In the diagram $m\angle 1=\underset{¯}{?}$, and$m\angle 2=\underset{¯}{?}$

The labelled diagram is:

From the diagram, it can be noticed that:

$\angle BAD=\angle BAC+\angle CAD$

Therefore, it can be obtained that:

$$\begin{gathered} \angle BAD=\angle BAC+\angle CAD \\ =50°+40° \\ =90° \end{gathered}$$

Therefore, the measure of the angle∠BAD is 90°.

The sum of the angles of a triangle is 180°.

Therefore, the sum of the angles of a triangleΔABD is 180°.

Therefore, it can be obtained that:

$$\begin{gathered} \angle BAD+\angle ABD+\angle ADB=180° \\ 90°+55°+\angle 2=180° \\ 145°+\angle 2=180° \\ \angle 2=180°-145° \\ \angle 2=35° \end{gathered}$$

Therefore, the measure of the angle∠2 is 35°.

The sum of the angles of a triangle is 180°.

Therefore, the sum of the angles of a triangleΔACD is 180°.

Therefore, it can be obtained that:

$$\begin{gathered} \angle CAD+\angle ACD+\angle ADC=180° \\ 40°+\angle 2+\angle 1=180° \\ 40°+35°+\angle 1=180° \\ 75°+\angle 1=180° \\ \angle 1=180°-75° \\ \angle 1=105° \end{gathered}$$

Therefore, the measure of the angle ∠1 is 105°.

In the diagram, $m\angle 1=105$and $m\angle 2=35$.