Question

$\overleftrightarrow{kl}$ is tangent to the circle at $m$. find the measure of major arc $\overarc{mon}$. write your answer as a whole number or a decimal. $m\overarc{mon} = \square ^\circ$

Explanation:

Step1: Use the tangent - radius theorem

Since \(\overleftrightarrow{KL}\) is tangent to the circle at \(M\) and \(OM\) is the radius, \(OM\perp KL\). So, \(\angle OML = 90^{\circ}\). Given \(\angle NML=58^{\circ}\), then \(\angle OMN=\angle OML-\angle NML = 90^{\circ}- 58^{\circ}=32^{\circ}\).

Step2: Use the property of isosceles triangle

\(OM = ON\) (radii of the same circle), so \(\triangle OMN\) is isosceles. Then \(\angle ONM=\angle OMN = 32^{\circ}\).

Step3: Find the measure of \(\angle MON\)

Using the angle - sum property of a triangle (\(\angle OMN+\angle ONM+\angle MON = 180^{\circ}\)), we have \(\angle MON=180^{\circ}-32^{\circ}-32^{\circ}=116^{\circ}\).

Step4: Calculate the measure of major arc \(\overparen{MON}\)

The measure of a full circle is \(360^{\circ}\). The measure of major arc \(\overparen{MON}\) is \(360^{\circ}-\text{measure of minor arc}\overparen{MN}\). Since the measure of the central angle \(\angle MON\) is equal to the measure of minor arc \(\overparen{MN}\), \(m\overparen{MON}=360^{\circ}-116^{\circ}\)

Answer:

\(244\)