As you can see in the picture, there is a triangle $ABC$ with $∠C=30°$ and $∠B=40°$. Now we assuming that $AB=CD$, try to find the exact value of $∠CAD$.
My attempt: Denote $∠CAD$ by $x$, we know that
$$\frac{\sin C}{AD}=\frac{\sin{x}}{CD},\quad\frac{\sin B}{AD}=\frac{\sin{(x+C)}}{AB}$$
Then we have ($∠C=30°=\frac{\pi}{6},∠B=40°=\frac{2\pi}{9}$)
$$\frac{\sin{\frac{\pi}{6}}}{\sin{x}}=\frac{\sin{\frac{2\pi}{9}}}{\sin{(x+\frac{\pi}{6})}}=\frac{AD}{AB}$$
Everything looks ok so far, but I have trouble solving the equation. What's more, Wolfram tells me that the answer is $x=\frac{5\pi}{18}$.
This exercise is in my sister's assignment, so I think this exercise should have a high-school (or high-school olympic) level answer.
More: The exercise appears in geometry part, so a pure geometric method will be better.
Best Answer
After asking a middle school math teacher, I got the answer as following.
Make an equilateral triangle $EBC$, and then connect $EA$ and make $DF/\!\!/EC$ which intersects $BE$ at point $F$. Now we have the picture above.
Notice that $∠BCA=∠ECA=30°$, we have $AB=AE$. Then from $∠ABC=40°$, we are able to get that $$∠EAB=180°-2∠AEB=180°-2(60°-∠ABC)=140°$$ From $DF/\!\!/EC$, we could obtain that $EF=CD=AB=AE$, which implies that $$∠EAF=∠EFA=\frac{180°-∠AEB}{2}=80°$$ Then $A,D,B,F$ are in a circle, which implies that $∠DAB=∠DFB=60°$ and the answer is clear.