Question

10. 22 and 21SupplementarySupp.11. In the figure below, l || m, n || p and m1 = 55°. Find the measure of each angle.22=23 =24 =25 626 =3 47 8→n910111227 =28 =p210 =212 =m

Answer 1

Two angles are supplementary if the sum of its measure is 180°. Usually a pair of supplementary angles looks like this:

In the picture you can see that there are several cases like this. Another geometric property that we are going to use is related to the fact that lines l and m are parallel just like lines n and p. This means that the angles on one interception are exactly the same as those in any other interception. This means that:

`\begin{gathered} \angle1=\angle3=\angle9=\angle11 \\ \angle5=\angle7 \\ \angle2=\angle4=\angle10=\angle12 \\ \angle6=\angle8 \end{gathered}`

Using this, the measure of angle 1 and identifying supplementary angles we should be able to find the 11 remaining measures. So let's start.

As I stated before :

`\angle1=\angle3=\angle9=\angle11`

And since the measure of 1 is 55° then angles 3, 9 and 11 also have a measure of 55°.

If you look at the picture representing supplementary angles you can see that angles 1 and 2 are supplementary. This means that:

`\begin{gathered} \angle1+\angle2=180^(\circ) \\ 55^(\circ)+\angle2=180^(\circ) \\ \angle2=180^(\circ)-55^(\circ) \\ \angle2=125^(\circ) \end{gathered}`

And since:

`\angle2=\angle4=\angle10=\angle12`

Then angles 2, 4, 10 and 12 all measure 125°.

Another pair of supplementary angles is 1 and 5. Repeating what we did with angles 1 and 2 we have that the measure of angle 5 is 125° and since:

`\angle5=\angle7`

Then the measure of angle 7 is also 125°.

Angles 2 and 6 are also supplementary, then:

`\begin{gathered} \angle6+\angle2=180^(\circ) \\ \angle6+125^(\circ)=180^(\circ) \\ \angle6=180^(\circ)-125^(\circ) \\ \angle6=55^(\circ) \end{gathered}`

And since:

`\angle6=\angle8`

Then the measure iof angles 6 and 8 is 55°.

In summary the measure of each angle is:

`\begin{gathered} \angle1=55^(\circ) \\ \angle2=125^(\circ) \\ \angle3=55^(\circ) \\ \angle4=125^(\circ) \\ \angle5=125^(\circ) \\ \angle6=55^(\circ) \\ \angle7=125^(\circ) \\ \angle8=55^(\circ) \\ \angle9=55^(\circ) \\ \angle10=125^(\circ) \\ \angle11=55^(\circ) \\ \angle12=125^(\circ) \end{gathered}`