Let PM be the perpendicular from the point P(1,2,3) to the x-y plane .lf $\overrightarrow{OP}$  makes an angle $\mathrm{\theta }$ with the positive direction of the z-axis and $\overrightarrow{OM}$ makes an angle $\mathrm{\varphi }$ with the positive direction of x-axis, where O is the origin and $\mathrm{\theta }$ and $\mathrm{\varphi }$ are acute angles, then

If P be (x, y, z), then from the figure, 

$\begin{array}{r}\mathrm{x}=\mathrm{rsin}\mathrm{\theta cos}\mathrm{\varphi },\mathrm{y}=\mathrm{rsin}\mathrm{\theta sin}\mathrm{\varphi }\text{ and }\mathrm{z}=\mathrm{rcos}\mathrm{\theta }\\ 1=\mathrm{rsin}\mathrm{\theta cos}\mathrm{\varphi },2=\mathrm{rsin}\mathrm{\theta sin}\mathrm{\varphi }\text{ and }3=\mathrm{rcos}\mathrm{\theta }\end{array}$

$\begin{array}{l}\Rightarrow 1^2+2^2+3^2={\mathrm{r}}^2\text{ or }\mathrm{r}=\pm \sqrt{14}\\ \therefore \sin \mathrm{\theta cos}\mathrm{\varphi }=\frac{1}{\sqrt{14}},\sin \mathrm{\theta sin}\mathrm{\varphi }\\ =\frac{2}{\sqrt{14}}\text{ and }\cos \mathrm{\theta }=\frac{3}{\sqrt{14}}\end{array}$

(neglecting negative sign as $\mathrm{\theta }$ and $\mathrm{\varphi }$ are acute) 

$\frac{\sin \mathrm{\theta sin}\mathrm{\varphi }}{\sin \mathrm{\theta cos}\mathrm{\varphi }}=\frac{2}{1}$

Also,  $\mathrm{tan\theta }=\frac{\sqrt{5}}{3}$