CG Board Class 9 Mathematics Chapter 9 Solutions: Straight Line and Angle (सरल रेखा और कोण) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 9 Mathematics Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 9 Mathematics Chapter 9 Solutions: Straight Line and Angle (सरल रेखा और कोण) in Bilingual Format (English & Hindi Medium).

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Q1

Activity Finding the number of line segments formed…


📝 Question:

Activity: Finding the number of line segments formed by n collinear points on a line. : n Completed Table / : 3 Points (P, Q, R): Names = PQ, PR, QR → 3 line segments 4 Points (P, Q, R, S): Names = PQ, PR, PS, QR, QS, RS → 6 line segments 5 Points (P, Q, R, S, T): Names = PQ, PR, PS, PT, QR, QS, QT, RS, RT, ST → 10 line segments 6 Points (P, Q, R, S, T, U): Total → 15 line segments 7 Points (P, Q, R, S, T, U, V): Total → 21 line segments General Formula / : For n points, number of line segments = 1 + 2 + 3 + … + (n-1) = (n(n-1))/(2) Can three collinear points make a triangle?

💡 Solution:

No. Three collinear points lie on the exact same straight line, so the area enclosed by them is zero and no triangle can be formed.

Q2

Try This Page 179 Q1 Draw two angles…


📝 Question:

Try This (Page 179) / Q.1:** Draw two angles which are not adjacent

💡 Solution:

(i) ∠ AOB and ∠ BOC: Adjacent* [They share a common vertex O and a common arm OB].

  • (ii) ∠ BOC and ∠ DOE: *Not adjacent [They do not share a common arm].
  • (iii) ∠ EOD and ∠ DOC: *Adjacent [They share a common vertex O and a common arm OD].

Q3

Where are two angles adjacent i When both…


📝 Question:

Where are two angles adjacent? (i) When both angles are obtuse (ii) Both are acute (iii) One angle is obtuse and the other is acute

💡 Solution:

Two angles can be adjacent in all three cases (i), (ii), and (iii) as long as they have a common vertex, a common arm, and non-overlapping interior regions.

Q4

Think and Discuss Page 181 Q1 Can two…


📝 Question:

Think and Discuss (Page 181) / * Q.1 Can two right angles be complementary angles? / ?

💡 Solution:

No. The sum of two right angles is 90°+ 90°= 180°, whereas complementary angles must sum to 90°. / , 180° , 90°

  • Q.2 Is each linear pair also supplementary angles? / ?

Yes. The sum of angles forming a linear pair is always 180°, so they are always supplementary. / , 180° ,

Q5

Try This Page 181 Q1 Which of the…


📝 Question:

Try This (Page 181) / Q.1 Which of the following angles are complementary or supplementary? Which ones are adjacent or linear pairs?

💡 Solution:

(i) ∠ 1 and ∠ 2:** Linear Pair / Supplementary / Adjacent

  • (ii) ∠ 5 and ∠ 6: Linear Pair / Supplementary / Adjacent
  • (iii) ∠ 6 and ∠ 3: Vertically Opposite Angles
  • (iv) ∠ 5 and ∠ 2: Vertically Opposite Angles
  • (v) ∠ 3 and ∠ 4: Linear Pair / Supplementary / Adjacent

Q6

Try This Page 183 Q1 Show all vertically…


📝 Question:

Try This (Page 183) / * Q.1 Show all vertically opposite angles in the figure

💡 Solution:

Pairs of vertically opposite angles are (∠ POQ, ∠ ROS), (∠ QOV, ∠ UOT), (∠ VOR, ∠ POT), (∠ SOU, ∠ ROP).

  • Q.2 In figure, if ∠ 2 = 110°, then find ∠ 1 and ∠ 4.
  • ∠ 1 + ∠ 2 = 180°(Linear pair) ⇒ ∠ 1 = 180°- 110°= {70°
  • ∠ 4 = ∠ 2 = {110° (Vertically opposite angles)

Q7

Table Activity Angles Formed by Transversal Page 188…


📝 Question:

Table Activity: Angles Formed by Transversal (Page 188) Completed Table / : 1. Alternate Interior Angles: ∠ c and ∠ e, ∠ d and ∠ f (Number of pairs = 2) 2. Interior Angles on Same Side of Transversal: ∠ d and ∠ e, ∠ c and ∠ f (Number of pairs = 2) 3. Alternate Exterior Angles: ∠ a and ∠ g, ∠ b and ∠ h (Number of pairs = 2) 4. Exterior Angles on Same Side of Transversal: ∠ a and ∠ h, ∠ b and ∠ g (Number of pairs = 2) 5. Corresponding Angles: ∠ a and ∠ e, ∠ b and ∠ f, ∠ d and ∠ h, ∠ c and ∠ g (Number of pairs = 4) 9. Think and Discuss (Page 188) / **Q.1 If corresponding angles are equal, are the lines parallel?

