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Pairs of Angles – Maharashtra Board Class 6 Solutions for Mathematics

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

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise-17

Solution 1:

  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°
  2. 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°
  3. 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°
  4. 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°
  5. 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°
  6. 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°
  7. 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°
  8. 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°
  9. 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°
  10. 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:

  1. 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°
  2. 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°
  3. 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°.
  4. 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°
  5. 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°
  6. 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°
  7. 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°
  8. 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°
  9. 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°
  10. 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:

  1. 69°, 111°
    69° + 111° = 180°
    ∴ The given pair is supplementary.
  2. 26°, 64°
    26° + 64° = 90°
    ∴ The given pair is complementary.
  3. 90°, 90°
    90° + 90° = 180°
    ∴ The given pair is supplementary.
  4. 50°, 40°
    50° + 40° = 90°
    ∴ The given pair is complementary.
  5. 163°, 17°
    163° + 17° = 180°
    ∴ The given pair is supplementary.
  6. 45°, 45°
    45° + 45° = 90°
    ∴ The given pair is complementary.
  7. 168°, 12°
    168° + 12°= 180°
    ∴ The given pair is supplementary.
  8. 35°, 55°
    35° + 55° = 90°
    ∴ The given pair is complementary.

Solution 4:

  1. m∠ AOC = 60°
  2. m∠ AOD = 120°
  3. m∠ BOC = 120°

Solution 5:

  1. ∠AOD
  2. 120°
  3. ∠ AOC and ∠ BOD
  4. 60°
  5. 30°
  6. 60°
  7. ∠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:

  1. m∠SWP = m∠SWJ + m∠JWP
    m∠SWJ = m∠SWP – m∠JWP
    = 90° – 37°
    = 53°
  2. m∠JGH + m∠JGP = 180° (linear pair)
    ∠JGH = 180° – m∠JGP
    = 180° – 110°
    = 70°
  3. m∠WYX + m∠WYZ = 180°
    m∠WYX = 180° – m∠WYZ
    = 180° – 132°
    = 48°
  4. 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:

  1. m∠DHP = m∠VHG = 85° (Vertically opposite angles have the same measure)
    Now, m∠VHG = m∠MGK = 85° (Corresponding angles)
    ∴ m∠MGK = 85°
  2. m∠VHD + m∠DHP = 180° (Linear pair)
    m∠VHD + 85° = 180°
    ∴ m∠VHD = 180 – 85
    ∴ m∠VHD = 95°
  3. m∠DHP + m∠PHG = 180° (Linear pair)
    85° + m∠PHG = 180°
    ∴ m∠PHG = 180 – 85
    ∴ m∠PHG = 95°
  4. m∠PHG + m∠HGS = 180° (Interior angles)
    95° + ∠HGS = 180°
    ∴ ∠HGS = 180 – 95
    ∴ ∠HGS = 85°

Solution 4:

  1. line AD ∥ line EH and line XY is the transversal
    ∴ m∠ABF = m∠EFB (Alternate angles)
    ∴ m∠EFB = 70°
  2. line AD ∥ line EH and line XY is the transversal
    m∠DBF = m∠GFY (Corresponding angles)
    ∴ m∠GFY = 70°
  3. line AD ∥ line EH and line PQ is the transversal
    m∠CGH = m∠BCG (Alternate angles)
    ∴ m∠BCG = 55°