Find angle B in the trigonometry problem

[neilparker62]

Very clever @neilparker62 :)

To show the triangles are similar, we need to show $de/da=da/db$. (These are corresponding sides around the same angle).
Now $da/db=da/ac=\sin(40)/\sin(80)$.
Meanwhile $de/da=\sin(30)/\sin(50)$.
We need to show $\sin(40)/\sin(80)=\sin(30)/\sin(50)$.

With $\sin(80)=2 \sin(40) \cos(40)$, and $\sin(30)=1/2$, and $\cos(40)=\sin(50)$, we have proven the triangles to be similar, and thereby angle $B=30$ degrees.

Yes I remembered now - my workings were with sin(100) rather than sin(80) but same thing obviously.