# Trigonometric vs Synthetic solutions for olympiad geometry problem...

In a triangle $ABC$ $\angle ABC=40^\circ$ , pick a point $D$ inside the triangle such that $\angle DCB=\angle ABD=\angle DBC=20^\circ$ and $|AB|=|AD|$, what is $\angle DCA=\alpha$

My Synthetic Solution If we take The symmetric of point $B$ as $B'$ and complete $B'$ to $C$ we have an equilateral triangle, from here $60-\alpha=50^\circ$ and $\alpha=10^\circ$....

My question is how do we get a trigonometric solution from here?

• I'm getting $ADC$ angle to be $200°$ Oct 1, 2017 at 13:04
• Hi, I've added The picture of the question in case İ misrepresented some parts of it Oct 1, 2017 at 13:37
• Would you please explain, why getting the answer is not enough? Oct 1, 2017 at 13:49
• I didn't mean anything bad, I already knew that the answer was $10^\circ$ not because I solved it myself, because it was formally posted, and I thought you might have found it that way due to my misrepresentation. That's all I meant. Oct 1, 2017 at 14:24

Let $\measuredangle ACD=x$ and $AB=a$.
Thus,$\measuredangle ADC=140^{\circ}$, $\measuredangle DAC=40^{\circ}-x$ and by the law of sines we obtain: $$\frac{\sin(40^{\circ}-x)}{\sin{x}}=\frac{DC}{AD}=\frac{a}{2a\sin{x}}=\frac{1}{2\sin10^{\circ}}.$$ Thus, $$\sin40^{\circ}\cot{x}-\cos40^{\circ}=\frac{1}{2\sin10^{\circ}},$$ which gives $$\cot{x}=\frac{1+2\sin10^{\circ}\cos40^{\circ}}{2\sin10^{\circ}\sin40^{\circ}}$$ and since $$\frac{1+2\sin10^{\circ}\cos40^{\circ}}{2\sin10^{\circ}\sin40^{\circ}}=\frac{1+\sin50^{\circ}-\sin30^{\circ}}{2\sin10^{\circ}\sin40^{\circ}}=\frac{\sin50^{\circ}+\sin30^{\circ}}{2\sin10^{\circ}\sin40^{\circ}}=$$ $$=\frac{2\sin40^{\circ}\cos10^{\circ}}{2\sin10^{\circ}\sin40^{\circ}}=\cot10^{\circ},$$ we obtain $$x=10^{\circ}$$ and we are done!