Find the missing angles in the following parallelogram :

In the figure, if $OP\parallel RS, \angle OPQ=110°$ and $\angle QRS=130°$, then $\angle PQR$ is equal to

In the given figure, if line $PQ$ and the line $RS$ intersect at $T$such that $\angle PRT=40°$, and $\angle RPT=95°$ and $\angle TSQ=75°$, then the value of $\angle SQT$ will be

In the adjoining figure, $ABC$ is right-angled at $B$. The point $D$ is on $AC$ such that $BD=BC$ and $BDEF$ is a parallelogram. If $\angle BEF=10°$, then $\angle ADE$ is equal to.

In the figure if $QT\perp PR, \angle TQR=40°$ and $\angle SPR=30°$. Then the value of $x$ and $y$ will be

$ABCD$ is a quadrilateral in which $AB=AC,AD=CD=13 \text{cm}, \angle BAC=20°\text{ and }\angle ADC=100°$. If $BC=12 \text{cm}$, then $AB$ is equal to 

In the given figure, if $BE\perp AC$, $\mathrm{\angle }\mathrm{EBC}={40}^{\circ }$ and $\mathrm{\angle }\mathrm{DAC}={30}^{\circ }$, then the value of $\angle x$ and $\angle y$ are:

In the figure, chord $BC$ is extended to $P$. Tangent from $P$ to the circle is $PA$. $AQ$ is the bisector of $\angle BAC$. If $\angle PAC=x$ and $\angle PCA=y$, then prove that $\angle BAC=y–x$.

In the given figure, if $PQ\parallel RS$, $\angle MXQ=135°$ and $MYR=40°$, then the value of $\angle XMY$ will be

In the figure, chord $BC$ is extended to $P$. Tangent from $P$ to the circle is $PA$. $AQ$ is the bisector of $\angle BAC$. Prove that $\angle PAQ=\frac{y+x}{2}$

$ABCD$ is a parallelogram where $AB=8 \mathrm{cm}, AD=4 \mathrm{cm}$ and $\angle B=120°$. What is the value of $\angle A$?

In the figure, $\angle B=90°, \angle C=44°$

What is the measure of $\angle A$?