Question

1. abcd is a rectangle

m∠1=
m∠2=
m∠3=
m∠4=
m∠5=
m∠6=
m∠7=

Explanation:

Step1: Find \(m\angle2\)

Vertical angles are equal. Since the angle opposite to \(\angle2\) is \(116^{\circ}\), \(m\angle2 = 116^{\circ}\)

Step2: Find \(m\angle1\)

In a rectangle, the diagonals are equal and bisect each other. \(\triangle BCD\) is isosceles. The sum of angles in a triangle is \(180^{\circ}\). \(\angle1=\angle4\). Also, \(\angle1 + \angle4+(180 - 116)^{\circ}=180^{\circ}\). So \(2\angle1=116^{\circ}\), \(m\angle1 = 58^{\circ}\)

Step3: Find \(m\angle3\)

\(\angle1+\angle3 = 90^{\circ}\) (angles in a rectangle). So \(m\angle3=90 - 58=32^{\circ}\)

Step4: Find \(m\angle4\)

\(m\angle4=m\angle1 = 58^{\circ}\) (proven in step 2)

Step5: Find \(m\angle5\)

\(\angle3\) and \(\angle5\) are alternate - interior angles (diagonals of rectangle, \(AD\parallel BC\)). So \(m\angle5=m\angle3 = 32^{\circ}\)

Step6: Find \(m\angle6\)

\(\angle5+\angle6 = 90^{\circ}\) (angles in a rectangle). So \(m\angle6=90 - 32 = 58^{\circ}\)

Step7: Find \(m\angle7\)

\(\angle6\) and \(\angle7\) are alternate - interior angles (diagonals of rectangle, \(AD\parallel BC\)). So \(m\angle7=m\angle6 = 58^{\circ}\)

Answer:

\(m\angle1 = 58^{\circ}\), \(m\angle2 = 116^{\circ}\), \(m\angle3 = 32^{\circ}\), \(m\angle4 = 58^{\circ}\), \(m\angle5 = 32^{\circ}\), \(m\angle6 = 58^{\circ}\), \(m\angle7 = 58^{\circ}\)