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Lines and Angles - Analyse intersecting lines and angle pairs using formal relationships

Grade 9CBSE

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

🔑Concepts

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When two lines intersect, the angles opposite each other at the vertex are called Vertically Opposite Angles and they are always equal. For lines ABAB and CDCD intersecting at OO, ∠AOC=∠BOD\angle AOC = \angle BOD and ∠AOD=∠BOC\angle AOD = \angle BOC.

Diagram showing two intersecting lines AB and CD meeting at point O forming vertically opposite angles.
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A Linear Pair of angles is formed when two adjacent angles are formed by a ray standing on a straight line. The sum of these angles is always 180∘180^\circ. If ray OCOC stands on line ABAB, then ∠AOC+∠BOC=180∘\angle AOC + \angle BOC = 180^\circ.

Ray OC standing on line AB creating a linear pair of angles.
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The sum of all angles formed around a single point is 360∘360^\circ, representing a complete rotation.

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If the sum of two adjacent angles is 180∘180^\circ, then the non-common arms of the angles form a straight line. This is the converse of the Linear Pair Axiom.

📐Formulae

Linear Pair: ∠1+∠2=180∘\angle 1 + \angle 2 = 180^\circ

Vertically Opposite Angles: ∠AOC=∠BOD\angle AOC = \angle BOD and ∠AOD=∠BOC\angle AOD = \angle BOC (for lines ABAB and CDCD intersecting at OO)

Sum of angles around a point: ∑θ=360∘\sum \theta = 360^\circ

Perpendicular Condition: l⊥m  ⟹  ∠(l,m)=90∘l \perp m \implies \angle(l, m) = 90^\circ

Parallel Condition: l∥m  ⟹  Distance (d) is constantl \parallel m \implies \text{Distance } (d) \text{ is constant}

💡Examples

Problem 1:

Two lines ABAB and CDCD intersect at point OO. If ∠AOC=55∘\angle AOC = 55^\circ, find the measures of ∠BOC\angle BOC, ∠BOD\angle BOD, and ∠AOD\angle AOD.

Solution:

  1. Since ABAB is a straight line and ray OCOC stands on it, ∠AOC\angle AOC and ∠BOC\angle BOC form a linear pair. Therefore, ∠AOC+∠BOC=180∘\angle AOC + \angle BOC = 180^\circ.
  2. Substitute the given value: 55∘+∠BOC=180∘  ⟹  ∠BOC=180∘−55∘=125∘55^\circ + \angle BOC = 180^\circ \implies \angle BOC = 180^\circ - 55^\circ = 125^\circ.
  3. ∠BOD\angle BOD and ∠AOC\angle AOC are vertically opposite angles, so ∠BOD=∠AOC=55∘\angle BOD = \angle AOC = 55^\circ.
  4. ∠AOD\angle AOD and ∠BOC\angle BOC are vertically opposite angles, so ∠AOD=∠BOC=125∘\angle AOD = \angle BOC = 125^\circ.

Explanation:

This problem uses the Linear Pair Axiom to find the adjacent supplement and the property of Vertically Opposite Angles to find the angles across the intersection.

Problem 2:

In a figure, lines XYXY and MNMN intersect at OO. If ∠POY=90∘\angle POY = 90^\circ and a:b=2:3a : b = 2 : 3 (where a=∠POMa = \angle POM and b=∠MOXb = \angle MOX), find the value of c=∠XONc = \angle XON.

Solution:

  1. Since XYXY is a line, ∠XOP+∠POY=180∘\angle XOP + \angle POY = 180^\circ. Given ∠POY=90∘\angle POY = 90^\circ, then ∠XOP=180∘−90∘=90∘\angle XOP = 180^\circ - 90^\circ = 90^\circ.
  2. We know ∠XOP=a+b\angle XOP = a + b, so a+b=90∘a + b = 90^\circ.
  3. Given a:b=2:3a : b = 2 : 3, let a=2xa = 2x and b=3xb = 3x. Then 2x+3x=90∘  ⟹  5x=90∘  ⟹  x=18∘2x + 3x = 90^\circ \implies 5x = 90^\circ \implies x = 18^\circ.
  4. Calculate bb: b=3×18∘=54∘b = 3 \times 18^\circ = 54^\circ.
  5. Since MNMN is a straight line, ∠MOX\angle MOX and ∠XON\angle XON form a linear pair. Thus, b+c=180∘b + c = 180^\circ.
  6. Substitute bb: 54∘+c=180∘  ⟹  c=180∘−54∘=126∘54^\circ + c = 180^\circ \implies c = 180^\circ - 54^\circ = 126^\circ.

