Find the Angle #4

This is probably not the most elegant solution, but it gets the job done.

First observe that we have four right angles at O since ABO is a 30-60-90 triangle.

For convenience , we let $OB=1$ ; then $OA=\tan60$ (all angles will be measured in degrees). Now $OD=(\tan10)(\tan60)$ and $OC=\tan20$ . If $\theta$ is the angle BDC, then $\tan{\theta}=\frac{OC}{OD}=\frac{\tan20}{(\tan10)(\tan60)}$ .

I remember a fun fact I learned in school long ago: $(\tan20)(\tan40)(\tan80)$ $=\tan60$ , which follows from $\prod_{k=1}^{(n-1)/2}\tan(\frac{k\pi}{n})=\sqrt{n}$ for odd $n$ , with $n=9$ .

Keeping in mind that $(\tan{t})(\tan(90-t))=1$ , we find that $\tan\theta=\frac{1}{\tan40}=\tan50$ and $\theta=\boxed{50^o}$ .

That's my solution. I don't think it's the best, but I liked it. Thank you for share this beautiful problem.

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Let $x=\angle{BDC}$ . Using Law of Sines on:

$\triangle{BCD}$ : $\frac{BC}{CD}=\frac{\sin x}{\sin20};\qquad\tag{1}$ $\triangle{ACD}$ : $\frac{CD}{AC}=\frac{\sin10}{\sin(x+80)};\qquad\tag{2}$ $\triangle{ACB}$ : $\frac{BC}{AC}=\frac{\sin30}{\sin80}.\qquad\tag{3}$ Dividing (3) by (2) and using (1), yields $\frac{\sin30\sin(x+80)}{\sin10\sin80}=\frac{\sin x}{\sin20},$ $\frac{\sin(x+80)}{\sin x}=\frac{\sin10\sin80}{\sin20\sin30}=\frac{\sin10\cos10}{2\sin10\cos10\sin30}=1.$ Angle $x$ is acute (right triangle ODC), then $\angle{BDC}=\boxed{50^\circ}$ .

$Angles~at~O~are~clearly~90^o s.~~~~~~ Let~x=\angle~BDC.\\ ~~~\\~~~\\ In~\Delta~DAO,~~\dfrac{AO}{OD}=Cot10.\\ In~\Delta~OAB,~~\dfrac{OB}{AO}=Tan30.\\ \implies~\dfrac{OB}{OD}=Tan80*Tan30..........(**)\\ ~~~\\~~~\\ In~\Delta~DAO,~~\dfrac{CO}{OD}=Tan(x).\\ In~\Delta~OAB,~~\dfrac{OB}{CO}=Cot20.\\ \implies~\dfrac{OB}{OD}=Tan(x)*Tan70..........(***)\\ ~~~\\~~~\\ By~(**)~and~(***)~~Tan(x)=\dfrac{Tan80*Tan30}{Tan70}.\\ ~~~\\~~~\\ \therefore~~x=\angle~BDC=Tan^{-1}\dfrac{Tan80*Tan30}{Tan70}=\large~~\color{#D61F06}{50^o}.$

i reflected the entire figure across BC, then extended the 2 AD segments so they'd meet and the entire figure becomes a kite.

then i first assumed that all the angles with vertex C are vertical angles, and filled out all the unknown ones, thinking that if they were not vertical angles, then at least 1 triangle would not add up to 180. turns out there was no such mistake. and eventually angle BDC is 50 degrees.

I also solved it using trigonometry (and a calculator) I tried to solve it geometrically as follows: I constructed <bde =,<bca =70 with e on ac so bcde are concylic Now if we can show that be =bd we are done but this is where I am stuck Maybe somebody can finish it from here