Given the above problem, the solution seems trivial. Shouldn't it just be $$\dfrac{180^\circ-40^\circ-90^\circ }{2}=25^\circ ?$$
However because this is an olympiad problem, I think I might have gotten the answer wrong. How do you work this problem out? What am I doing wrong?
Considering the following picture, we use the fact that the sum of inner angles of a triangle is 180° to deduce the values 70° then 50° shown below. We then have the relations \begin{align*} b_1&=a_1\tan10=a_2\tan x\\ b_1+b_2&=a_1\tan20=a_2\tan50 \end{align*} from which we deduce $\displaystyle\tan x=\frac{\tan50\tan10}{\tan20}\cdot$
It remains to simplify this expression.
Let's use the formula $\displaystyle\tan\alpha\tan\beta
=\frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}
=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{\cos(\alpha-\beta)+\cos(\alpha+\beta)}$,
with $(\alpha,\beta)=(50,10)$.
We get: $\displaystyle\tan50\tan10=\frac{\cos40-\cos60}{\cos40+\cos60} =\frac{2\cos40-1}{2\cos40+1}=\frac{4\cos^220-3}{4\cos^220-1}\cdot$
Since $\cos(3\alpha)=\cos\alpha(4\cos^2\alpha-3)$, we multiply numerator & denominator by $\cos20$ to obtain $\displaystyle\tan50\tan10=\frac{\cos60}{\cos20(4\cos^220-1)}$, and $\displaystyle\frac{\tan50\tan10}{\tan20} =\frac{\cos60}{\sin20(4\cos^220-1)}\cdot$
Finally, using the formula $2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$ we have \begin{eqnarray*} \sin20(4\cos^220-1)&=&2\cos20(2\sin20\cos20)-\sin20\\ &=&2\cos20\sin40-\sin20\\ &=&(\sin60+\sin20)-\sin20\\ &=&\sin60. \end{eqnarray*}
Conclusion: $\displaystyle\tan x =\frac{\tan50\tan10}{\tan20} =\frac{\cos60}{\sin60} =\tan30$, and $x=30$.
P.S.: the trigonometric computations are rather convoluted; they probably can be simplified but I didn't find out how…