Lines and Angles

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 1800

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 = 1800

5.

с + b = 1800

 

 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 = 1800

So from 1 and 2  we have :

2x+ x+ 3x = 180  ⤇    6x = 180   ⤇    x =300

 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= 1800

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

From 1 and 2 ⤇ z+ z+z +y=1800   ⤇ 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  L1 // L2     
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.

 

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