What Are The Different Types Of Angles

What Are The Different Types Of Angles

Angle:
Two rays with a common end point form an angle.
What Are The Different Types Of Angles 1OA, OB are rays & O is end point.

Types of Angles

In geometry, angles can be classified according to the size (or magnitude) of the angle.

  • Right angle: An angle whose measure is 90°, is called a right angle. Two lines that meet at a right angle are said to be perpendicular.
  • Acute angle: An angle whose measure is less than 90°, is called an acute angle.
  • Obtuse angle: An angle whose measure is greater than 90° and less than 180°, is called an obtuse angle.
  • Straight angle: An angle whose measure is 180° is called a straight angle.
  • Reflex angle: An angle whose measure is more than 180° but less than 360°, is called a reflex angle.
  • Complete angle: An angle whose measure is 360° is called a complete angle.
  • Zero angle: An angle whose measure is 0° is called a zero angle.

Measure of an angle:
The amount of turning from OA to OB is called the measure of ∠AOB, written as m ∠AOB. An angle is measured in degrees denoted by ‘°’.
What Are The Different Types Of Angles 2

  • An angle of 360°:
    If a ray OA starting from its original position OA, rotates about O, in the anticlockwise direction and after making a complete revolution it comes back to its original position, we say that it has rotated through 360 degrees, written as 360°.
    What Are The Different Types Of Angles 3
    This complete rotation is divided into 360 equal parts. Each part measures 1°.
    1° = 60 minutes, written as 60′.
    1′ = 60 seconds, written as 60”.
    We use a protractor to measure an angle.
  • Bisector of an angle:  
    A ray OC is called the bisector of ∠AOB, if m∠AOC = m∠BOC.
    What Are The Different Types Of Angles 4
    In this case, ∠AOC = ∠BOC = 1/2 ∠AOB.
  • Complementary Angles:
    Two angles are said to be complementary if the sum of their measures is 90°, and each angle is said to be complement of each other.
    For example, (45°, 45°), (10°, 80°), (20°, 70°), (30°, 60°).
    Note:
    (a) If two angles are complement of each other then each angle is an acute angle, but any two acute angles need not be complementary, for example, 20° and 60° are acute angles but are not complement of each other.
    (b) Two obtuse angles and two right angles cannot be complement of each other.
  • Supplementary Angles:
    Two angles are said to be supplementary, if the sum of their measures is 180°.
    Two supplementary angles are called the supplement of each other.
    For example, (10°, 170°), (20°, 160°), (30°, 150°), (40°, 140°), (50°, 130°) etc. are all pairs of supplementary angles.
    Note:
    (a) Two acute angles cannot be supplement of each other.
    (b) Two right angles are always supplementary.
    (c) Two obtuse angles cannot be supplement of each other.
  • Adjacent Angles:
    Two angles are called adjacent angles, if
    (i) they have the same vertex,
    (ii) they have a common arm and
    (iii) their non-common arms are on either side of the common arm.
    adjacent-angles
    In the given figure, ∠AOC and ∠BOC are adjacent angles having the same vertex O, a common arm OC and their non-common arms OA and OB on either side of OC.
  • Linear Pair Angles
    Two adjacent angles are said to form a linear pair if their non-common arms are two opposite rays.
    In figure, ∠POQ and ∠QOR form a linear pair as their non-common arms OP and OR are two opposite rays i.e., POR is a line.
    What Are The Different Types Of Angles 6
    Note:
    (a) Two linear pair angles can also be adjacent angles but it is not necessary that two adjacent angles will be linear pair angles.
    (b) A pair of supplementary angles forms a linear pair when placed adjacent to each other.
  • Vertically Opposite Angles
    Two angles formed by two intersecting lines having no common arm are called vertically opposite angles.
    In figure, two lines PQ and RS are intersecting at point O. We observe that with the intersection of these lines, four angles have been formed.
    What Are The Different Types Of Angles 5
    ∠POR and ∠SOQ form a pair of vertically opposite angles, while ∠POS and ∠ROQ form another pair of vertically opposite angles.
    Note:
    Vertically opposite angles are always equal.
    Important Facts:
    (1) The sum of all angles formed on the same side of a line at a given point on a line is 180°.
    (2) The sum of all angles around a point is 360°.

