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Geometry - Lines and Angles: Pairs of angles (Complementary, Supplementary, Adjacent, Vertically Opposite)

Grade 7ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Complementary Angles: Two angles are said to be complementary if the sum of their measures is 90∘90^{\circ}. Each angle is called the complement of the other. When placed adjacent to each other, they form a right angle.

Diagram showing two adjacent angles x and y forming a 90 degree right angle.
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Supplementary Angles: Two angles are supplementary if their sum is 180∘180^{\circ}. They form a straight line when placed together as a linear pair. One angle is the supplement of the other.

Diagram of a linear pair where angles A and B lie on a straight line, summing to 180 degrees.
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Adjacent Angles: Two angles are adjacent if they have a common vertex, a common arm, and their non-common arms lie on opposite sides of the common arm.

Two angles sharing a common vertex V and a middle arm.
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Vertically Opposite Angles: When two lines intersect at a point, the angles formed opposite to each other are called vertically opposite angles. They are always equal.

Intersecting lines forming two pairs of vertically opposite angles (1=2 and 3=4).

📐Formulae

Complementary Condition: ∠A+∠B=90∘\angle A + \angle B = 90^{\circ}

Supplementary Condition: ∠A+∠B=180∘\angle A + \angle B = 180^{\circ}

Measure of Complement: Complement of x=(90∘−x)\text{Complement of } x = (90^{\circ} - x)

Measure of Supplement: Supplement of x=(180∘−x)\text{Supplement of } x = (180^{\circ} - x)

Vertically Opposite Angles: If lines ABAB and CDCD intersect at OO, then ∠AOC=∠BOD\angle AOC = \angle BOD and ∠AOD=∠BOC\angle AOD = \angle BOC

Sum of Angles in a Linear Pair: θ1+θ2=180∘\theta_1 + \theta_2 = 180^{\circ}

💡Examples

Problem 1:

Find the measure of an angle which is 24∘24^{\circ} less than its supplement.

Solution:

  1. Let the angle be xx.
  2. Its supplement is (180∘−x)(180^{\circ} - x).
  3. According to the question: x=(180∘−x)−24∘x = (180^{\circ} - x) - 24^{\circ}.
  4. Simplify: x=156∘−xx = 156^{\circ} - x.
  5. Move xx to one side: x+x=156∘  ⟹  2x=156∘x + x = 156^{\circ} \implies 2x = 156^{\circ}.
  6. Solve for xx: x=156∘2=78∘x = \frac{156^{\circ}}{2} = 78^{\circ}.

Explanation:

We use the definition of supplementary angles (sum = 180∘180^{\circ}) and set up a linear equation based on the given relationship.

Problem 2:

In an 'X' shape formed by two intersecting lines, one of the angles is 2x+10∘2x + 10^{\circ} and its vertically opposite angle is 3x−20∘3x - 20^{\circ}. Find the value of xx and the measure of these angles.

Solution:

  1. Since vertically opposite angles are equal, we set the expressions equal to each other: 2x+10∘=3x−20∘2x + 10^{\circ} = 3x - 20^{\circ}.
  2. Subtract 2x2x from both sides: 10∘=x−20∘10^{\circ} = x - 20^{\circ}.
  3. Add 20∘20^{\circ} to both sides: x=30∘x = 30^{\circ}.
  4. Substitute xx back into either expression: 2(30∘)+10∘=60∘+10∘=70∘2(30^{\circ}) + 10^{\circ} = 60^{\circ} + 10^{\circ} = 70^{\circ}.
  5. Therefore, both vertically opposite angles are 70∘70^{\circ}.

Explanation:

We apply the property that vertically opposite angles are always equal to solve for the unknown variable xx and then calculate the specific angle measure.

Problem 3:

In the given figure, three lines intersect at a point OO. If ∠AOC=40∘\angle AOC = 40^{\circ} and ∠BOE=70∘\angle BOE = 70^{\circ}, find the value of ∠COD\angle COD.

Three lines intersecting at point O with angles marked 40 and 70 degrees.

Solution:

  1. ∠AOC\angle AOC and ∠BOD\angle BOD are vertically opposite angles. Therefore, ∠BOD=∠AOC=40∘\angle BOD = \angle AOC = 40^{\circ}.
  2. Points C,O,DC, O, D do not lie on a straight line, but let's look at the straight line AOBAOB (assuming ABAB is a line). If ABAB and CDCD are intersecting lines, ∠AOC+∠COE+∠EOB=180∘\angle AOC + \angle COE + \angle EOB = 180^{\circ}.
  3. However, based on the vertically opposite property: ∠COD\angle COD is vertically opposite to ∠FOE\angle FOE (if we name the lines).
  4. Let's solve for xx assuming AOBAOB is a straight line: 40∘+∠COE+70∘=180∘40^{\circ} + \angle COE + 70^{\circ} = 180^{\circ}.
  5. ∠COE=180∘−110∘=70∘\angle COE = 180^{\circ} - 110^{\circ} = 70^{\circ}.
  6. Since ∠COD\angle COD is a straight line, ∠COD=180∘\angle COD = 180^{\circ}.

Explanation:

We use the property that angles on a straight line sum up to 180∘180^{\circ} and vertically opposite angles are equal to find missing measures.

Problem 4:

Find the value of xx in the following figure where OPOP and OQOQ form a linear pair.

A straight line with a ray dividing it into two angles labeled 2x+10 and 3x-20.

Solution:

  1. Since the angles form a linear pair, their sum is 180∘180^{\circ}.
  2. The equation is: (2x+10)∘+(3x−20)∘=180∘(2x + 10)^{\circ} + (3x - 20)^{\circ} = 180^{\circ}.
  3. Combine like terms: 5x−10=1805x - 10 = 180.
  4. Add 10 to both sides: 5x=1905x = 190.
  5. Divide by 5: x=38x = 38.
  6. The angles are: 2(38)+10=76+10=86∘2(38) + 10 = 76 + 10 = 86^{\circ} 3(38)−20=114−20=94∘3(38) - 20 = 114 - 20 = 94^{\circ}

Explanation:

A linear pair consists of adjacent angles whose sum is always 180∘180^{\circ}. By setting up an algebraic equation, we can solve for the unknown variable xx.