\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}}{2}\Rightarrow B=45^0\)
\(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{1}{2}\Rightarrow A=60^0\)
\(\Rightarrow C=180^0-\left(A+B\right)=75^0\)
\(h_a=\frac{bc.sinA}{a}=\frac{2.\left(\sqrt{3}+1\right)sin60^0}{\sqrt{6}}=\frac{\sqrt{6}+\sqrt{2}}{2}\)