The angles which a vector $\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\sqrt{2}\mathrm{k}$ makes with X, Y and Z axes respectively are

$\overrightarrow{\mathrm{R}}=\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\sqrt{2}\widehat{\mathrm{k}}$

comparing the given vector with $\overrightarrow{\mathrm{R}}={\mathrm{R}}_{\mathrm{x}}\widehat{\mathrm{i}}+{\mathrm{R}}_{\mathrm{y}}\widehat{\mathrm{j}}+{\mathrm{R}}_{\mathrm{s}}\widehat{\mathrm{k}}$

${\mathrm{R}}_{\mathrm{x}}=1,{\mathrm{R}}_{\mathrm{y}}=\sqrt{2}\mathrm{and}\left|\overrightarrow{\mathrm{R}}\right|=\sqrt{{\mathrm{R}}_{\mathrm{x}}^2+{\mathrm{R}}_{\mathrm{y}}^2+{\mathrm{R}}_{\mathrm{z}}^2}=2$

$\cos \mathrm{\alpha } = \frac{{\mathrm{R}}_{\mathrm{x}}}{\mathrm{R}} = \frac{1}{2} \Rightarrow \mathrm{\alpha } = {60}^0$

$\mathrm{cos\beta } = \frac{{\mathrm{R}}_{\mathrm{y}}}{\mathrm{R}} = \frac{1}{2} \Rightarrow \mathrm{\beta } = {60}^0$

$\mathrm{cos\gamma } = \frac{{\mathrm{R}}_{\mathrm{z}}}{\mathrm{R}}= \frac{1}{\sqrt{2}} \Rightarrow \mathrm{\gamma } = {45}^0$