Question

bc bisects ∠abd
find
m∠1=
m∠2=
m∠3=
m∠4=
m∠5=
m∠6=

Explanation:

Step1: Find \(m\angle1\)

In the left - hand triangle, using the triangle angle - sum theorem (\(m\angle A+m\angle1 + m\angle2=180^{\circ}\)). First, find \(m\angle2\). Since \(\angle2\) and the \(49^{\circ}\) angle are vertical angles, \(m\angle2 = 49^{\circ}\). Then \(m\angle1=180^{\circ}-68^{\circ}-49^{\circ}=63^{\circ}\).

Step2: Find \(m\angle5\)

In the right - hand triangle, using the triangle angle - sum theorem (\(m\angle5 + 43^{\circ}+m\angle6=180^{\circ}\)). First, find \(m\angle6\). Since \(\angle6\) and \(\angle2\) are vertical angles, \(m\angle6 = 49^{\circ}\). Then \(m\angle5=180^{\circ}-43^{\circ}-49^{\circ}=88^{\circ}\).

Step3: Find \(m\angle3\)

Since \(BC\) bisects \(\angle ABD\), and \(\angle ABD=m\angle2 + m\angle4\). Also, \(\angle ABD\) and \(m\angle5\) are vertical angles, so \(m\angle ABD = 88^{\circ}\). Then \(m\angle3=m\angle4=\frac{88^{\circ}}{2}=44^{\circ}\).

Step4: Summarize

• \(m\angle1 = 63^{\circ}\) (from the left - hand triangle angle - sum)
• \(m\angle2 = 49^{\circ}\) (vertical angles)
• \(m\angle3 = 44^{\circ}\) (angle - bisector property)
• \(m\angle4 = 44^{\circ}\) (angle - bisector property)
• \(m\angle5 = 88^{\circ}\) (vertical angles and right - hand triangle angle - sum)
• \(m\angle6 = 49^{\circ}\) (vertical angles)

Answer:

\(m\angle1 = 63^{\circ}\), \(m\angle2 = 49^{\circ}\), \(m\angle3 = 44^{\circ}\), \(m\angle4 = 44^{\circ}\), \(m\angle5 = 88^{\circ}\), \(m\angle6 = 49^{\circ}\)