Which has greater dipole moment: cis-1,2-dichloroethylene or 1,1-dichloroethylene?
Dipole moment is a vector entity. Any applying dipole moment can be replaced by two perpendicular components:



For convenience to identify, I put $a$, $b$, $c$, and $d$ as the magnitudes of applying dipole moments of relevant $\ce{C-Cl}$ bonds. Yet they are theoretically identical: $a = b = c = d = \mu$. 

Let's consider cis-1,2-dichloroethene first. The horizontal components of vectors $a$ and $b$, $a \sin 30^\circ$ and $-b \sin 30^\circ$, will cancel each other (negative sing is because of opposite direction of the vector). However, vertical components are in the same direction, thus they add together: $a \cos 30^\circ + b \cos 30^\circ = 2\mu \cos 30^\circ = 2\mu \times \frac{\sqrt{3}}{2} = \sqrt{3}\mu$. This is the net dipole moment of cis-1,2-dichloroethene.

Now, consider 1,1-dichloroethene. The vertical components of vectors $c$ and $d$, $-c \cos 30^\circ$ and $d \cos 30^\circ$, will cancel each other (again, negative sing is because of opposite direction of the vector). Here, horizontal components are in the same direction, thus they add together: $c \sin 30^\circ + d \sin 30^\circ = 2\mu \sin 30^\circ = 2\mu \times \frac{1}{2} = \mu$. This is the net dipole moment of 1,1-dichloroethene.

Therefore it is clearly, $\sqrt{3}\mu \gt \mu$. Realistically, $\mu_{1,1}$ is smaller than $\mu_{1,2}$ because of increasing electron density of two chlorine atom attached carbon in 1,1-dichloroethene compared to that of one chlorine atom attached carbon in 1,2-dichloroethene. This can be easily show using MaxW's values: $\mu_{1,1} = \mu = \pu{1.3 D}$ and hence, $\mu_{1,2} = \sqrt{3}\mu = \pu{1.3 D} \times \sqrt{3} = \pu{2.25 D}$, which is actually $\pu{1.9 D}$.