Types Of Angles Example Problems With Solutions

Example 1:    Find the measure of an angle which is 20° more than its complement.
Soluton:    Let the measure of the required angle be x°.
Then, measure of its complement = (90 – x)°.
∴  x – (90 – x) = 20  ⇔  2x = 110  ⇔  x = 55
Hence, the measure of the required angle is 55°.

Example 2:    Find the measure of an angle which is 40° less than its supplement.
Soluton:    Let the measure of the required angle be x°.
Then, measure of its supplement = (180 – x)°.
∴  (180 – x) – x = 40  ⇔  2x = 140  ⇔  x = 70
Hence, the measure of the required angle is 70°.

Example 3:    Find the measure of an angle, if six times its complement is 12° less than twice its supplement.
Soluton:    Let the measure of the required angle be x°.
Then, measure of its complement = (90 – x)°.
Measure of its supplement = (180 – x)°.
∴  6(90 – x) = 2(180 – x) – 12
⇔  540 – 6x = 360 – 2x – 12
⇔  4x = 192  ⇔  x = 48.
Hence the measure of the required angle is 48°.

Example 4:    Convert 180° in degree, minute & second.
Soluton:    180° = 179° 59′ 60”.

Example 5:    Find the measure of the supplement of an angle of 87°28’43”.
Soluton:    We may write, 180° = 179°59’60”.
∴supplement of an angle of (87°28’43”)
= an angle of [180° – (87°28’43”)]
= an angle of [179°59’60” – 87°28’43”]
= an angle of (92°31’17”).
Hence, the measure of the required angle
= (92°31’17”).

Example 6:    If ∠A = 36°27’46” and ∠B = 28°43’39”,
find ∠A + ∠B.
Soluton:    ∠A + ∠B = 36°27’46” + 28°43’39”
= 64°70’85”
= 64°71’25”
= 65°11’25”

Example 7:    Find the complement of each of the following angles :
(i) 58°       (ii) 16°
(iii) 2/3 of a right angle       (iv) 46° 30′
Soluton:    (i) 90° – 58° = 32°
(ii) 90° – 16° = 74°
(iii) 90° – 2/3 (90°)
= 90° – 60° = 30°
(iv) 90° – 46° 30′
= 89° 60′ – 46° 30′
= 43° 30′

Example 8:    Find the measure of an angle which is complement of itself.
Soluton:    Let the measure of the angle be xº, Then,
Then, the measure of its complement is given to be xº.
Since, the sum of the measures of an angle and its complement is 90º
xº + xº = 90º
⇒ 2xº = 90º
⇒ xº = 45º

Example 9:    Find the measure of an angle which forms a pair of supplementary angles with itself.
Soluton:    Let the measure of the angle be xº. Then,
xº + xº = 180º
⇒ 2xº = 180º
⇒ xº = 90º

Example 10:    An angle is equal to five times its complement. Determine its measure.
Soluton:    Let the measure of the given angle be x degrees. Then, its complement is (90 – x)º.
It is given that :
Angle = 5 × Its complement
⇒ x = 5(90 – x)
⇒ x = 450 – 5x
⇒ 6x = 450
⇒ x = 75
Thus, the measure of the given angles is 75º.

Example 11:    An angle is equal to one-third of its supplement. Find its measure.
Soluton:    Let the measure of the required angle be x degrees. Then,
Its supplement = 180º – x. It is given that:
Angle = 1/3 (Its supplement)
⇒ x = 1/3 (180º – x)
⇒ 3x = 180º – x
⇒ 4x = 180º
⇒ x = 45º
Thus, the measure of the given angle is 45º.

