and the $20$ degree angle. – Matti P. Nov 8 '18 at 9:06