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Lines and Angles - Classify rays and angles and compute angle measures accurately

Grade 9CBSE

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

🔑Concepts

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A Ray is a part of a line that starts at a fixed point (endpoint) and extends infinitely in one direction. An Angle is formed when two rays originate from the same endpoint (vertex).

An angle formed by two rays OA and OB starting from vertex O.
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Angles are classified by their measure: Acute (0∘<θ<90∘0^\circ < \theta < 90^\circ), Right (θ=90∘\theta = 90^\circ), Obtuse (90∘<θ<180∘90^\circ < \theta < 180^\circ), Straight (θ=180∘\theta = 180^\circ), and Reflex (180∘<θ<360∘180^\circ < \theta < 360^\circ).

A reflex angle showing a rotation greater than 180 degrees.
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When two lines intersect, the Vertically Opposite Angles are equal. These angles are formed opposite to each other at the vertex.

Two intersecting lines showing vertically opposite angles are equal.
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Two angles are Adjacent if they have a common vertex, a common arm, and their non-common arms are on different sides of the common arm.

📐Formulae

Sum of Complementary Angles: ∠A+∠B=90∘\angle A + \angle B = 90^\circ

Sum of Supplementary Angles: ∠A+∠B=180∘\angle A + \angle B = 180^\circ

Linear Pair Axiom: ∠x+∠y=180∘\angle x + \angle y = 180^\circ

Reflex Angle calculation: Reflex ∠A=360∘−∠A\text{Reflex } \angle A = 360^\circ - \angle A

💡Examples

Problem 1:

Find the measure of an angle which is 24∘24^\circ more than its complement.

Solution:

  1. Let the measure of the required angle be xx.
  2. Its complement will be (90∘−x)(90^\circ - x).
  3. According to the problem: x=(90∘−x)+24∘x = (90^\circ - x) + 24^\circ
  4. Simplify the equation: x=114∘−xx = 114^\circ - x
  5. 2x=114∘2x = 114^\circ
  6. x=114∘2=57∘x = \frac{114^\circ}{2} = 57^\circ

Explanation:

We use the definition of complementary angles (sum is 90∘90^\circ) to set up a linear equation based on the given condition.

Problem 2:

In a linear pair, the ratio of two adjacent angles is 2:32:3. Find the measure of both angles.

Solution:

  1. Let the two angles be 2x2x and 3x3x.
  2. Since they form a linear pair, their sum is 180∘180^\circ.
  3. 2x+3x=180∘2x + 3x = 180^\circ
  4. 5x=180∘5x = 180^\circ
  5. x=180∘5=36∘x = \frac{180^\circ}{5} = 36^\circ
  6. First angle: 2×36∘=72∘2 \times 36^\circ = 72^\circ
  7. Second angle: 3×36∘=108∘3 \times 36^\circ = 108^\circ

Explanation:

Using the Linear Pair Axiom, we sum the ratio-based components to 180∘180^\circ to find the common multiplier xx, then calculate individual angles.

Problem 3:

In the given figure, lines ABAB and CDCD intersect at OO. If ∠AOC+∠BOE=70∘\angle AOC + \angle BOE = 70^\circ and ∠BOD=40∘\angle BOD = 40^\circ, find ∠BOE\angle BOE and reflex ∠COE\angle COE.

Intersecting lines AB and CD with an additional ray OE from O.

Solution:

  1. ∠AOC=∠BOD\angle AOC = \angle BOD (Vertically opposite angles).
  2. Since ∠BOD=40∘\angle BOD = 40^\circ, then ∠AOC=40∘\angle AOC = 40^\circ.
  3. Given ∠AOC+∠BOE=70∘\angle AOC + \angle BOE = 70^\circ, so 40∘+∠BOE=70∘  ⟹  ∠BOE=30∘40^\circ + \angle BOE = 70^\circ \implies \angle BOE = 30^\circ.
  4. ABAB is a straight line, so ∠AOC+∠COE+∠BOE=180∘\angle AOC + \angle COE + \angle BOE = 180^\circ.
  5. 40∘+∠COE+30∘=180∘  ⟹  ∠COE=110∘40^\circ + \angle COE + 30^\circ = 180^\circ \implies \angle COE = 110^\circ.
  6. Reflex ∠COE=360∘−110∘=250∘\angle COE = 360^\circ - 110^\circ = 250^\circ.

Explanation:

We use the properties of vertically opposite angles to find ∠AOC\angle AOC, then use the given sum to find ∠BOE\angle BOE. Finally, we use the straight line property to find ∠COE\angle COE and subtract it from 360∘360^\circ for the reflex angle.

Problem 4:

In the figure, ∠PQR=∠PRQ\angle PQR = \angle PRQ. Prove that ∠PQS=∠PRT\angle PQS = \angle PRT.

Triangle PQR on a straight line ST extending from S through Q and R to T.

Solution:

  1. SQRSQR is a straight line, so ∠PQS+∠PQR=180∘\angle PQS + \angle PQR = 180^\circ (Linear pair).
  2. Similarly, QRTQRT is a straight line, so ∠PRQ+∠PRT=180∘\angle PRQ + \angle PRT = 180^\circ (Linear pair).
  3. Therefore, ∠PQS+∠PQR=∠PRQ+∠PRT\angle PQS + \angle PQR = \angle PRQ + \angle PRT.
  4. Since it is given that ∠PQR=∠PRQ\angle PQR = \angle PRQ, we can subtract this equal value from both sides.
  5. Hence, ∠PQS=∠PRT\angle PQS = \angle PRT.

Explanation:

This proof relies on the Linear Pair Axiom which states that angles on a straight line add up to 180∘180^\circ. If inner angles are equal, their supplementary outer angles must also be equal.