💡 Solution:

Yes, by the converse of the Corresponding Angles Axiom, the lines are parallel. / ,

  • Q.2 If alternate interior angles are equal, are the lines parallel?

Yes, if alternate interior angles are equal, the lines are parallel. / ,

Q8

Try This Page 189 In the given figures…


📝 Question:

Try This (Page 189) / In the given figures, find the values of ∠ 1 and ∠ 2 (l ∥ m)

💡 Solution:

(i)** Given angle 120°

  • ∠ 1 = {120° (Corresponding angle)
  • ∠ 2 = 180°- 120°= {60° (Linear pair with ∠ 1)
  • (ii) Given angle 50°
  • ∠ 1 = 180°- 50°= {130° (Linear pair)
  • ∠ 2 = {50° (Vertically opposite to 50°/ Alternate interior)
  • (iii) Given angle 70°
  • ∠ 1 = 180°- 70°= {110° (Linear pair)
  • ∠ 2 = {70° (Alternate interior angle)
  • (iv) Given angle 80°
  • ∠ 1 = 180°- 80°= {100° (Co-interior angle)
  • ∠ 2 = {80° (Alternate interior angle)

Q9

Solved Examples Summary Examples 1 to 16 Example…


📝 Question:

Solved Examples Summary (Examples 1 to 16) Example 1: ∠ AOC = 2x + 20°, ∠ BOC = 2x form a linear pair. (2x + 20°) + 2x = 180° ⇒ 4x = 160° ⇒ x = 40°. ∠ AOC = {100°, ∠ BOC = {80°. Example 2: ∠ AOC : ∠ COB = 7 : 8. 7x + 8x = 180° ⇒ 15x = 180° ⇒ x = 12°. ∠ AOC = {84°, ∠ COB = {96°, ∠ BOD = {84°, ∠ AOD = {96°. Example 3: ∠ COD = 90°, ∠ BOE = 72°, ∠ AOE = 3x, ∠ AOC = x, ∠ BOD = y. 3x + 72°= 180° ⇒ x = 36°. x + 90°+ y = 180° ⇒ 36°+ 90°+ y = 180° ⇒ y = 54°. ∠ AOC = {36°, ∠ BOD = {54°, ∠ AOE = {108°. Example 4: Prove ∠ ROT = 90°where OR, OT bisect linear pair ∠ POS, ∠ SOQ. ∠ ROS = x/2, ∠ SOT = 90°- x/2 ⇒ ∠ ROT = x/2 + 90°- x/2 = {90°. Example 5: Prove ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°. Example 6: Given figure with transversal: x = {70°, y = {110°, z = {70°, a = {115°, b = {65°, c = {65°. Example 7: PQ ∥ RS, ∠ MXQ = 135°, ∠ MYR = 40° ⇒ ∠ XMY = 45°+ 40°= {85°. Example 8: CD ∥ AB, ∠ BAE = 105°, ∠ AEC = 25°, ∠ DCE = x ⇒ x = {130°. Example 10: CD ∥ AB, EF ∥ CD, EA ⊥ AB, ∠ BEF = 55° ⇒ x = {125°, y = {125°, z = {35°. Example 11: Angles (2x + 1)°, (3x + 6)°, (4x – 16)° ⇒ 9x – 9 = 180° ⇒ x = 21°. Angles are {43°, 69°, 68°. Example 12: AB ∥ QR ⇒ ∠ APB = {42°, ∠ AQR = {38°, ∠ QRP = {100°. Example 15: BO, CO bisect ∠ B, ∠ C of △ ABC with ∠ A = 40° ⇒ ∠ BOC = 90°+ frac{40°{2} = {110°. Example 16: Prove ∠ BOC = 90°- 1/2∠ BAC for exterior angle bisectors. Part 2: End-of-Chapter Exercise 9.1 Q.1 In the figure alongside, ∠ POR and ∠ QOR form a linear pair. If x – y = 80°, find the values of x and y

💡 Solution:

  • Since ∠ POR and ∠ QOR form a linear pair
  • x + y = 180°—— (1)
  • Given: x – y = 80°—— (2)
  • Adding equations (1) and (2)
  • 2x = 260° ⇒ x = {130°
  • Substituting x = 130°in (1)
  • 130°+ y = 180° ⇒ y = {50°

