# Related Angles – Interior, Exterior, Corresponding Angles

Related angles are nothing but the pairs of angles and assigned with specific names that we come across. Related angles have some conditions to mention. Learn the detailed concept of Related angles with images and examples in this article. Improve your preparation level by reading the entire concept without missing any subtopic. We have given complete information about Lines and Angles on our website for free of cost.

## Different Types of Related Angles

Check different types of Related Angles along with examples to clearly understand the concept. We have explained each of them with definitions, solved examples, etc. Refer to the following modules and get a grip on it.

### Complementary Angles

If the sum of the measures of two angles is about 90Â°, those angles are called complementary angles.

#### Facts of Complementary Angles

• The two right angles never complement each other.
• Also, the two obtuse angles never complement each other.
• Two complementary angles are always acute but there is no possibility of vice versa.

Example:

Let us take two angles which are complementary angles. If one angle is a, then the other angle is 90Â° – a.
An angle of 40Â° and another angle of 50Â° are complementary angles of each other.

Also, complement of 40Â° is 90Â° – 40Â° = 50Â°.
And complement of 50Â° is 90Â° – 50Â° = 40Â°

The âˆ XOY is 40Â° and âˆ MON is 50Â°. By adding two angles âˆ XOY and âˆ MON, we get 90Â° as they are the Complementary Angles.

Therefore, the sum of the âˆ XOY and âˆ MON is 90Â°
âˆ XOY + âˆ MON = 90Â°.

### Supplementary Angles

Supplementary Angles are the angles when the sum of the measures of two angles is 180Â°. If the sum of the two angles forms a straight angle, then those angles are called Supplementary Angles. If one angle is a, then the other angle is 180Â° – a.

Example:

Let us take two angles which are Supplementary angles. If one angle is x, then the other angle is 180Â° – x.
An angle of 110Â° and another angle of 70Â° are supplementary angles of each other. Also, a supplement of 110Â° is 180Â° – 110Â° = 70Â°.
And the supplement of 60Â° is 180Â° – 70Â° = 110Â°.

From the above figure, the âˆ XOY is 110Â° and âˆ MON is 70Â°. By adding two angles âˆ XOY and âˆ MON, we get 180Â° as they are the Supplementary Angles.

Therefore, the sum of the âˆ XOY and âˆ MON is 180Â°
âˆ XOY + âˆ MON = 180Â°.

The two angles are called to be Adjacent angles when they have a common arm, a common vertex, and also the non-common arms present on the opposite side of the common arm.

From the above figure, âˆ ABD and âˆ CBD are adjacent angles with the common arm BD. The B is the common vertex and BA, BC is opposite sides of BD.

### Linear Pair

When two adjacent angles form a linear pair of angles with the non-common arms are two opposite rays. In other words, the sum of two adjacent angles is 180Â°.

From the above figure, the âˆ XOY and âˆ XOZ are two adjacent angles. By adding two angles âˆ XOY and âˆ XOZ, we get 180Â°.

Therefore, the sum of the âˆ XOY and âˆ XOZ is 180Â°
âˆ XOY + âˆ XOZ = 180Â°.

### Vertically Opposite Angles

The arms of the lines are opposite in direction and both lines are interesting to each other in Vertically opposite angles. The pair of vertically opposite angles are equal.

From the above figure, the âˆ MOR and âˆ SON and âˆ MOS and âˆ RON are pairs of vertically opposite angles.

### Theorems on Related Angles

1. If a ray stands on a line, then the sum of adjacent angles formed is 180Â°.

Given: A ray BE standing on (AC) âƒ¡ such that âˆ ABE and âˆ CBE are formed.

Construction: Draw BD âŠ¥ AC.

Proof: Take the angle ABE.
Now âˆ ABE = âˆ ABD + âˆ DBE â€¦â€¦â€¦â€¦â€¦. (1)
Take the angle CBE.
Also âˆ CBE = âˆ CBD – âˆ DBE â€¦â€¦â€¦â€¦â€¦. (2)
Now, add equation 1 and equation 2.
âˆ ABE + âˆ CBE = âˆ ABD + âˆ CBD + âˆ DBE – âˆ DBE
âˆ DBE – âˆ DBE = 0
= âˆ ABD + âˆ CBD
âˆ ABD = 90Â°; âˆ CBD = 90Â°
Substitute âˆ ABD and âˆ CBD values in âˆ ABD + âˆ CBD
= 90Â° + 90Â°
= 180Â°

2. The sum of all the angles around a point is equal to 360Â°.

Given: A point O and rays OA, OB, OC, OD, OE which make angles around O.

Construction: Draw OD opposite to ray OA

Proof: Since, OB stands on DA therefore
âˆ AOB + âˆ BOD = 180Â°
âˆ BOD = âˆ BOC + âˆ COD
Substitute âˆ BOD = âˆ BOC + âˆ COD in âˆ AOB + âˆ BOD = 180Â°
âˆ AOB + (âˆ BOC + âˆ COD) = 180Â°
âˆ AOB + âˆ BOC + âˆ COD = 180Â° â€¦â€¦â€¦â€¦â€¦. (i)
Again OE stands on DA, therefore
âˆ DOE + âˆ EOA = 180Â°
âˆ EOA = âˆ EOF + âˆ FOA
Substitute âˆ EOA = âˆ EOF + âˆ FOA in âˆ DOE + âˆ EOA = 180Â°
âˆ DOE + (âˆ EOF + âˆ FOA) = 180Â°
âˆ DOE + âˆ EOF + âˆ FOA = 180Â° â€¦â€¦â€¦â€¦â€¦. (ii)
Now, add euqtion (i) and equation (ii)
âˆ AOB + âˆ BOC + âˆ COD + âˆ DOE + âˆ EOF + âˆ FOA
= 180Â° + 180Â°
= 360Â°

3. If two lines intersect, then vertically opposite angles are equal.

Given: MN and AB intersect at point O.

Proof: OB stands on MN.
Therefore, âˆ MOB + âˆ BON = 180Â° â€¦â€¦â€¦â€¦â€¦. (i)
MO stands on AB
âˆ MOB + âˆ MOA = 180Â° â€¦â€¦â€¦â€¦â€¦. (ii)
From (i) and (ii),
âˆ MOB + âˆ BON = âˆ MOB + âˆ MOA
âˆ BON + âˆ MOA
Similarly, âˆ MOB = âˆ AON can be proved.

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