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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?enter image description here

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  • $\begingroup$ I'm getting $ADC$ angle to be $200°$ $\endgroup$
    – dEmigOd
    Oct 1, 2017 at 13:04
  • $\begingroup$ Hi, I've added The picture of the question in case İ misrepresented some parts of it $\endgroup$ Oct 1, 2017 at 13:37
  • $\begingroup$ Would you please explain, why getting the answer is not enough? $\endgroup$
    – dEmigOd
    Oct 1, 2017 at 13:49
  • $\begingroup$ 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. $\endgroup$ Oct 1, 2017 at 14:24

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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!

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  • $\begingroup$ Hello, I've added a picture to the body, in case I misrepresented the question. $\endgroup$ Oct 1, 2017 at 13:38
  • $\begingroup$ @Deniz Tuna Yalçın I fixed my post. See now. $\endgroup$ Oct 1, 2017 at 13:55
  • $\begingroup$ @Deniz Tuna Yalçın You are welcome! $\endgroup$ Oct 1, 2017 at 14:22

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