Find $\angle CAD$ in following figure.

Question

In $\triangle ABC$, $\angle ACB =20°$ and $\angle CAB = \angle CBA=80°$

$D$ is a point on $CB$ such that $CD=AB$

Find $\angle CAD$

I assumed $\angle CAD=\theta$ and $AB=CD=x$

Applying sine rule in $\triangle CAD$ and $\triangle ABD$

$$\frac {x} {\sin\theta}=\frac {AD} {\sin 20^\circ}$$

$$\frac {x} {\sin(\theta+20°)}=\frac {AD} {\sin 80^\circ}$$

So,

$$\frac {\sin(\theta+20)} {\sin\theta}=\frac {\sin 80^\circ} {\sin 20^\circ}=\frac {1} {2\sin 10^\circ}$$

I don't think this equation for $\theta$ can be solved without using calculator. If anyone can solve it, please show me the method. (Please do not guess.)

If not solvable, then please provide me any other solution. (If the solution is pure geometric and contains construction, then please also tell me your intuition behind your construction.)

Answer 1

Continue by solving the equation

$$\frac {\sin(\theta+20^\circ)} {\sin\theta}=\frac {1} {2\sin 10^\circ}.$$

as follows

\begin{align} \cot \theta & = \frac1{2\sin 10^\circ\sin 20^\circ}-\cot 20^\circ = \frac{\sin(10^\circ+ 20^\circ)}{\sin 10^\circ\sin 20^\circ}-\cot 20^\circ =\cot 10^\circ \end{align}

Thus, $\theta = 10^\circ$.

Answer 2

Let $a=AB, b=BD, \alpha=∠BDA, \beta=∠BAD$
$b = CB - CD = {a/2 \over \sin(10°)} - a$

For ΔABD, we now have SAS, and use Law of Tangents:
$${a+b \over a-b} = {\tan{\alpha+\beta \over 2} \over \tan{\alpha-\beta \over 2}}$$ $$ {1 \over 4\sin(10°)-1} = {\tan{50°} \over \tan{\alpha-\beta \over 2}}$$ $\begin{align} \tan{\alpha-\beta \over 2} &= (4\sin(10°)-1) \tan(50°) \cr &= {4\sin(10°)\sin(50°) - \sin(50°) \over \cos(50°) }\cr &= {2\cos(40°) - 2\cos(60°) - \cos(40°) \over \sin(40°)}\cr &= {\cos(40°) - 1 \over \sin(40°)}\cr &= {-2\sin^2(20°) \over 2\sin(20°)\cos(20°)}\cr &= \tan(-20°) \end{align}$ $$\beta = {\alpha+\beta \over 2} - {\alpha-\beta \over 2} = 50° + 20° = 70°$$ $$ ∠CAD = ∠CAB - \beta = 80° - 70° = 10° $$