Question

In the diagram, $ACDE$ is a trapezium with $AC\parallel ED.$ Given that $\angle EAB={52}^{\circ },\angle CDR={126}^{\circ }$ and $\angle PBC={90}^{\circ }$ and $EQ\parallel DR.$ Then $\angle BCD$ equals

• A. ${36}^{\circ }$
• B. ${74}^{\circ }$
• C. ${54}^{\circ }$
• D. ${38}^{\circ }$

Answer 1

Answer: (B)

${74}^{\circ }$

From the given figure, extend ED to M such that
$\angle RDM=\angle QED={52}^{\circ }$ (Corresponding angles for $QE\parallel RD$)
Now, $\angle MDC={126}^o-\angle RDM$
$\angle MDC={126}^o-{52}^o={74}^{\circ }$
Now, $\angle BCD=\angle MC={74}^{\circ }$ (Corresponding angles for $ED\parallel AC$)