Question

City planners want to design a park between parallel streets, Main Street and Willow Lane, in the shape of a trapezoid. There are two paths of equal length on the east and west sides of the park. The border of the park makes a 60° angle between Willow Lane and the east path. A trapezoid is shown. The west path is the left side and the east path is the right side and they are congruent. Main street is the top side and willow lane is the bottom side and they are parallel. The angle formed by willow lane with the east path is 60 degrees.

Answer 1

Question Completion

• What is the angle between Main Street and the west path?
• What is the angle between the west path and Willow Lane?

Answer:

$(a)120^\circ\\(b)60^\circ$

Step-by-step explanation:

In the diagram BC(Main Street) is parallel to AD(Willow Lane)

Therefore, ABCD is a Trapezoid.

Since in a trapezoid, adjacent angles are supplementary

Therefore:

$\angle C+ \angle D=180^\circ\\\angle C+ 60^\circ=180^\circ\\\angle C=180^\circ- 60^\circ\\\angle C=120^\circ$

Since $AB \cong CD$, ABCD is an Isosceles Trapezoid.

In an Isosceles trapezoid, the base angles are congruent:  

Therefore:$\angle A \cong \angle D and \angle B \cong \angle C$

Therefore:

$\angle A \cong \angle D=60^\circ\\\angle B \cong \angle C=120^\circ$

(a)

The angle between Main Street and the west path is $\angle CBA = 120^\circ$

(b)

The angle between the west path and Willow Lane is $\angle BAD =60^\circ$