# Measure of an Interior Angle

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 $\angle ACD$ ?

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 $\angle ACD$.
• @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. – coffeemath May 25 '14 at 16:50