Question

Find the measure of each numbered angle.

Short answer

The angles are:

$\begin{array}{l}\text{m}\angle \mathbf{1}={\mathbf{75}}^{\circ }\\ \text{m}\angle \mathbf{2}={\mathbf{40}}^{\circ }\\ \text{m}\angle \mathbf{3}={\mathbf{40}}^{\circ }\\ \text{m}\angle \mathbf{4}={\mathbf{65}}^{\circ }\end{array}$

Step-by-step solution

- Consider the diagram. 

Here, $AB\parallel CD$.

- Show the calculation.

Since, $AB\parallel CD$,

Here, $m\angle 4=m\angle C=65°$, since they are corresponding angles, where AC is the transversal and AB is parallel to CD.

$m\angle 3+m\angle AED=180°$ …… [linear angles]

That implies,

$m\angle 3+140°=180°$

Subtract $140°$ from each side, then,

$\begin{array}{c}m\angle 3={180}^{\circ }-{140}^{\circ }\\ m\angle 3={40}^{\circ }\end{array}$

Now, $m\angle 2=m\angle 3=40°$, since they are alternate interior angles, where AE is the transversal and AB is parallel to CD.

$\begin{array}{c}m\angle 1+m\angle 3+m\angle C={180}^{\circ }\\ m\angle 1+{40}^{\circ }+{65}^{\circ }={180}^{\circ }\\ m\angle 1+{105}^{\circ }={180}^{\circ }\end{array}$

Subtract $105°$ from each side, then,

$\begin{array}{c}m\angle 1={180}^{\circ }-{105}^{\circ }\\ m\angle 1={75}^{\circ }\end{array}$

- State the conclusion.

Therefore,

$\begin{array}{l}m\angle 1={75}^{\circ }\\ m\angle 2={40}^{\circ }\\ m\angle 3={40}^{\circ }\\ m\angle 4={65}^{\circ }\end{array}$