Question

5.

1. (overline{bc}) bisects (angle acd)

Explanation:

Step1: Find \(m\angle1\)

In a right - triangle, the sum of two non - right angles is \(90^{\circ}\). Given one angle is \(43^{\circ}\) and the sum of \(\angle1\) and \(12^{\circ}\) with \(43^{\circ}\) in a right - triangle.
\(m\angle1=90^{\circ}-(43^{\circ} + 12^{\circ})=35^{\circ}\)

Step2: Find \(m\angle2\)

Since the triangle is a right - triangle, \(\angle2 = 43^{\circ}\) (alternate interior angles or property of the given triangle structure)

Step3: Find \(m\angle3\)

In a right - triangle, \(\angle3=90^{\circ}-(43^{\circ}+35^{\circ}) = 12^{\circ}\)

Step4: For the second problem (7)

First, in \(\triangle ABC\), since \(\angle A = 48^{\circ}\) and \(\angle C=90^{\circ}\), then \(\angle1=42^{\circ}\) (because \(m\angle1=90^{\circ}-48^{\circ}\))
Since \(BC\) bisects \(\angle ACD\), \(\angle2 = 42^{\circ}\)
In \(\triangle BCD\), \(\angle3=90^{\circ}-42^{\circ}=48^{\circ}\)
Using the angle - sum property of a triangle (\(m\angle3 + m\angle4+83^{\circ}=180^{\circ}\)), \(m\angle4=180^{\circ}-(48^{\circ}+83^{\circ}) = 49^{\circ}\)

Answer:

For problem 5: \(m\angle1 = 35^{\circ}\), \(m\angle2 = 43^{\circ}\), \(m\angle3 = 12^{\circ}\)
For problem 7: \(m\angle1 = 42^{\circ}\), \(m\angle2 = 42^{\circ}\), \(m\angle3 = 48^{\circ}\), \(m\angle4 = 49^{\circ}\)