Question

In figure, $AB\parallel DE\parallel PQ$ and $\angle ABC={130}^o,\angle CDE={110}^o$. Find $\angle BCD$.

• A. ${60}^o$
• B. ${30}^o$
• C. ${110}^o$
• D. ${130}^o$

Answer 1

Answer: (A)

$\because PQ\parallel DE$
$\therefore \angle CDE+\angle DCQ={180}^o$ (co-interior angles are supplementary)
${110}^o+\angle DCQ={180}^o(\angle CDE={110}^o)$
$\angle DCQ={70}^o$
Similarly, $AB\parallel PQ$
$\therefore \angle ABC+\angle PCB={180}^o$ (co-interior angles are supplementary)
${130}^o+\angle PCB={180}^o(\angle ABC={130}^o)$
$\angle PCB={50}^o$
$\angle PCB+\angle BCD+\angle DCQ={180}^o$ (Angle on same side of straight line)
${50}^o+\angle BCD+{70}^p={180}^o$
$\angle BCD={180}^o-{50}^o-{70}^o$
$={60}^o$