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**Hint:**Use the property of the triangle that the sum of the angles of the triangle is $180^\circ $in the triangle AQM, it gives:

\[\angle AQM{\text{ }} + \angle QAM + \angle AMQ{\text{ }} = {\text{ }}180^\circ \]

After that use the property of the sum of the linear pairs of angles, which says that if we have a pair of linear angles then their sum is$180^\circ $.

**Complete step by step solution:**

It is given in the problem that AMN, QRM, and PRN are all straight lines and we have to find the value of$\left( {\alpha + \beta } \right)$.

First, applying the angle sum property in triangle AQM,

We know that the sum of the angles in a triangle is always$180^\circ $. That is,

\[\angle AQM{\text{ }} + \angle QAM + \angle AMQ{\text{ }} = {\text{ }}180^\circ \]

Substitute the values of the given angles$\angle AQM = \alpha $and$\angle QAM = 55^\circ $into the equation.

\[\alpha {\text{ }} + 55^\circ + \angle AMQ{\text{ }} = {\text{ }}180^\circ \]

\[\angle AMQ{\text{ }} = {\text{ }}180^\circ {\text{ }}-{\text{ }}55^\circ {\text{ }} - {\text{ }}\alpha \]

On simplifying the above equation, we get

\[\angle AMQ = 125^\circ {\text{ }} - {\text{ }}\alpha \]

As we have given that AMN is a straight line, we know that the linear pair of angles have the sum$180^\circ $, then

$\angle $AMQ + $\angle $RMN = 180°

Now, substitute the values into the equation:

\[\;125^\circ {\text{ }}-{\text{ }}\alpha {\text{ }} + \angle RMN{\text{ }} = {\text{ }}180^\circ \]

\[\angle RMN{\text{ }} = {\text{ }}55^\circ {\text{ }} + {\text{ }}\alpha \]

It is also given to us in the problem that PRN is a straight line, so we know that the linear pair of angles have the sum $180^\circ $, then

\[\angle PRM{\text{ }} + \angle MRN{\text{ }} = {\text{ }}180^\circ \]

\[125^\circ {\text{ }} + \angle MRN{\text{ }} = {\text{ }}180^\circ \]

\[\angle MRN{\text{ }} = {\text{ }}55^\circ \]

We again use the property of the sum of the angles of the triangle MRN. Then we have,

\[\angle MRN{\text{ }} + \angle RMN + \angle RNM{\text{ }} = {\text{ }}180^\circ \]

Substitute the values into the equation:

\[55^\circ + 55^\circ + \alpha + \beta = 180^\circ \]

\[\alpha + \beta = 70^\circ \]

**$\therefore$The required sum of angles have the value\[\alpha + \beta = 70^\circ \].**

**Note:**

Try to find all the angles of the triangle RMN in terms of $\alpha $using the angle sum property and linear pair of angles property of the triangle.

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