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In $\triangle{ABC}$, given $\angle{A}=80^\circ$, $\angle{B}=\angle{C}=50^\circ$, D is a point in $\triangle{ABC}$, which $\angle{DBC}=20^\circ,\angle{DCB}=40^\circ$. Then how to find find $\angle{DAC}$?

thanks.

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I tried the geometric method but could not succeed. However the trigonometric route yields the result. Here it is: Extend $CD$ to meet $AB$ at $G$. Angle $B$ is $50$. Angle $BCD$ is $40$. So $CG$ is perpendicular to $AB$. Let $AH$ be the right bisector of angle $A$ resting on $BC$ at $H$. Let the required angle $DAC$ be $k$ degrees. $$ GD = AG\tan(80-k) = BG\tan(30), $$ $$ \tan(80-k) = \frac{GD}{AG} = {BG\tan(30)}{AG}. $$ $$ BG = BC\cos(50) $$ $$ BC = 2BH = 2AB\cos(50). $$ So, $$ BG = 2AB\cos(50) \cos(50) $$ $$ AG = AC \cos(80) = AB \cos(80) $$ So, $$ \tan(80-k) = \frac{BG\tan(30)}{AG} = \frac{2\cos )(50) \cos (50) \tan (30)}{\cos(80)} $$ $$ 80-k = \arctan\left(\frac{2\cos (50) \cos (50) \tan (30)}{\cos (80)}\right) = 70^\circ $$ Thus the required $\angle{DAC} = 10^\circ$

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The regular 18-polygon has the nice property that each node sees all other nodes separated with $10^\circ$. Embed $\triangle{ABC}$ in this polygon as shown below.

Solution

Now use this polygon to construct the two equilateral triangles as illustrated, and consider the intersection of their drawn bisectors. This intersection is the center of a circle through A and C, and it is easy to verify that this center is $D$. Obviously $\triangle{ADC}$ is an isosceles, so $\angle{DAC} = \angle{DCA} = 10^\circ$

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