Let $M$ be the intersection of two diagonal, $AC$ and $BD$, of a rhombus $ABCD$, where angle $A<90^\circ$. Construct $O$ on segment $MC$ so that $OB<OC$ and let $t=\frac{MA}{MO}$, provided that $O \neq M$. Construct a circle that has $O$ as centre and goes through $B$ and $D$. Let the intersections between the circle and $AB$ be $B$ and $X$. Let the intersections between the circle and $BC$ be $B$ and $Y$. Let the intersections of $AC$ with $DX$ and $DY$ be $P$ and $Q$, respectively. Express $\frac{OQ}{OP}$ in terms of $t$.
1. **Identify the angles and properties of the rhombus:**
   - Let $\alpha = \angle DAB = \angle BCD$ and $\beta = \angle ABC = \angle CDA$.
   - Since $ABCD$ is a rhombus, $\alpha + \beta = 180^\circ$.

2. **Establish the cyclic nature of quadrilateral $XBYD$:**
   - Since $XBYD$ is cyclic, the sums of the opposite angles are equal to $180^\circ$:
     \[
     \angle BYD = 180^\circ - \angle BXD
     \]
     \[
     \angle XDY = 180^\circ - \angle XBY = 180^\circ - \beta = \alpha
     \]
   - Thus, $\angle XDY \equiv \angle PDQ = \alpha$ is constant.

3. **Apply the sine theorem to triangles $\triangle BDX$ and $\triangle BDY$:**
   - For $\triangle BDX$:
     \[
     \frac{DX}{BC} = \frac{\sin{\widehat{XBD}}}{\sin{\widehat{BXD}}} = \frac{\sin{\frac{\beta}{2}}}{\sin{\widehat{BXD}}}
     \]
   - For $\triangle BDY$:
     \[
     \frac{DY}{BC} = \frac{\sin{\widehat{YBD}}}{\sin{\widehat{BYD}}} = \frac{\sin{\frac{\beta}{2}}}{\sin{(180^\circ - \widehat{BXD})}} = \frac{\sin{\frac{\beta}{2}}}{\sin{\widehat{BXD}}}
     \]

4. **Conclude that $DX = DY$:**
   - Since $\frac{DX}{BC} = \frac{DY}{BC}$, it follows that $DX = DY$.
   - Therefore, $\triangle XYD$ is isosceles, and the perpendicular bisector of $XY$ passes through $D$ and is identical with the bisector of $\angle XDY \equiv \angle PDQ$.

5. **Determine the position of $O$:**
   - Since $(O)$ is the circumcircle of the isosceles triangle $\triangle XYD$, the circumcenter $O$ lies on the perpendicular bisector and the angle bisector.

6. **Relate $MO$ and $MA$:**
   - Denote $\phi = \angle MDO$. Then:
     \[
     MO = MD \tan \phi = MA \tan{\frac{\alpha}{2}} \tan \phi
     \]
   - Given $t = \frac{MA}{MO}$:
     \[
     t = \frac{1}{\tan{\frac{\alpha}{2}} \tan \phi} = \cot{\frac{\alpha}{2}} \cot \phi
     \]

7. **Use the angle bisector theorem in $\triangle DPQ$:**
   - The angle bisector $DO$ divides the opposite side of $\triangle DPQ$ in the ratio of the adjacent sides $OP$ and $OQ$:
     \[
     \frac{OQ}{OP} = \frac{QD}{PD} = \frac{\cos{\left(\frac{\alpha}{2} - \phi\right)}}{\cos{\left(\frac{\alpha}{2} + \phi\right)}}
     \]

8. **Simplify the ratio using trigonometric identities:**
   - Using the sum-to-product identities:
     \[
     \frac{\cos{\left(\frac{\alpha}{2} - \phi\right)}}{\cos{\left(\frac{\alpha}{2} + \phi\right)}} = \frac{\cos{\frac{\alpha}{2}} \cos \phi + \sin{\frac{\alpha}{2}} \sin \phi}{\cos{\frac{\alpha}{2}} \cos \phi - \sin{\frac{\alpha}{2}} \sin \phi}
     \]
   - Substitute $\cot{\frac{\alpha}{2}} \cot \phi = t$:
     \[
     \frac{\cos{\frac{\alpha}{2}} \cos \phi + \sin{\frac{\alpha}{2}} \sin \phi}{\cos{\frac{\alpha}{2}} \cos \phi - \sin{\frac{\alpha}{2}} \sin \phi} = \frac{\cot{\frac{\alpha}{2}} \cot \phi + 1}{\cot{\frac{\alpha}{2}} \cot \phi - 1} = \frac{t + 1}{t - 1}
     \]

The final answer is $\boxed{\frac{t + 1}{t - 1}}$