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  1. In the given, AB || CD. Then X is equal to:












  1. View Hint View Answer Discuss in Forum

    Through O, draw a line l parallel to both AB and CD. Then
    ∠1 = 45° (alt. ∠S)
    and ∠2 = 30° (alt. ∠S)
    ∴ ∠BOC = ∠1 + ∠2 = 45° + 30° = 75°

    Correct Option: E

    Through O, draw a line l parallel to both AB and CD. Then
    ∠1 = 45° (alt. ∠S)
    and ∠2 = 30° (alt. ∠S)
    ∴ ∠BOC = ∠1 + ∠2 = 45° + 30° = 75°
    So, X = 360° – ∠BOC = 360° – 75° = 285°
    Hence X = 285°.

  1. An angle is equal to one-third of its supplement. Its measure is equal to :










  1. View Hint View Answer Discuss in Forum

    Let the measured of the required angle be P degree.
    Then, its supplement = 180 – P
    Now use the formula,

    Angle = 1 ( its supplement )
    3

    Correct Option: C

    Let the measured of the required angle be P degree.
    Then, its supplement = 180 – P
    Now use the formula,

    Angle = 1 ( its supplement )
    3

    P = ( 180 - P )
    3

    3P + P 180°
    ⇒ P = 45°

  1. In the adjoining figure, AB || CD, t is the traversal, EG and FG are the bisectors of ∠BEE and ∠DFE respectively, then ∠EGF is equal to :












  1. View Hint View Answer Discuss in Forum

    AB || CD and t transversal intersects them at E and F
    ∠BEF + ∠EFD = 180° (co-interior angles)

    ⇒1∠BEF+ 1 ∠EFD= 180/2
    22

    ⇒ ∠FEG + ∠EFG = 90°
    In Δ GEF
    ∠EGF + ∠FEG + ∠EFG = 180°

    Correct Option: A

    AB || CD and t transversal intersects them at E and F
    ∠BEF + ∠EFD = 180° (co-interior angles)

    ⇒1∠BEF+ 1 ∠EFD= 180/2
    22

    ⇒ ∠FEG + ∠EFG = 90°
    In Δ GEF
    ∠EGF + ∠FEG + ∠EFG = 180°
    ∴ ∠EGF + 90° = 180°
    ∴ ∠EGF = 90°.
    The above result can be restated as :
    If two parallel lines are cut by a traversal, then the bisectors of the interior angles on the same side of the traversal intersect each other at right angles.

  1. Find the measure of an angle, if six times its complement is 12° less than twice its supplement :










  1. View Hint View Answer Discuss in Forum

    Let, the measure of the required angled be A°.
    Then, measure of its complement = ( 90 – A )° measure of its supplement = (180 – A)°
    According to question,
    6(90° – A) = 2(180° – A) –12°

    Correct Option: A

    Let, the measure of the required angled be A°.
    Then, measure of its complement = ( 90 – A )° measure of its supplement = (180 – A)°
    According to question,
    6(90° – A) = 2(180° – A) –12°
    ⇒ 540° – 6A = 360° – 2A – 12°
    ⇒ 4A = 192°
    ⇒ A = 48°.

  1. In fig., AB || CD, ∠a is equal to:












  1. View Hint View Answer Discuss in Forum

    CD || AB (Given)
    Produce RQ to meet AB in S
    ∠CRS = ∠PSR (at. int. ∠s)
    But ∠CRS = 55°
    ∴ ∠PSR = 55°


    Correct Option: A

    CD || AB (Given)
    Produce RQ to meet AB in S
    ∠CRS = ∠PSR (at. int. ∠s)
    But ∠CRS = 55°
    ∴ ∠PSR = 55°
    Now in QSP
    ∠QSP + ∠QPS + ∠PQS = 180°
    55° + 38° + ∠SQP = 180°
    ∴ ∠SQP = 180° – 93° = 87°
    But angle a and ∠PQS are linear
    ∠a = 180° – 87°
    ∠a = 93°