Find an unknown angle in a quadrilateral.

Question

In a quadrilateral $ABCD$, $BC=CD=AD$, $\angle ADC=96^{\circ}$ and $\angle BCD=48^{\circ}$. Find $\angle ABC$.

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It seems like there is very little clue given in this problem, and simple angle chasing doesn't really work. We have $$\angle CBD=\angle BDC=66^{\circ}$$ and $$\angle ADB=30^{\circ}$$ All we need is $\angle ABD$, which to me is impossible to find. It's very likely that some auxiliary lines, triangles, or circles should be constructed. However, I'm not clever enough to construct them correctly, so I post this problem.

The answer, according to GeoGebra, is $\color{blue}{\angle ABC=162^{\circ}}$.

Answer 1

WLOG, we can assume $BC=CD=AD=1$.

Triangle $BCD$ being isosceles (as you have observed), cosine rule gives $BD^2=2-2 \cos 48°$.

Apply to triangle $ABD$ :

1. at first, cosine rule (again) in order to get length $AB$.

2. then sine rule in this way :

$$\frac{1}{\sin (\angle ABD)} = \frac{AB}{\sin(30°)}$$

giving angle $\angle ABD$.

Answer 2

Set up a coordinate system so that the three equal double-slashed line segments have a length of "1", point $A$ is at the origin, and point $D$ is on the $x$ axis at $(1, 0)$. Then:

$$C = (1 + \cos{84°}, \sin{84°}) \approx (1.1045284632676535, 0.9945218953682733)$$ $$B = (1 + \cos{84°} - \cos{36°}, \sin{84°} - \sin{36°}) \approx (0.2955114688927061, 0.40673664307580015)$$

This gives us $\vec{AB} = (1 + \cos{84°} - \cos{36°}, \sin{84°} - \sin{36°})$ with a direction of $\tan^{-1}(\frac{\sin{84°} - \sin{36°}}{1 + \cos{84°} - \cos{36°}}) = 54°$, and $\vec{BC} = (\cos{36°}, \sin{36°})$, which obviously has a direction of 36°. So, around the corner from $\vec{AB}$ to $\vec{BC}$ is an 18° right turn. $\angle ABC$ is the supplement of that: $180° - 18° = 162°$, Q.E.D.

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Given the simple whole-number angles, it seems like there should be an alternative solution with no trig functions, but I just haven't found one yet.