Check the picture below.
so the height of the pole is "p", and the angle of depression to the foot of the pole is really "44 + a".
so let's use the 44° angle first
[tex]\tan(44^o )=\cfrac{\stackrel{opposite}{30-p}}{\underset{adjacent}{25}}\implies 25\tan(44^o)=30-p\implies 25\tan(44^o)+p=30 \\\\\\ p=30-25\tan(44^o)\implies \boxed{p\approx 9}[/tex]
now, let's use the red triangle, namely the angle "44 + a"
[tex]\tan(44+a)=\cfrac{\stackrel{opposite}{30}}{\underset{adjacent}{25}}\implies \tan(44+a)=\cfrac{6}{5} \\\\\\ 44+a=\tan^{-1}\left( \cfrac{6}{5} \right) \implies 44+a\approx 50^o[/tex]
Make sure your calculator is in Degree mode.
