How do you calculate the dipole moment of water?

Let's suppose we put water on the xy-plane like so:

The dipole moment is calculated by looking up the dipole moment contributions from each $\text{O}-\text{H}$ bond, which are polar, and summing them to get the net dipole. Each contribution is $\text{1.5 D}$ (debyes).

The net dipole points through oxygen down the y-axis in the negative direction.

Note that the dipole projection along the x directions cancel each other out; if the left dipole contribution was pointing to +x, the right contribution points to -x.

As a result, to calculate the net dipole, determine the projection of each dipole in the y direction, and then double it, since both $\text{OH}$ bonds are identical.

The $\text{H}-\text{O}-\text{H}$ bond angle of water is pretty much ${104.4776}^{\circ }$ (it is okay to use ${104.5}^{\circ }$).

Take the angle used in the projection to be from the vertical until each $\text{OH}$ bond and you will get ${104.4776}^{\circ }\text{/}2$, then imagine two right triangles.

With that, and the fact that $\cos \theta =\cos (-\theta )$:

${\mu }_{\text{y,left contribution}}={\mu }_{\text{OH}}\times \cos \left({52.2388}^{\circ }\right)$

$=\text{1.5 D}\times 0.612=0.9187$

${\mu }_{\text{y,right contribution}}={\mu }_{\text{OH}}\times \cos (-{52.2388}^{\circ })$

$=\text{1.5 D}\times 0.612=0.9187$

Finally, we sum them up because they are both in the same y direction:

${\mu }_{\text{tot}}={\mu }_{\text{y,left contribution}}+{\mu }_{\text{y,right contribution}}$

$=0.9187+0.9187=\text{1.837 D}$

which is pretty close to the actual $\text{1.85 D}$.

Likely the error was either from the referenced $\text{OH}$ dipole moment or the calculated bond angle (via Hartree-Fock theory using a cc-pVQZ basis set, but you don't need to know that).