Question

In the figure below, $A{A}^{\prime }$ is parallel to $C{C}^{\prime }$. The measure $w$ of $\angle A^{\prime }AB$ is equal to ${135}^{\circ }$ and the measure $z$ of $\angle C^{\prime }CB$ is equal to ${147}^{\circ }$. Find $\angle ABC$.

• A. ${100}^{\circ }$
• B. ${78}^{\circ }$
• C. ${80}^{\circ }$
• D. ${110}^{\circ }$

Answer 1

Answer: (B)

${78}^{\circ }$

Draw $B{B}^{\prime }$ parallel to $A{A}^{\prime }$ and $C{C}^{\prime }$as shown in the figure below.

Note that angle $ABC$ is given by

$\angle ABC=\angle AB{B}^{\prime }+\angle CB{B}^{\prime }$

$\angle w^{\prime }$ and $\angle AB{B}^{\prime }$ are alternate interior angles and their measure are equal.

$\angle AB{B}^{\prime }=\angle w^{\prime }$

$\angle z^{\prime }$ and $\angle CB{B}^{\prime }$ are alternate interior angles and their sizes are equal.

$\angle CB{B}^{\prime }=\angle z^{\prime }$

Angles $w$ and $w^{\prime }$ are supplementary which gives

$w^{\prime }={180}^{\circ }-w$

$={180}^{\circ }-{135}^{\circ }$

$={45}^{\circ }$

Angles $z$ and $z^{\prime }$ are also supplementary which gives

$z^{\prime }={180}^{\circ }-z$

$={180}^{\circ }-{147}^{\circ }$

$={33}^{\circ }$

We now substitute $\angle AB{B}^{\prime }$ by $w^{\prime }$ and $\angle CB{B}^{\prime }$ by $z^{\prime }$ in $\angle ABC=\angle AB{B}^{\prime }+\angle CB{B}^{\prime }$ found above.

$\angle ABC=w^{\prime }+z^{\prime }$

$={45}^{\circ }+{33}^{\circ }$

$={78}^{\circ }$