1In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Answer & Reference:
Given: ∠AOC + ∠BOE = 70° and ∠BOD = 40°
Since AB is a straight line, ∠AOC + ∠COE + ∠BOE = 180°
⇒ 70° + ∠COE = 180° ⇒ ∠COE = 110°
Reflex ∠COE = 360° - 110° = 250°
Also, CD is a straight line: ∠COE + ∠BOE + ∠BOD = 180°
⇒ 110° + ∠BOE + 40° = 180° ⇒ ∠BOE = 30°
(Based on Exercise 6.1, Question 1, Page Reprint 2025-26)
Fig. 6.13: Intersecting lines AB and CD at O
2In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a:b = 2:3, find c.
Answer & Reference:
Given: ∠POY = 90° and a:b = 2:3
Let a = 2x, b = 3x. Since XY is straight: a + b + 90° = 180° ⇒ x = 18°
So a = 36°, b = 54°. Since MN is straight: b + c = 180° ⇒ c = 126°
(Based on Exercise 6.1, Question 2, Page Reprint 2025-26)
Fig. 6.14: Intersecting lines XY and MN at O
3In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Fig. 6.15: ST is a straight line with point P above
4In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
Answer & Reference:
Sum of angles around point O: x + y + w + z = 360°
Given x + y = w + z ⇒ 2(x + y) = 360° ⇒ x + y = 180°
So AOB is a straight line (Linear Pair Axiom).
(Based on Exercise 6.1, Question 4, Page Reprint 2025-26)
Fig. 6.16: Angles around point O
5In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = ½(∠QOS - ∠POS).
6It 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.
7In Fig. 6.23, if AB || CD, CD || EF and y:z = 3:7, find x.
Answer & Reference:
AB || CD || EF ⇒ x = z (alt. int. angles)
x + y = 180° (int. angles), y:z = 3:7 ⇒ y = 3k, z = 7k = x
7k + 3k = 180° ⇒ k = 18° ⇒ x = 126°
(Based on Exercise 6.2, Question 1, Page Reprint 2025-26)
Fig. 6.23: AB || CD || EF with transversal
8In Fig. 6.24, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Answer & Reference:
AB || CD ⇒ ∠AGE = ∠GED = 126° (alt. int.)
EF ⊥ CD ⇒ ∠FED = 90° ⇒ ∠GEF = 126° - 90° = 36°
In △GEF: ∠FGE = 180° - 90° - 36° = 54°
(Based on Exercise 6.2, Question 2, Page Reprint 2025-26)
Fig. 6.24: AB || CD, EF ⊥ CD
9In Fig. 6.25, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
Answer & Reference:
Draw AB || ST through R.
∠QRA = 180° - 110° = 70°, ∠SRB = 180° - 130° = 50°
∠QRS = 70° + 50° = 120°
(Based on Exercise 6.2, Question 3, Page Reprint 2025-26)
Fig. 6.25: PQ || ST with transversal QR and RS
10In Fig. 6.26, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Answer & Reference:
AB || CD ⇒ x = ∠APQ = 50° (alt. int.)
∠APR = ∠PRD ⇒ 50° + y = 127° ⇒ y = 77°
(Based on Exercise 6.2, Question 4, Page Reprint 2025-26)
Fig. 6.26: AB || CD with transversals PQ and PR
11In Fig. 6.27, 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.
Answer & Reference:
Draw BE ⊥ PQ, CF ⊥ RS. Since PQ || RS, BE || CF.
By reflection: ∠ABE = ∠EBC, ∠BCF = ∠FCD
Alternate angles: ∠EBC = ∠BCF ⇒ ∠ABE = ∠FCD
⇒ ∠ABC = ∠BCD ⇒ AB || CD
(Based on Exercise 6.2, Question 5, Page Reprint 2025-26)
Fig. 6.27: Parallel mirrors PQ and RS with reflections
12State the key properties studied in this chapter.
Answer & Reference:
The key properties studied in this chapter are:
If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice-versa. This property is called the Linear Pair Axiom.
If two lines intersect each other, then the vertically opposite angles are equal.
Lines which are parallel to a given line are parallel to each other.
(Based on Section 6.6 Summary, Page Reprint 2025-26)