  • Solution / :
  • Lines PQ and RS intersect at O ⇒ ∠ POR = ∠ QOS = 40°(Vertically opposite angles).
  • Given ∠ POR + ∠ QOT = 70° ⇒ 40°+ ∠ QOT = 70° ⇒ ∠ QOT = {30°.
  • On line PQ: ∠ POR + ∠ ROT + ∠ QOT = 180°* 70°+ ∠ ROT = 180° ⇒ ∠ ROT = 110°.
  • Reflex ∠ ROT = 360°- 110°= {250°.
  • Solution / :
  • Since ∠ NOB = 90°on line AB ⇒ ∠ NOA = 90° ⇒ x + y = 90°.
  • Let x = 2k and y = 3k ⇒ 2k + 3k = 90° ⇒ 5k = 90° ⇒ k = 18°.
  • Therefore, y = 3 × 18°= 54°.
  • Since LM is a straight line, y + z = 180°(Linear pair).
  • 54°+ z = 180° ⇒ z = 180°- 54°= {126°.
  • Solution / :
  • On the line AB:
  • ∠ ECA + ∠ ECD = 180°(Linear pair) —— (1)
  • ∠ EDB + ∠ EDC = 180°(Linear pair) —— (2)
  • From (1) and (2): ∠ ECA + ∠ ECD = ∠ EDB + ∠ EDC.
  • Since ∠ ECD = ∠ EDC, subtracting this equal angle from both sides:
  • ∠ ECA = ∠ EDB. Hence proved.
  • Solution / :
  • Sum of all angles around point O is 360°:
  • a + b + c + d = 360°* Since c + d = a + b: (a + b) + (a + b) = 360° ⇒ 2(a + b) = 360° ⇒ a + b = 180°.
  • Since the adjacent angles a and b sum to 180°, their non-common arms form a straight line.
  • Therefore, LOM is a straight line. Hence proved.
  • Solution / :
  • Since OC ⊥ AB ⇒ ∠ AOC = ∠ BOC = 90°.
  • ∠ BOD = ∠ BOC + ∠ COD = 90°+ ∠ COD —— (1)
  • ∠ AOD = ∠ AOC – ∠ COD = 90°- ∠ COD —— (2)
  • Subtracting equation (2) from (1):
  • ∠ BOD – ∠ AOD = (90°+ ∠ COD) – (90°- ∠ COD) = 2∠ COD
  • ∠ COD = 1/2(∠ BOD – ∠ AOD). Hence proved.
  • Solution / :
  • Since AX is a straight line, ∠ ABC + ∠ CBX = 180° ⇒ 64°+ ∠ CBX = 180° ⇒ ∠ CBX = 116°.
  • Ray BY bisects ∠ CBX ⇒ ∠ CBY = ∠ YBX = frac{116°{2} = 58°.
  • ∠ ABY = ∠ ABC + ∠ CBY = 64°+ 58°= {122°.
  • Reflex ∠ YBX = 360°- ∠ YBX = 360°- 58°= {302°.
  • (i) 2x + 3x = 180°(Co-interior / Linear pair) ⇒ 5x = 180° ⇒ {x = 36°.
  • y = 3x = 3(36°) = {108° (Vertically opposite / Corresponding).
  • (ii) 4x – 23°+ 3x = 180° ⇒ 7x = 203° ⇒ {x = 29°.
  • y = 3x = 3(29°) = {87°.
  • (iii) 2y + 15°+ 3y – 10°= 180° ⇒ 5y + 5°= 180° ⇒ 5y = 175° ⇒ {y = 35°.
  • x = 180°- (2y + 15°) = 180°- 85°= {95°.
  • Solution / :
  • Since AB ∥ CD ⇒ x + y = 180°(Co-interior angles).
  • Since AB ∥ EF ⇒ x = z (Alternate interior angles).
  • Therefore, z + y = 180°.
  • Let y = 3k and z = 7k ⇒ 3k + 7k = 180° ⇒ 10k = 180° ⇒ k = 18°.
  • z = 7 × 18°= 126° ⇒ {x = 126°.
  • Solution / :
  • Since AB ∥ CD ⇒ ∠ AGE = ∠ GED = {126° (Alternate interior angle).
  • ∠ GEF = ∠ GED – ∠ FED = 126°- 90°= {36°.
  • Since AB is a straight line, ∠ AGE + ∠ FGE = 180° ⇒ 126°+ ∠ FGE = 180° ⇒ ∠ FGE = {54°.
  • Solution / :
  • Draw a line RU ∥ ST passing through R.
  • ∠ SRU + ∠ RST = 180° ⇒ ∠ SRU = 180°- 130°= 50°.
  • Since PQ ∥ RU ⇒ ∠ QRU = ∠ PQR = 110°(Alternate interior angle).
