Lines and Angles

The chapter on Lines and Angles introduces fundamental geometric concepts. It builds upon the basic understanding of points and lines, exploring their properties and relationships when they interact.

Key Topics

• Basic Geometric Terms: Lines, line segments, rays, collinear points, angles (acute, obtuse, right, straight, reflex, complementary, supplementary, adjacent, linear pair, vertically opposite angles).
• Intersecting Lines: Properties of angles formed when two lines intersect.
• Parallel Lines: Definition, properties of corresponding angles, alternate interior angles, interior angles on the same side of the transversal, and exterior angles.

Core Ideas

• Exploring the properties of parallel lines and the angles formed by a transversal.
• Applying these concepts to solve geometric problems.

This chapter lays the groundwork for more complex geometric concepts and proofs introduced in subsequent chapters.

Exercise 6.1

1. In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Ans : 

Given:

• Lines AB and CD intersect at point O.
• ∠AOC + ∠BOE = 70°
• ∠BOD = 40°

To find:

• ∠BOE
• Reflex ∠COE

Solution

Step 1: Find ∠AOC

Since ∠AOC and ∠BOD are vertically opposite angles, they are equal. Therefore, ∠AOC = ∠BOD = 40°.

Step 2: Find ∠BOE

We know that ∠AOC + ∠BOE = 70°. Substituting the value of ∠AOC, we get: 40° + ∠BOE = 70° So, ∠BOE = 70° – 40° = 30°.

Step 3: Find Reflex ∠COE

We know that ∠AOC + ∠COE + ∠BOE = 180° (angles on a straight line). 

Substituting the values of ∠AOC and ∠BOE, we get: 

40° + ∠COE + 30° = 180° 

∠COE = 180° – 70° = 110° 

Therefore, reflex ∠COE = 360° – ∠COE = 360° – 110° = 250°.

2. In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.

Ans : 

Since XOY is a straight line.

∴ b+a+∠POY= 180°

But ∠POY = 90° [Given]

∴ b + a = 180° – 90° = 90° …(i)

Also a : b = 2 : 3 ⇒ b = 3a/2 …(ii)

Now from (i) and (ii), we get

3a/2 + A = 90°

⇒ 5a/2= 90°

⇒ a = 90∘5×2=36∘

 = 36°

From (ii), we get

b = 3/2 x 36° = 54°

Since XY and MN interstect at O,

∴ c = [a + ∠POY] [Vertically opposite angles]

or c = 36° + 90° = 126°

Thus, the required measure of c = 126°.

3. In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Ans :

We are given a figure where ∠PQR = ∠PRQ. We need to prove that ∠PQS = ∠PRT.

Proof:

Step 1: Identify the relevant angles

• We are given that ∠PQR = ∠PRQ.
• We need to prove that ∠PQS = ∠PRT.

Step 2: Apply the concept of linear pairs

• Angles PQS and PQR form a linear pair. Therefore, ∠PQS + ∠PQR = 180°.
• Angles PRT and PRQ form a linear pair. Therefore, ∠PRT + ∠PRQ = 180°.

Step 3: Establish equality

• Since ∠PQR = ∠PRQ (given), we can substitute ∠PQR for ∠PRQ in the second equation.
• This gives us ∠PRT + ∠PQR = 180°.
• Now we have two equations:
  • ∠PQS + ∠PQR = 180°
  • ∠PRT + ∠PQR = 180°
• Both equations are equal to 180°, so the left-hand sides must be equal.
• Therefore, ∠PQS = ∠PRT.

4. In figure, if x + y = w + ⇒, then prove that AOB is a line.

Ans : 

We are given a figure where four lines intersect at a point O, forming angles x, y, z, and w. We are also given that x + y = w + z. We need to prove that AOB is a straight line.

Proof

Step 1: Analyze the given information

We know that x + y = w + z.

Step 2: Consider the angles around point O

The sum of all angles around a point is 360 degrees. Therefore, x + y + w + z = 360 degrees.

Step 3: Substitute the given information

Since x + y = w + z, we can substitute w + z for x + y in the previous equation. 

So, (w + z) + (w + z) = 360 degrees. 

This simplifies to 2(w + z) = 360 degrees. 

Therefore, w + z = 180 degrees.

Step 4: Conclusion

Since the sum of angles w and z, which are adjacent angles with a common arm OA, is 180 degrees, we can conclude that AOB is a straight line.

Hence proved.

5. In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

Ans : 

We are given a line PQ with ray OR perpendicular to it. Another ray OS lies between OP and OR. We need to prove that ∠ROS = 1/2 (∠QOS – ∠POS).

Proof:

Step 1: Analyze the given information

POQ is a straight line.

OR is perpendicular to PQ, so ∠ROP = ∠ROQ = 90°.

.

Step 2: Introduce variables

Let ∠POS = a, ∠ROS = b, and ∠QOS = c.

Step 3: Form equations

Since POQ is a straight line, ∠POS + ∠ROS + ∠QOS = 180°. 

Therefore, a + b + c = 180°.

Since OR is perpendicular to PQ, ∠ROP = ∠ROQ = 90°. 

