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Triangle $ABC$ has $AC = BC$ , $\angle ACB = 96^\circ$ . $D$ is a point in $ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$ . What is the measure (in degrees) of $\angle ACD$ ?

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If we put $AB=1$ and apply the low of sines, we get $$AC=\frac{\sin 42^\circ}{\sin 96^\circ}=\frac{\sin 42^\circ}{\cos 6^\circ}, \\ AD = \frac{\sin 30^\circ}{\sin 132^\circ}=\frac{1}{2\cos 42^\circ}.$$ But then it follows that $AC=AD$ from $2 \sin 42^\circ \cos 42^\circ=\sin 84^\circ = \cos 6^\circ$. So $\triangle ADC$ is isosceles with the angle $24^\circ$ at $A$, which means the two equal angles are each $78^\circ$, and one of these is the desired $\angle ACD$.

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  • $\begingroup$ WoW ... thank you very much great explanation :) $\endgroup$ – LordAhmed May 25 '14 at 12:17
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    $\begingroup$ @LordAhmed I noticed that in this setup $AD/BD$ is the golden ratio $\phi=(1+\sqrt{5})/2.$ Maybe something using this could finish the problem without using the law of sines, though I haven't had success with that yet. $\endgroup$ – coffeemath May 25 '14 at 16:50
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    $\begingroup$ yeah indeed AD/BD = (1+5√)/2 and that is the golden ratio but to get to this point we have to use the law of sines ... I don't know much about the golden ratio but I think all that we can prove from that statement is that BD/AD = AD/AD+BD $\endgroup$ – LordAhmed May 26 '14 at 3:42

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