Question

In the given figure, AB is perpendicular to CD; AP // CD, $\angle CBP={138}^{\circ }$. Find the value of$\angle APB$.

• A. ${36}^o$
• B. ${42}^o$
• C. ${56}^o$
• D. none of the above

Answer 1

The correct option is B ${42}^o$

$\angle CBP={138}^{\circ }$

Since, AB is perpendicular CD, $\angle ABC={90}^{\circ }$

$\angle CBP=\angle ABC+\angle ABP$

${138}^{\circ }={90}^{\circ }+\angle ABP$

$\angle ABP={138}^{\circ }-{90}^{\circ }={48}^{\circ }$

Since, AP || CD, hence $\angle BAP=\angle ABC={90}^{\circ }$ (Corresponding angles)

In $△$ ABP,

$\angle BAP+\angle ABP+\angle APB={180}^{\circ }$

${90}^{\circ }+{48}^{\circ }+\angle APB={180}^{\circ }$ (Angle Sum Property)

$\angle APB={42}^{\circ }$