Therefore, b + c = 90°.

Step 4: Solve for b

From equation 2, we get c = 90 – b.

Substituting c in equation 1, we get a + b + (90 – b) = 180°.

Simplifying, we get a + 90 = 180°, so a = 90°.

Step 5: Prove the required relation

We know that c – a = (90 – b) – 90 = -b.

Therefore, b = (c – a)/2.

Substituting b with ∠ROS, c with ∠QOS, and a with ∠POS, we get

 ∠ROS = 1/2 (∠QOS – ∠POS).

Hence, proved.

6. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Ans : 

XYP is a straight line.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180°

⇒ 64° + ∠ZYQ + ∠QYP = 180°

[∵ ∠XYZ = 64° (given)]

⇒ 64° + 2∠QYP = 180°

[YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ]

⇒ 2∠QYP = 180° – 64° = 116°

⇒ ∠QYP = 116∘2 = 58°

∴ Reflex ∠QYP = 360° – 58° = 302°

Since ∠XYQ = ∠XYZ + ∠ZYQ

⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP]

⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°]

Exercise 6.2

1. In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Ans : 

1. Identify the relationship between angles:
   • Since AB is parallel to CD, and CD is parallel to EF, we can conclude that AB is also parallel to EF.
   • Therefore, angle x and angle z are corresponding angles.
2. Corresponding Angles Theorem:
   • So, angle x = angle z.
3. Use the given ratio:
   • We are given that y:z = 3:7.
   • Let y = 3k and z = 7k for some constant k.
4. Find the value of x:
   • Since x = z, we can substitute z with 7k in the equation x + y = 180° (angles on a straight line).
   • This gives us x + 3k = 180°.
   • But we also know that x = 7k, so substituting x with 7k, we get 7k + 3k = 180°.
   • Solving for k, we get 10k = 180°, so k = 18.
   • Therefore, x = 7k = 7 * 18 = 126°.

Therefore, the value of angle x is 126°.

2. In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Ans : 

Step 1: Identify the Relationship between Angles

• Since AB is parallel to CD and EF is a transversal, we can use the properties of parallel lines.
• Angle AGE and angle GED are alternate interior angles.
• Angle GEF and angle GED form a right angle triangle (since EF is perpendicular to CD).
• Angle FGE and angle CEG are alternate interior angles.

Step 2: Find the Angles

• Angle AGE: As angle AGE and angle GED are alternate interior angles, they are equal. Therefore, ∠AGE = 126°.
• Angle GEF: Since EF is perpendicular to CD, angle GED is a right angle (90°). So, ∠GEF = ∠GED – ∠FED = 126° – 90° = 36°.
• Angle FGE: As angle FGE and angle CEG are alternate interior angles, they are equal. Since angle CEG is the complement of angle GEF, ∠FGE = 90° – ∠GEF = 90° – 36° = 54°.

Therefore, ∠AGE = 126°, ∠GEF = 36°, and ∠FGE = 54°.

3. In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.

Ans : 

1. Analyze the angles formed:
   • Since PQ || ST and XY || ST, we can say that PQ || XY.
   • Now, angle PQR and angle QRX form a pair of interior angles on the same side of the transversal QR.
   • Similarly, angle RST and angle SRY form a pair of interior angles on the same side of the transversal RS.
2. Calculate the angles QRX and SRY:
   • So, angle PQR + angle QRX = 180°. Therefore, angle QRX = 180° – 110° = 70°.
   • Similarly, angle RST + angle SRY = 180°. Therefore, angle SRY = 180° – 130° = 50°.
3. Find angle QRS:
   • Angles QRX, SRY, and QRS together form a straight line.
   • So, angle QRX + angle SRY + angle QRS = 180°.
   • Substituting the values we found, we get: 70° + 50° + angle QRS = 180°.
   • Therefore, angle QRS = 180° – 70° – 50° = 60°.

Hence, the measure of angle QRS is 60°.

4. In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Ans : 

∴ ∠APQ = ∠PQR

[Alternate interior angles]

⇒ 50° = x [ ∵ ∠APQ = 50° (given)]

∴ ∠APR = ∠PRD [Alternate interior angles]

⇒ ∠APR = 127°

⇒ ∠APQ + ∠QPR = 127°

⇒ 50° + y = 127° [ ∵ ∠APQ = 50° (given)]

⇒ y = 127°- 50° = 77°

Thus, x = 50° and y = 77°.

5. In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Ans :

Draw ray BL ⊥PQ 

and CM ⊥ RS

∵ PQ || RS ⇒ BL || CM

[∵ BL || PQ and CM || RS]

Now, BL || CM and BC is a transversal.

∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]

Since, angle of incidence = Angle of reflection

∠ABL = ∠LBC and ∠MCB = ∠MCD

⇒ ∠ABL = ∠MCD …(2) [By (1)]

Adding (1) and (2), we get

∠LBC + ∠ABL = ∠MCB + ∠MCD

⇒ ∠ABC = ∠BCD

i. e., a pair of alternate interior angles are equal.

∴ AB || CD.