Explanation:

The solution first uses the linear pair property on line XYXY to isolate the sum of aa and bb, then uses ratios to find their specific values, and finally applies the linear pair property on line MNMN to find cc.

Problem 3:

In the given figure, lines PQPQ and RSRS intersect each other at point OO. If ∠POR:∠ROQ=5:7\angle POR : \angle ROQ = 5 : 7, find all the angles.

Two intersecting lines PQ and RS meeting at O where POR and ROQ are adjacent angles.

Solution:

Let ∠POR=5x and ∠ROQ=7x\text{Let } \angle POR = 5x \text{ and } \angle ROQ = 7x Since PQ is a line, ∠POR+∠ROQ=180∘ (Linear Pair)\text{Since } PQ \text{ is a line, } \angle POR + \angle ROQ = 180^\circ \text{ (Linear Pair)} 5x+7x=180∘5x + 7x = 180^\circ 12x=180∘  ⟹  x=15∘12x = 180^\circ \implies x = 15^\circ ∠POR=5×15∘=75∘\angle POR = 5 \times 15^\circ = 75^\circ ∠ROQ=7×15∘=105∘\angle ROQ = 7 \times 15^\circ = 105^\circ ∠QOS=∠POR=75∘ (Vertically Opposite Angles)\angle QOS = \angle POR = 75^\circ \text{ (Vertically Opposite Angles)} ∠POS=∠ROQ=105∘ (Vertically Opposite Angles)\angle POS = \angle ROQ = 105^\circ \text{ (Vertically Opposite Angles)}

Explanation:

We use the ratio to express the adjacent angles in terms of xx. Since they form a linear pair on the straight line PQPQ, their sum is 180∘180^\circ. Solving for xx gives the specific measures. Finally, we apply the property of vertically opposite angles to find the remaining two angles.

Problem 4:

In the figure, ray OSOS stands on a line POQPOQ. Ray OROR and ray OTOT are angle bisectors of ∠POS\angle POS and ∠SOQ\angle SOQ, respectively. If ∠POS=x\angle POS = x, find ∠ROT\angle ROT.

Line POQ with ray OS, and bisectors OR and OT of the resulting linear pair.

Solution:

Given ∠POS=x\text{Given } \angle POS = x Since POQ is a line, ∠POS+∠SOQ=180∘ (Linear Pair)\text{Since } POQ \text{ is a line, } \angle POS + \angle SOQ = 180^\circ \text{ (Linear Pair)} ∠SOQ=180∘−x\angle SOQ = 180^\circ - x OR bisects ∠POS  ⟹  ∠ROS=12∠POS=x2\text{OR bisects } \angle POS \implies \angle ROS = \frac{1}{2} \angle POS = \frac{x}{2} OT bisects ∠SOQ  ⟹  ∠SOT=12∠SOQ=180∘−x2=90∘−x2\text{OT bisects } \angle SOQ \implies \angle SOT = \frac{1}{2} \angle SOQ = \frac{180^\circ - x}{2} = 90^\circ - \frac{x}{2} ∠ROT=∠ROS+∠SOT\angle ROT = \angle ROS + \angle SOT ∠ROT=x2+90∘−x2=90∘\angle ROT = \frac{x}{2} + 90^\circ - \frac{x}{2} = 90^\circ

Explanation:

By using the Linear Pair property, we determine the expression for ∠SOQ\angle SOQ. Since OROR and OTOT are bisectors, they halve their respective angles. Adding these two half-angles shows that the angle between the bisectors of a linear pair is always a right angle.