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**Pairs of Angles – Maharashtra Board Class 6 Solutions for Mathematics (English Medium)**

MathematicsGeneral ScienceMaharashtra Board Solutions

### Exercise-17

**Solution 1:**

- 37°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 37°.

Thus, its complement measures = 90° – 37° = 53° - 48°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 48°.

Thus, its complement measures = 90° – 48° = 42° - 55°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 55°.

Thus, its complement measures = 90° – 55° = 35° - 79°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 79°.

Thus, its complement measures = 90° – 79° = 11° - 68°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 68°.

Thus, its complement measures = 90° – 68° = 22° - 10°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 10°.

Thus, its complement measures = 90° – 10° = 80° - 25°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 25°.

Thus, its complement measures = 90° – 25° = 65° - 40°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 40°.

Thus, its complement measures = 90° – 40° = 50° - 89°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 89°.

Thus, its complement measures = 90° – 89° = 1° - 17°

The sum of the measures of complementary angles is 90°.

The measure of one of them is 17°.

Thus, its complement measures = 90° – 17° = 73°

**Solution 2:**

- 65°

Sum of the measures of supplementary angles is 180°. Measure of one of the angles is 65°.

Thus, its supplement measures = 180° – 65°= 115° - 24°

Sum of the measures of supplementary angles is 180°. Measure of one of the angles is 24°.

Thus, its supplement measures = 180° – 24°= 156° - 90°

Sum of the measures of supplementary angles is 180°. Measure of one of the angles is 90°.Thus, its supplement measures = 180° – 90°= 90°. - 47°

Sum of the measures of supplementary angles is 180° Measure of one of the angles is 47°

Thus, its supplement measures = 180° – 47°= 133° - 79°

Sum of the measures of supplementary angles is 180°. Measure of one of the angles is 79°

Thus, its supplement measures = 180° – 79° = 101° - 58°

Sum of the measures of supplementary angles is 180°. Measure of one of the angles is 58°

Thus, its supplement measures = 180° – 58°= 122° - 154°

Sum of the measures of supplementary angles is 180°. Measure of one of the angles is 154°

Thus, its supplement measures = 180° – 154°= 26° - 125°

Sum of the measures of supplementary angles is 180°. Measure of one of the angles is 125°

Thus, its supplement measures = 180° – 125°= 55° - 140°

Sum of the measures of supplementary angles is 180° Measure of one of the angles is 140°

Thus, its supplement measures = 180° – 140°= 40° - 165°

Sum of the measures of supplementary angles is 180°. Measure of one of the angles is 165°

Thus, its supplement measures = 180° – 165°= 15°

**Solution 3:**

- 69°, 111°

69° + 111° = 180°

∴ The given pair is supplementary. - 26°, 64°

26° + 64° = 90°

∴ The given pair is complementary. - 90°, 90°

90° + 90° = 180°

∴ The given pair is supplementary. - 50°, 40°

50° + 40° = 90°

∴ The given pair is complementary. - 163°, 17°

163° + 17° = 180°

∴ The given pair is supplementary. - 45°, 45°

45° + 45° = 90°

∴ The given pair is complementary. - 168°, 12°

168° + 12°= 180°

∴ The given pair is supplementary. - 35°, 55°

35° + 55° = 90°

∴ The given pair is complementary.

**Solution 4:**

- m∠ AOC = 60°
- m∠ AOD = 120°
- m∠ BOC = 120°

**Solution 5:**

- ∠AOD
- 120°
- ∠ AOC and ∠ BOD
- 60°
- 30°
- 60°
- ∠AOD and ∠BOC

**Solution 6:**

Given angle | 60° | 45° | 78° | 10° | 25° | 80° | 37° |

Complement of the angle | 30° | 45° | 12° | 80° | 65° | 10° | 53° |

**Solution 7:**

Given angle | 32° | 90° | 110° | 137° | 165° | 129° | 65° |

Supplement of the angle | 148° | 90° | 70° | 43° | 15° | 51° | 115° |

**Solution 8:**

- m∠SWP = m∠SWJ + m∠JWP

m∠SWJ = m∠SWP – m∠JWP

= 90° – 37°

= 53° - m∠JGH + m∠JGP = 180° (linear pair)

∠JGH = 180° – m∠JGP

= 180° – 110°

= 70° - m∠WYX + m∠WYZ = 180°

m∠WYX = 180° – m∠WYZ

= 180° – 132°

= 48° - m∠QOR = m∠POQ + m∠POR

m∠POR = m∠QOR – m∠POQ

= 90° – 45°

= 45°

### Exercise-18

**Solution 1:**

A line which cuts two other lines at two different points is called a transversal.

In figure (1) line z intersects the line t and line p at two distinct points.

Hence, line z a transversal of the lines t and p.

**Solution 2:**

Pairs of corresponding angles: ∠KGB and ∠GMP, ∠BGM and ∠PMT, ∠KGS and ∠GMV, ∠SGM and ∠VMT.

Pairs of alternate angles: ∠SGM and ∠GMP, ∠BGM and ∠GMV.

Pairs of interior angles: ∠BGM and ∠GMP and ∠SGM and ∠VMG

**Solution 3:**

- m∠DHP = m∠VHG = 85° (Vertically opposite angles have the same measure)

Now, m∠VHG = m∠MGK = 85° (Corresponding angles)

∴ m∠MGK = 85° - m∠VHD + m∠DHP = 180° (Linear pair)

m∠VHD + 85° = 180°

∴ m∠VHD = 180 – 85

∴ m∠VHD = 95° - m∠DHP + m∠PHG = 180° (Linear pair)

85° + m∠PHG = 180°

∴ m∠PHG = 180 – 85

∴ m∠PHG = 95° - m∠PHG + m∠HGS = 180° (Interior angles)

95° + ∠HGS = 180°

∴ ∠HGS = 180 – 95

∴ ∠HGS = 85°

**Solution 4:**

- line AD ∥ line EH and line XY is the transversal

∴ m∠ABF = m∠EFB (Alternate angles)

∴ m∠EFB = 70° - line AD ∥ line EH and line XY is the transversal

m∠DBF = m∠GFY (Corresponding angles)

∴ m∠GFY = 70° - line AD ∥ line EH and line PQ is the transversal

m∠CGH = m∠BCG (Alternate angles)

∴ m∠BCG = 55°