7) Find the missing angles.

Solution:

In the given figure, we have two triangles, \(\triangle AOB\) and \(\triangle COD\).

We are given that \(\angle AOB = 137^{\circ}\).

Angles \(\angle AOB\) and \(\angle COD\) are vertically opposite angles, so \(\angle COD = \angle AOB = 137^{\circ}\).

We are given marks on sides OA and OB, and on sides OC and OD, indicating that \(OA = OB\) and \(OC = OD\).

In \(\triangle AOB\), since \(OA = OB\), it is an isosceles triangle. The angles opposite the equal sides are equal, so \(\angle OAB = \angle OBA\).

The sum of angles in \(\triangle AOB\) is \(180^{\circ}\).

\(\angle OAB + \angle OBA + \angle AOB = 180^{\circ}\)

\(\angle OAB + \angle OBA + 137^{\circ} = 180^{\circ}\)

\(\angle OAB + \angle OBA = 180^{\circ} - 137^{\circ} = 43^{\circ}\)

Since \(\angle OAB = \angle OBA\), let \(x = \angle OAB = \angle OBA\).

\(x + x = 43^{\circ}\)

\(2x = 43^{\circ}\)

\(x = 21.5^{\circ}\)

So, \(\angle OAB = 21.5^{\circ}\) and \(\angle OBA = 21.5^{\circ}\).

Similarly, in \(\triangle COD\), since \(OC = OD\), it is an isosceles triangle. The angles opposite the equal sides are equal, so \(\angle OCD = \angle ODC\).

The sum of angles in \(\triangle COD\) is \(180^{\circ}\).

\(\angle OCD + \angle ODC + \angle COD = 180^{\circ}\)

\(\angle OCD + \angle ODC + 137^{\circ} = 180^{\circ}\)

\(\angle OCD + \angle ODC = 180^{\circ} - 137^{\circ} = 43^{\circ}\)

Since \(\angle OCD = \angle ODC\), let \(y = \angle OCD = \angle ODC\).

\(y + y = 43^{\circ}\)

\(2y = 43^{\circ}\)

\(y = 21.5^{\circ}\)

So, \(\angle OCD = 21.5^{\circ}\) and \(\angle ODC = 21.5^{\circ}\).

The diagram also shows that A, O, C are collinear and B, O, D are collinear. This means AC and BD are straight lines intersecting at O.

The angle marked '0' is at vertex O, which is \(\angle AOD\) and \(\angle BOC\).

\(\angle AOD\) and \(\angle BOC\) are vertically opposite angles to each other. Also, \(\angle AOD\) and \(\angle AOB\) are supplementary angles as they form a straight line AC.

\(\angle AOD + \angle AOB = 180^{\circ}\)

\(\angle AOD + 137^{\circ} = 180^{\circ}\)

\(\angle AOD = 180^{\circ} - 137^{\circ} = 43^{\circ}\)

Similarly, \(\angle BOC + \angle AOB = 180^{\circ}\), so \(\angle BOC = 180^{\circ} - 137^{\circ} = 43^{\circ}\).

The angle marked '0' is likely referring to \(\angle AOD\) or \(\angle BOC\). Let's assume it is asking for \(\angle AOD\) and \(\angle BOC\).

Answer: The missing angles are \(\angle OAB = 21.5^{\circ}\), \(\angle OBA = 21.5^{\circ}\), \(\angle OCD = 21.5^{\circ}\), \(\angle ODC = 21.5^{\circ}\), \(\angle AOD = 43^{\circ}\), \(\angle BOC = 43^{\circ}\).