## Difficult Trigonometry Problem

The problem is to find
$x$
in the problem below.

$OAB+ABO=90^o \rightarrow AOB=90^0$
and in fact all the angles at
$O$
are right angles. Let
$A)=x$
.

$EO=x tan40, \: BO=xtan30, \: OC=(xtan30)tan10$

Then
$tan \alpha = \frac{(xtan30)tan10}{tan40}=0.1213 \rightarrow \alpha= tan^{-1}(0.1213)=6.92^o$
.