Example 12:    Two supplementary angles are in the ratio 2 : 3. Find the angles.
Soluton:    Let the two angles be 2x and 3x in degrees. Then,
2x + 3x = 180º
⇒ 5x = 180º
⇒ x = 36º
Thus, the two angles are 2x = 2 × 36º = 72º
and 3x = 3 × 36º = 108º

Example 13:    Write the complement of the following angles: 30º 20´
Soluton:    Complement of
30º20′ = 90º – 30º 20′
= 90º – (30º + 20′)
= (89º – 30º) + (1º – 20′)
= 59º + (60′ – 20′)          [∵ 1º = 60′]
= 59º + 40′ = 59º 40′

Example 14:    Find the supplement of the following angles : 134º 30′ 26”
Soluton:    Supplement of an angle of 134º 30′ 26”
= 180º – (134º 30′ 26”)
= (179º – 134º) + (1º – 30′ 26”)
= 45º + (60′ – (30′ + 26”))       [∵ 1º = 60′]
= 45º + (59′ – 30′) + (1′ – 26”)
= 45º + 29′ + 34” = 45º29′ 34”

Linear Pair Of Angles

Linear Pair Of Angles

Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays.
Linear-Pair-Of-Angles
In the adjoining figure, ∠AOC and ∠BOC are two adjacent angles whose non-common arms OA and OB are two opposite rays, i.e., BOA is a line
∴ ∠AOC and ∠BOC form a linear pair of angles.

Theorem 1:
Prove that the sum of all the angles formed on the same side of a line at a given point on the line is 180°.
Given: AOB is a straight line and rays OC, OD and OE stand on it, forming ∠AOC, ∠COD, ∠DOE and ∠EOB.
Linear-Pair-Of-Angle-theorem-1
To prove: ∠AOC + ∠COD + ∠DOE + ∠EOB = 180°.
Proof: Ray OC stands on line AB.
∴ ∠AOC + ∠COB = 180°
⇒ ∠AOC + (∠COD + ∠DOE + ∠EOB) = 180°
[∵ ∠COB = ∠COD + ∠DOE + ∠EOB]
⇒ ∠AOC + ∠COD + ∠DOE + ∠EOB = 180°.
Hence, the sum of all the angles formed on the same side of line AB at a point O on it is 180°.

Theorem 2:
Prove that the sum of all the angles around a point is 360°.
Given: A point O and the rays OA, OB, OC, OD and OE make angles around O.
To prove: ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
Construction: Draw a ray OF opposite to ray OA.
Proof: Since ray OB stands on line FA,
Linear-Pair-Of-Angle-theorem-2
we have, ∠AOB + ∠BOF = 180°   [linear pair]
∴ ∠AOB + ∠BOC + ∠COF = 180°             ….(i)
[∵ ∠BOF = ∠BOC + ∠COF]
Again, ray OD stands on line FA.
∴ ∠FOD + ∠DOA = 180° [linear pair]
or ∠FOD + ∠DOE + ∠EOA = 180°               …(ii)
[∵ ∠DOA = ∠DOE + ∠EOA]
Adding (i) and (ii), we get,
∠AOB + ∠BOC + ∠COF + ∠FOD + ∠DOE + ∠EOA = 360°
∴ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
[∵ ∠COF + ∠FOD = ∠COD]
Hence, the sum of all the angles around a point O is 360°.

Linear Pair Of Angles Example Problems With Solutions

Example 1:    In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ∠AOC, ∠COD and ∠BOD.
Linear-Pair-Of-Angles-Example-1
Solution:    (3x + 7)° + (2x – 19)° + x° = 180′ (linear pair)
⇒ 6x – 12) = 180°
⇒ 6x = 192°
⇒ x = 32°
∴ ∠AOC = 3x + 7 = 3(32) + 7 = 96 + 7 = 103°
∠COD = 2x – 19 = 2(32) – 19 = 64 – 19 = 45°
∠BOD = x° = 32°.

