Answer
$\frac{8\sqrt 2-9\sqrt 3}{5}$
Work Step by Step
1. $h(\alpha+\beta)=tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta} $
2. From the given figures, we have $tan\alpha=\frac{\sqrt 3}{3}, tan\beta=-2\sqrt 2$
3. Thus $h(\alpha+\beta)=\frac{\frac{\sqrt 3}{3}-2\sqrt 2}{1+\frac{\sqrt 3}{3}(2\sqrt 2)}=\frac{\sqrt 3-6\sqrt 2}{3+2\sqrt 6}=\frac{\sqrt 3-6\sqrt 2}{3+2\sqrt 6}\times\frac{3-2\sqrt 6}{3-2\sqrt 6}
=\frac{3\sqrt 3-6\sqrt 2-18\sqrt 2+24\sqrt 3}{-15}=\frac{27\sqrt 3-24\sqrt 2}{-15}=\frac{8\sqrt 2-9\sqrt 3}{5}$
