Question

1. find the measure of each missing angle.

m∠1 = ______
m∠2 = ______
m∠3 = ______
m∠4 = ______
m∠5 = ______

Explanation:

Step1: Find \(m\angle1\)

In a right - angled triangle with two equal sides (legs), the two non - right angles are equal. Since the triangle with \(\angle1\) and \(\angle5\) is a right - angled isosceles triangle, \(m\angle1 = 45^{\circ}\) (because the sum of angles in a triangle is \(180^{\circ}\), and \(180^{\circ}-90^{\circ}=90^{\circ}\), then divided equally for the two non - right angles: \(\frac{90^{\circ}}{2}=45^{\circ}\)).

Step2: Find \(m\angle2\)

The larger triangle has a \(35^{\circ}\) angle and is also a right - angled isosceles triangle (by the side markings). The sum of angles in a triangle is \(180^{\circ}\). In the larger right - angled triangle (with the \(35^{\circ}\) angle), the non - right angles: one is \(35^{\circ}\), and since \(m\angle1 + m\angle2+35^{\circ}=90^{\circ}\) (the two non - \(90^{\circ}\) angles in the whole large right - angled triangle), \(m\angle2=90^{\circ}- 35^{\circ}-45^{\circ}=10^{\circ}\).

Step3: Find \(m\angle5\)

As it is a right - angled isosceles triangle (by side markings), \(m\angle5 = 45^{\circ}\) (sum of angles in a triangle \(180^{\circ}\), \(180^{\circ}-90^{\circ}=90^{\circ}\), divided equally: \(\frac{90^{\circ}}{2}=45^{\circ}\)).

Step4: Find \(m\angle4\)

\(m\angle4=180^{\circ}-(m\angle1 + m\angle2 + 35^{\circ})\) (linear pair with the angle in the larger triangle). First, \(m\angle1 + m\angle2+35^{\circ}=45^{\circ}+10^{\circ}+35^{\circ}=90^{\circ}\), so \(m\angle4 = 90^{\circ}\).

Step5: Find \(m\angle3\)

In the larger right - angled triangle (by side markings), \(m\angle3 = 35^{\circ}\) (by the angle - side relationship in an isosceles right - angled triangle - like figure, the non - right angles are equal when the legs are equal, and here by the overall structure and angle sums).

Answer:

\(m\angle1 = 45^{\circ}\)
\(m\angle2 = 10^{\circ}\)
\(m\angle3 = 35^{\circ}\)
\(m\angle4 = 90^{\circ}\)
\(m\angle5 = 45^{\circ}\)