Answer
$\alpha\approx 42.22^{o}$
Work Step by Step
We find $\beta$ first:$\displaystyle \quad \tan\beta=\frac{21}{50}$
(the right triangle with legs 21 and 50)
$\displaystyle \beta=\tan^{-1}\frac{21}{50}$
The third angle in the triangle containing $\alpha$ and $\beta$ is $(180-65)^{o}=115^{o}$
The sum of interior angles of a triangle is $180^{o}$
$\displaystyle \alpha+\tan^{-1}\frac{21}{50}+115^{o}=180^{o}$
$\displaystyle \alpha=65^{o}-\tan^{-1}\frac{21}{50}$
$\alpha\approx 42.22^{o}$
