#### Answer

$40^{0}$

#### Work Step by Step

$\cot \alpha =\tan \theta \Rightarrow \alpha +\theta =\dfrac {\pi }{2}\Rightarrow \cot \left( \theta -10\right) =\tan \left( 2\theta -20\right) \Rightarrow \left( \theta -10\right) +\left( 2\theta -20\right) =\dfrac {\pi }{2}=90\Rightarrow \theta =40^{0}$