Question

1. in the diagram below of triangle mno, ∠m and ∠o are bisected by

overline{ms} and overline{or}, respectively. segments ms and or intersect at t, and
m∠n = 40°. if m∠tmr = 28°, the measure of angle ots is
a. 40° b. 50° c. 60° d. 70°

Explanation:

Step1: Find \(m\angle OMN\)

Since \(MS\) bisects \(\angle OMN\) and \(m\angle TMR = 28^{\circ}\), then \(m\angle OMN=2\times28^{\circ} = 56^{\circ}\)

Step2: Find \(m\angle MON\)

In \(\triangle MNO\), using the angle - sum property of a triangle (\(m\angle MNO + m\angle MON+m\angle OMN=180^{\circ}\)). Given \(m\angle MNO = 40^{\circ}\) and \(m\angle OMN = 56^{\circ}\), then \(m\angle MON=180^{\circ}-(40^{\circ}+ 56^{\circ})=84^{\circ}\)

Step3: Find \(m\angle MOT\)

Since \(OR\) bisects \(\angle MON\), then \(m\angle MOT=\frac{1}{2}m\angle MON=\frac{1}{2}\times84^{\circ} = 42^{\circ}\)

Step4: Find \(m\angle OTS\)

\(\angle OTS\) is an exterior angle of \(\triangle MOT\). By the exterior - angle theorem (\(m\angle OTS=m\angle MOT + m\angle TMO\)). We know \(m\angle TMO = 28^{\circ}\) and \(m\angle MOT = 42^{\circ}\), so \(m\angle OTS=42^{\circ}+28^{\circ}=70^{\circ}\)

Answer:

d. \(70^{\circ}\)