Answer
$\angle A+\angle B+\angle C=\dfrac{\pi}{2}$
Work Step by Step
Here, we have $\tan \angle A=\dfrac{1}{3}$ and $\tan \angle B=\dfrac{1}{2}$
and $\tan \angle C=\dfrac{1}{1}=1$
$\tan (\angle A+\angle B+\angle C)=\dfrac{\tan(\angle A+\angle B)+\tan \angle C}{1-\tan (\angle A+\angle B) \tan \angle C}$
or, $\tan (\angle A+\angle B+\angle C)=\dfrac{\tan\angle A+\tan \angle B}{1-\tan \angle A \tan \angle B}$
or, $\tan (\angle A+\angle B+\angle C)=\dfrac{(\dfrac{1}{3})+(\dfrac{1}{2})}{1-(\dfrac{1}{3})(\dfrac{1}{2})}$
Thus, $(\angle A+\angle B+\angle C)=\tan^{-1}(\infty)=\dfrac{\pi}{2}$
