Question

In the parallelogram ABCD, find x.​

• A. ${30}^{\circ }$
• B. ${40}^{\circ }$
• C. ${50}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C

${50}^{\circ }$

From the figure,
$\angle DBC$ = ${75}^{\circ }$ (Vertically opposite angles)
$\angle CDB$ = ${30}^{\circ }$ = $\angle DBA$ (Alternate angles)
$\angle ABC$ = $\angle DBA$ + $\angle DBC$ = ${30}^{\circ }$ + ${75}^{\circ }$ = ${105}^{\circ }$

In $\Delta ABC$, $\angle CAB$ + $\angle ABC$ + $\angle ACB$ = ${180}^{\circ }$ (Angle sum property of a triangle)
${25}^{\circ }$ + ${105}^{\circ }$ +$x$ = ${180}^{\circ }$
130° + $x$ = 180°
$x={180}^{\circ }$ - ${130}^{\circ }$
$x={50}^{\circ }$

Hence,$x$ = 50°