We are given a circle with center $O$. A line $AM$ is tangent to the circle at point $A$. Another line $DM$ intersects the circle at point $D$. $OC$ is a radius of the circle, and we are given that $\angle OCA = 50^\circ$ and $\angle OBC = 20^\circ$. We are asked to find the measure of $\angle M$.

1. Problem Description

We are given a circle with center $O$. A line $AM$ is tangent to the circle at point $A$. Another line $DM$ intersects the circle at point $D$. $OC$ is a radius of the circle, and we are given that $\angle OCA = 50^\circ$ and $\angle OBC = 20^\circ$. We are asked to find the measure of $\angle M$.

2. Solution Steps

First, we can find $\angle ACB$. Since $OC$ and $OB$ are radii of the circle, $OC = OB$. Triangle $OCB$ is an isosceles triangle, so $\angle OCB = \angle OBC = 20^\circ$. Therefore, $\angle ACB = \angle ACO + \angle OCB = 50^\circ + 20^\circ = 70^\circ$.

Next, we recognize that $\angle BAC$ is the angle between a tangent $AM$ and a chord $AC$. By the tangent-chord theorem, $\angle BAC = \angle ACB = 70^\circ$.

Now, let's find $\angle ABC$. We know that $\angle OBC = 20^\circ$. Since $OA$ and $OB$ are radii, $\angle OAB = \angle OBA$. Also $\angle OAC = \angle OCA = 50^\circ$.

$\angle OAB = 90^\circ$ because $AM$ is tangent to the circle at point $A$. Since $OA = OB$, we have $\angle OBA = \angle OAB = 90^\circ$. Therefore $\angle ABC = \angle OBA - \angle OBC = 90^\circ - 20^\circ = 70^\circ$.

In triangle $ABC$, we have $\angle BAC = 70^\circ$ and $\angle ABC = 70^\circ$. Thus, $\angle ACB = 180^\circ - (\angle BAC + \angle ABC) = 180^\circ - (70^\circ + 70^\circ) = 180^\circ - 140^\circ = 40^\circ$. However, we found above that $\angle ACB = 70^{\circ}$.

Let us revisit finding $\angle BAC$. Since $OA \perp AM$, $\angle OAC + \angle CAM = 90^{\circ}$. $\angle CAM = 90^\circ - \angle OAC = 90^\circ - 50^\circ = 40^\circ$.

Now, $\angle ABC = 20^\circ$. In triangle $AMC$, $\angle ACM = \angle ACB = 70^{\circ}$.

In triangle $AMC$, $\angle MAC = 40^{\circ}$, $\angle MCA = 70^{\circ}$. $\angle AMC = 180^{\circ} - (40^{\circ} + 70^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.

In triangle $ACM$, we know that $\angle CAM = 90 - 50 = 40^{\circ}$. We also know that $\angle ACB = 50 + 20 = 70^{\circ}$, so $\angle ACM = 70^{\circ}$. Then $\angle AMC = 180 - 40 - 70 = 70^{\circ}$.

3. Final Answer

$70^\circ$