Example 2:    In figure, OA, OB are opposite rays and ∠AOC + ∠BOD = 90°. Find ∠COD.
Linear-Pair-Of-Angles-Example-2
Solution:    Since OA and OB are opposite rays. Therefore, AB is a line. Since ray OC stands on line AB.
∴ ∠AOC + ∠COB = 180°
⇒ ∠AOC + ∠COD + ∠BOD = 180°
[∵ ∠COB = ∠COD + ∠BOD]
⇒ (∠AOC + ∠BOD) + ∠COD = 180°
⇒ 90° + ∠COD = 180°
[∵ ∠AOC + ∠BOD = 90° (Given)]
⇒ ∠COD = 180° – 90° = 90°

Example 3:    In figure, OP bisects ∠BOC and OQ, ∠AOC. Show that ∠POQ = 90°.
Linear-Pair-Of-Angles-Example-3
Solution:    According to question, OP is bisector of ∠BOC
Linear-Pair-Of-Angles-Example-3-1

Example 4:    In figure OA and OB are opposite rays :
Linear-Pair-Of-Angles-Example-4
(i) If x = 75, what is the value of y ?
(ii) If y = 110, what is the value of x ?
Solution:    Since ∠AOC and ∠BOC form a linear pair.
Therefore, ∠AOC + ∠BOC = 180º
⇒ x + y = 180º        …(1)
(i) If x = 75, then from (i)
75 + y = 180º
y = 105º.
(ii) If y = 110 then from (i)
x + 110 = 180
⇒ x = 180 – 110 = 70.

Example 5:    In figure ∠AOC and ∠BOC form a linear pair. Determine the value of x.
Linear-Pair-Of-Angles-Example-5
Solution:    Since ∠AOC and ∠BOC form a linear pair.
∴ ∠AOC + ∠BOC = 180º
⇒ 4x + 2x = 180º
⇒ 6x = 180º
⇒ x = 180/6 = 30º
Thus, x = 30º

Example 6:    In figure OA, OB are opposite rays and
∠AOC + ∠BOD = 90º. Find ∠COD.
Linear-Pair-Of-Angles-Example-6
Solution:    Since OA and OB are opposite rays. Therefore, AB is a line. Since ray OC stands on line AB. Therefore,
∠AOC + ∠COB = 180º      [Linear Pairs]
⇒ ∠AOC + ∠COD + ∠BOD = 180º
[∵ ∠COB = ∠COD + ∠BOD]
⇒ (∠AOC + ∠BOD) + ∠COD = 180º
⇒ 90º + ∠COD = 180º
[∵ ∠AOC + ∠BOD = 90º (Given)]
⇒ ∠COD = 180º – 90º
⇒ ∠COD = 90º

Example 7:    In figure ray OE bisects angle ∠AOB and OF is a ray opposite to OE. Show that
∠FOB = ∠FOA.
Linear-Pair-Of-Angles-Example-7
Solution:    Since ray OE bisects angle AOB. Therefore,
∠EOB = ∠EOA ….(i)
Now, ray OB stands on the line EF.
∴ ∠EOB + ∠FOB = 180º     …(ii)       [linear pair]
Again, ray OA stands on the line EF.
∴ ∠EOA + ∠FOA = 180º    ….(iii)
Form (ii) and (iii), we get
∠EOB + ∠FOB = ∠EOA + ∠FOA
⇒ ∠EOA + ∠FOB = ∠EOA + ∠FOA
[∵ ∠EOB = ∠EOA (from (i)]
⇒ ∠FOB = ∠FOA.

