Value of one angle of a quadrilateral, when three sides are quadrilateral are given

Question

In quadrilateral $ABCD$, $AB=BC=CD$. The value of $∠BAC$ is $40º$ and the value of $∠CAD$ is $30º$ what is the value of $∠ADC$?

Answer 1

First note that $\triangle ABC$ is isosceles so $\angle BCA = 40°$ and $\angle ABC = 100°$

Construct $Q$ as the other point of an equilateral triangle on $CD$, towards $A$. Then since $\angle CAD=30°$, $\angle CQD=60°$ and $Q$ is on the bisector of $CD$, $Q$ is the centre of circle $ACD$.

Thus $|AQ|=|CQ|=|CD|$ and $\triangle ACQ$ is isosceles, sharing $AC$ with $\triangle ABC$ and the edges $AB$, $BC$, $AQ$, $CQ$ are the same length. So $\triangle ABC$ and $\triangle ACQ$ are congruent. Now there are two possible cases:

• As per the diagram, $\triangle ABC$ and $\triangle ACQ$ are mirror images in the shared edge $AC$. Thus $\angle BCD=40+40+60 = 140°$ and $\angle CDA=360-140-100-70 = \fbox{50°}$

• $\triangle ABC$ and $\triangle ACQ$ are the same triangle, that is, $Q$ is coincident with $B$, giving $\angle BCD = 60°$ and thus $\angle CDA = \fbox{130°}$

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(As another intuition on the two results: Given the measures and angles, you can construct $ABC$ and the ray from $A$ towards $D$; then the "swinging gate" of $CD$ can place $D$ in two places along that ray)

Answer 2

For the triangle $ABC$ we have $\angle BCA=\angle BAC =40º$

Sine rule for the triangle $ABC$

$$\frac{AC}{\sin 100º}=\frac{BC}{\sin 40º}$$

Sine rule for the triange $ACD$

$$\frac{AC}{\sin x}=\frac{DC}{\sin 30º}$$

But $BC=DC$ so

$$\frac{\sin x}{\sin 30º}=\frac{\sin 100º}{\sin 40º}$$

Furthermore

$$\sin 100º=\sin 80º=2\sin 40º \cos 40º$$ then

$$\sin x=\cos40º=\sin 50º =\sin 130º\rightarrow x=50º \quad \text{or} \quad x=130º$$