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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 $\angle{DAC}$?


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I wonder if this isn't the same kind of problem as… – Gerry Myerson Sep 15 '11 at 1:35
(didn't do a careful check, but...) possible duplicate of Finding an angle within an 80-80-20 isosceles triangle – Aryabhata Sep 15 '11 at 2:15

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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