Example 8:    In figure OE bisects ∠AOC, OF bisects ∠COB and OE ⊥OF. Show that A, O, B are collinear.
Linear-Pair-Of-Angles-Example-8
Solution:    Since OE and OF bisect angles AOC and COB respectively. Therefore,
∠AOC = 2∠EOC ….(i)
and ∠COB = 2∠COF ….(ii)
Adding (i) and (ii), we get
∠AOC + ∠COB = 2∠EOC + 2∠COF
⇒ ∠AOC + ∠COB = 2(∠EOC + ∠COF)
⇒ ∠AOC + ∠COB = 2(∠EOF)
⇒ ∠AOC + ∠COB = 2 × 90º
[∵ OE ⊥ OF ∴ ∠EOF = 90º]
⇒ ∠AOC + ∠COB = 180º
But ∠AOC and ∠COB are adjacent angles.
Thus, ∠AOC and ∠COB are adjacent supplementary angles. So, ∠AOC and ∠COB form a linear pair. Consequently OA and OB are two opposite rays. Hence, A, O, B are collinear.

Example 9:    If ray OC stands on line AB such that
∠AOC = ∠COB, then show that
∠AOC = 90º.
Solution:    Since ray OC stands on line AB. Therefore,
∠AOC + ∠COB = 180º [Linear pair] …(i)
Linear-Pair-Of-Angles-Example-9
But ∠AOC = ∠COB     (Given)
∴ ∠AOC + ∠ OC = 180º
⇒ 2∠AOC = 180º
⇒ ∠AOC = 90º

Example 10:    In fig if ∠AOC + ∠BOD = 70º, find ∠COD.
Linear-Pair-Of-Angles-Example-10
Solution:    ∠AOC + ∠COD + ∠BOD = 180º
or   (∠AOC + ∠BOD) + ∠COD = 180º
or   70º + ∠COD = 180º
or   ∠COD = 180º – 70º
or   ∠COD = 110º

Example 11:    In fig. find the value of y.
Linear-Pair-Of-Angles-Example-11
Solution:    2y + 3y + 5y = 180º
⇒ 10y = 180º
⇒ y = 180°/10º = 18º

 

Vertically Opposite Angles

Vertically Opposite Angles

Two angles are called a pair of vertically opposite angles, if their arms form two pairs of opposite rays.
Let two lines AB and CD intersect at a point O. Then, two pairs of vertically opposite angles are formed.
(i) ∠AOC and ∠BOD            (ii) ∠AOD and ∠BOC
Vertically-opposite-angles

Theorem 1:
If two lines intersect then the vertically opposite angles are equal.
Given: Two lines AB and CD intersect at a point O.
To prove: (i) ∠AOC = ∠BOD, (ii) ∠AOD = ∠BOC
Proof: Since ray OA stands on line CD, we have:
Vertically-opposite-angles
∠AOC + ∠AOD = 180°     [linear pair].
Again, ray OD stands on line AB.
∴ ∠AOD + ∠BOD = 180°     [linear pair]
∴ ∠AOC + ∠AOD = ∠AOD + ∠BOD   [each equal to 180°]
∴ ∠AOC = ∠BOD
Similarly, ∠AOD = ∠BOC

Vertically Opposite Angles Example Problems With Solutions

Example 1:    Two lines AB and CD intersect at O. If ∠AOC = 50°, find ∠AOD, ∠BOD and ∠BOC.
Vertically-Opposite-Angles-Example-1
Solution:    ∠AOD + ∠AOC = 180° (linear pair)
∠AOD + 50° = 180°
∠AOD = 130°
Also ∠BOD = ∠AOC
(vertically opposite angles)
& ∠BOC = ∠AOD = 130°
(vertically opposite angles)
∵ 130°, 50°, 130°.

