Angle Measurement

Question

What is the value of $x$ ?

I know that it will be less than 20.

Answer 1

Lets first work out $|AD|$ using the isosceles triangle $ACD$: \begin{align} |AD|&=2z\sin(70°),\\ \end{align} where I define $z\equiv |AB|=|AC|=|CD|$.

Form the isosceles triangle $ABC$ we can work out $|BC|:$ \begin{align} |BC|&=2z\sin(10°).\\ \end{align}

For the triangle $ABD$ one now knows the angle at $A$ which is $20°+20°=40°$ and the two sides $|AB|=z$ and $|AD|$ from the equation above. One can calculate the missing side with the cosine law/rule: \begin{align} |BD|^2&=z^2 +(2z\sin(70°))^2-2z(2z\sin(70°))\sin(40°).\\ \end{align} At last we can use the sine law/rule in the triangle $BCD$ to get $x$ using that the missing angle at $B$ is $360°-140°-80°=140°$: \begin{align} \frac{|BD|}{\sin(140°)}&=\frac{|BC|}{\sin(x)},\\ \sin(x)&=\frac{|BC|\sin(140°)}{|BD|},\\ \sin(x)&=\frac{\sin(10°)\sin(140°)}{\sqrt{1/4+\sin(70°)^2-\sin(70°)\sin(40°)}}. \end{align}

This results in an angle $x\sim8.8°$, which looks plausible to me. I might have screwed up a factor two or a half at some point/angle but I am sure that this way works. There might be another way to get to this or to simplify the last equation but I stopped there.

Answer 2

Let $l$ be the line bisecting $\angle ACD$, and reflect the figure across this line. Notice that the reflection preserves $C$ and switches $A$ and $D$. Let $B'$ be the image of $B$ under this reflection.

Since $\triangle ABC$ is isoceles, with $\angle ACB = \angle ABC$ and $\angle BAC = 20^{\circ}$, it follows that $\angle ACB = 80^{\circ}$, and hence $\angle DCB' = 80^{\circ}$. Thus, \begin{align} \angle ACD + \angle ACB + \angle BCB' + \angle DCB' &= 360^{\circ} \\\implies 140^{\circ} + 80^{\circ} + \angle BCB' + \angle 80^{\circ} &= 360^{\circ}\\\implies \angle BCB' &= 60^{\circ}.\end{align} Now, $BC = B'C$ since $B'$ and $C$ are the images of $B$ and $C$ under the reflection. Hence, $\triangle BCB'$ is a triangle with $BC = B'C$ and $\angle BCB' = 60^{\circ}$, so it must be equilateral (by SAS congruency). Thus, $BB' = BC$. Since $AB = DB'$ due to reflection and $AB = CD$ by hypothesis, it follows that $CD = DB'$. Thus, by SSS congruence, we have $\triangle BCD\cong \triangle BB'D$, so $\angle CDB = \angle B'DB$. In particular, $\angle CDB$ bisects $\angle CDB'$. By reflection, $\angle CDB' = \angle CAB = 20^{\circ}$, so $\angle CDB = \boxed{10^{\circ}}$.

(Thanks to @zar for the picture!)