Measuring resistance of a capacitor - unexpected results
You need to take into account that the voltages across the capacitor and the resistor are \$90^{\circ}\$ out of phase. The impedance of a capacitor is 

$$Z = \frac{1}{j\cdot\omega \cdot C}$$

where \$j{\equiv}\sqrt{-1}\$ is the imaginary unit. This makes all the difference. You need to use phasors and complex math.

Your circuit is simple enough, that you can solve it with a trick. Since the voltages are \$90^{\circ}\$ out of phase you can use the property
$$\left | V_C \right |^2=\left | V_{\text{M}2} \right |^2 - \left | V_{\text{M}1} \right |^2$$