Example 2:    Two lines AB and CD intersect at a point O such that ∠BOC + ∠AOD = 280°, as shown in the figure. Find all the four angles.
Vertically-Opposite-Angles-Example-2
Solution:    ∠AOC = ∠BOD = x   (Let)
(vertically opposite angles)
∵ ∠AOC + (∠AOD + ∠BOC) + ∠BOD = 360°
⇒ x + 280° + x = 360°
⇒ 2x = 80°
⇒ x = 40°
∵ ∠AOC = ∠BOD = x° = 40°.
and ∠BOC = ∠AOD = 280°/2 = 140°.

Example 3:    In Fig., lines l1 and l2 intesect at O, forming angles as shown in the figure. If a = 35º, find the values of b, c, and d.
Vertically-Opposite-Angles-Example-3
Solution:    Since lines l1 and l2 intersect at O.
Therefore,
∠a = ∠c           [Vertically opposite angles]
⇒ ∠c = 35º     [∵ ∠a = 35º]
Clearly, ∠a + ∠b = 180º
[Since ∠a and ∠b are angles of a linear pair]
⇒ 35º + ∠b = 180º
⇒ ∠b = 180º – 35º
⇒ ∠b = 145º
Since ∠b and ∠d are vertically opposite angles. Therefore,
∠d = ∠b ⇒ ∠d = 145º       [∵ ∠b = 145º]

Example 4:    In Fig., determine the the value of y.
Vertically-Opposite-Angles-Example-4
Solution:    Since ∠COD and ∠EOF are vertically opposite angles. Therefore,
∠COD = ∠EOF ⇒ ∠COD = 5yº
[∵ ∠EOF = 5yº (Given)]
Now, OA and OB are opposite rays.
∵ ∠AOD + ∠DOC + ∠COB = 180º
⇒ 2yº + 5yº + 5yº = 180º
⇒ 12yº = 180º
⇒ yº = 180º/12 = 15.
Thus, yº = 15.

Example 5:    In Fig., AB and CD are straight lines and OP and OQ are respectively the bisectors of angles BOD and AOC. Show that the rays OP and OQ are in the same line.
Vertically-Opposite-Angles-Example-5
Solution:    In order to prove that OP and OQ are in the same line, it is sufficient to prove that
∠POQ = 180º.
Now, OP is the bisector of ∠AOC
⇒ ∠1 = ∠6           …(i)
and, OQ is the bisector of ∠AOC
⇒ ∠3 = ∠4        ….(ii)
Clearly, ∠2 and ∠5 are vertically opposite angles.
∵ ∠2 = ∠5        ….(iii)
We know that the sum of the angles formed at a point is 360º.
Therefore,
∠1 +∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360º
⇒ (∠1 + ∠6) + (∠3 + ∠4) + (∠2 + ∠5) = 360º
⇒ 2∠1 + 2∠3 + 2∠2 = 360º
[Using (i), (ii) and (iii)]
⇒ 2(∠1 + ∠3 + ∠2) = 360º
⇒ ∠1 + ∠2 + ∠3 = 180º ⇒ ∠POQ = 180º
Hence, OP and OQ are in the same straight line.

Example 6:    In Fig., two staright lines PQ and RS intersect each other at O. If ∠POT = 75º, find the values of a, b and c.
Vertically-Opposite-Angles-Example-6
Solution:    Since OR and OS are in the same line. Therefore,
∠ROP + ∠POT + ∠TOS = 180º
⇒ 4bº + 75º + bº = 180º ⇒ 5bº + 75º = 180º
⇒ 5bº = 105º ⇒ bº = 21
Since PQ and RS intersect at O. Therefore,
∠QOS = ∠POR
[Vertically oppsostie angles]
⇒ a = 4b
⇒ a = 4 × 21 = 84         [∵ b = 21]
Now, OR and OS are in the same line. Therefore.
∠ROQ + ∠QOS = 180º       [Linear pair]
⇒ 2c + a = 180
⇒ 2c + 84 = 180          [∵ b = 84]
⇒ 2c = 96
⇒ c = 48
Hence, a = 84, b = 21 and c = 48