  • ∠ QRS = ∠ QRU – ∠ SRU = 110°- 50°= {60°.
  • Solution / :
  • Since AB ∥ CD and PQ is transversal ⇒ x = ∠ APQ = {50° (Alternate interior angle).
  • Since AB ∥ CD and PR is transversal ⇒ ∠ APR = ∠ PRD = 127°.
  • 50°+ y = 127° ⇒ y = 127°- 50°= {77°.
  • Solution / :
  • Corresponding angle x = {80°.
  • Co-interior angle y = 180°- 80°= {100°.
  • (i) x = {59°, y = {60°
  • (ii) x = {40°, y = {40°
  • (iii) x = {18°, y = {60°
  • (iv) x = {20°, y = {63°
  • (i) x = {140°
  • (ii) x = {100°
  • (iii) x = {250°
  • Solution / :
  • Draw normals BN ⊥ PQ and CM ⊥ RS.
  • Since PQ ∥ RS ⇒ BN ∥ CM.
  • By laws of reflection: ∠ 1 = ∠ 2 and ∠ 3 = ∠ 4.
  • Since BN ∥ CM with transversal BC ⇒ ∠ 2 = ∠ 3 (Alternate interior angles).
  • Multiplying by 2: 2∠ 2 = 2∠ 3 ⇒ ∠ 1 + ∠ 2 = ∠ 3 + ∠ 4 ⇒ ∠ ABC = ∠ BCD.
  • Since alternate interior angles ∠ ABC and ∠ BCD are equal, AB ∥ CD. Hence proved.
  • Solution / :
  • ∠ ACB + ∠ BCE = 180°(Linear pair) ⇒ ∠ ACB = 180°- 163°= {17°.
  • Exterior angle ∠ BCE = ∠ BAC + ∠ ABC ⇒ 163°= 34°+ ∠ ABC ⇒ ∠ ABC = {129°.
  • ∠ DBC + ∠ ABC = 180°(Linear pair) ⇒ ∠ DBC = 180°- 129°= {51°.
  • (i) x = {40°, y = {70°
  • (ii) x = {50°, y = {20°
  • (iii) x = {51°, y = {35°
  • (iv) x = {30°, y = {75°
  • Solution / :
  • Using angle relationships in parallel lines and angle bisector property:
  • {∠ 1 = 60°.
  • Solution / :
  • ∠ QPR = 180°- 135°= 45°(Linear pair)
  • ∠ PQR = 180°- 110°= 70°(Linear pair)
  • In △ PQR: ∠ PRQ = 180°- (45°+ 70°) = 180°- 115°= {65°.
  • Solution / :
  • In △ XYZ: ∠ XZY = 180°- (62°+ 54°) = 180°- 116°= 64°.
  • ZO bisects ∠ XZY ⇒ ∠ OZY = frac{64°{2} = {32°.
  • YO bisects ∠ XYZ ⇒ ∠ OYZ = frac{54°{2} = 27°.
  • In △ OYZ: ∠ YOZ = 180°- (27°+ 32°) = 180°- 59°= {121°.
  • Solution / :
  • Since AB ∥ DE ⇒ ∠ CED = ∠ BAC = 35°(Alternate interior angle).
  • In △ CDE: ∠ DCE = 180°- (53°+ 35°) = 180°- 88°= {92°.
  • Solution / :
  • In △ PRT: ∠ PTR = 180°- (95°+ 40°) = 180°- 135°= 45°.
  • Vertically opposite angle ∠ STQ = ∠ PTR = 45°.
  • In △ STQ: ∠ SQT = 180°- (75°+ 45°) = 180°- 120°= {60°.
  • Solution / :
  • Exterior angle property for △ BCD: x + 28°= 65° ⇒ {x = 37°? (Or alternate angle x + 28°= 65°).
  • In right triangle ABC: y = 90°- x = 90°- 37°= {53°.
  • Solution / :
  • In △ ABC, exterior ∠ ACD = ∠ BAC + ∠ ABC.
  • Dividing by 2: 1/2∠ ACD = 1/2∠ BAC + 1/2∠ ABC —— (1)
  • In △ BEC, exterior ∠ ECD = ∠ BEC + ∠ EBC.
  • Since EC bisects ∠ ACD ⇒ ∠ ECD = 1/2∠ ACD.
  • Since EB bisects ∠ ABC ⇒ ∠ EBC = 1/2∠ ABC.
  • So 1/2∠ ACD = ∠ BEC + 1/2∠ ABC —— (2)
  • Comparing (1) and (2):
  • 1/2∠ BAC + 1/2∠ ABC = ∠ BEC + 1/2∠ ABC
  • ∠ BEC = 1/2∠ BAC. Hence proved.

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