Lets talk about **lines and angles**

When two lines cross, four angles are formed. “Vertical” angles are equal and look like this:

1.

2. When two pair of vertical anles are formed, four pairs of adjacent angel (side by side) are also formed. Adjacent angles add up to 180^{0}

3. When two parallel lines are crossed by another line, all acute angles are equal, and all obtuse angles are equal. Also, every acute angle is supplementary to every obtuse angle.

To help yourself to see the relationships between angles is parallel line system, you might try looking for there special pictures:

4.

a + b = 180^{0}

5.

с + b = 180^{0}

** 1. Problem**

The figure above shows the intersection of three lines. Find **x**

**Solution**

1. Because ∠AOB and x vertical, so ∠AOB = x.

2. Because ∠AOC , ∠AOB and ∠BOD – adjacent, so they sum = 180^{0}

So from 1 and 2 we have :

2x+ x+ 3x = 180 ⤇ 6x = 180 ⤇ x =30^{0}

** 2. Problem
**

The figure above shows a parallelogram with one side extended. If z = 40, than what would be y ?

**Solution**

1. Looking on the picture we can conclude that: z+ z + z+∠BOC= 180^{0}

2. Because opposite angles in parallelogram are congruent , ∠BOC = y

From 1 and 2 ⤇ z+ z+z +y=180^{0 }⤇ 3 z+y =180 , substitute z by 40 we will get 120 + y =180 ⤇ y = 60 ^{0}

**3. Problem**

In the figure above , if , then what value will be a + b ?

**Solution**

Because * ⤇ ∠*AOC = b and *∠*DOB= a and looking on the picture we can see that:

a + b= 180 + *∠* AOB = 180 + 130= 310

4**. Problem**

In the figure above, if *L _{1}* // L

_{2 }

then what is the velue of n in terms of m ?

**Solution**

First of all, we notes that angles *∠* OBA = *∠*NOC as corresponding anles.

*∠* EON = *∠* BOA =n and *∠ *DOB = *∠ *NOC =m. Let’s look on the picture above. What we can tell about angle *∠*AOE ?

*∠*AOE =180 = *∠*BOA + *∠*BOD + *∠*DOE = n + m + m+5 = 180, hence

n + 2m + 5 = 180 and we will get : n = 175 – 2m.

5**. Problem**

In the diagram above, if , then what would be the **x** ?

**Solution**

Let’s sketch additional lines :

In ∆CDP : *∠* CDP = 180 -50 = 130 and *∠* CPD = *∠* ORA. The angle *∠* ORA will find from ∆ORA :

Noticing, that *∠* ROA = 180-40 =140, then 140 + 30 + *∠* ORA = 180 , so *∠* ORA = 10.

So in ∆CDP : 130 + 10 +x = 180 ⤇ x = 40.