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2 August, 08:44

In triangle ΔABC, ∠C is a right angle and CD is the height to AB, Find the angles in ΔCBD and ΔCAD if: m∠A = α

m∠DBC =

m∠DCB =

m∠CDB =

m∠ACD =

m∠ADC =

+1
Answers (1)
  1. 2 August, 08:52
    0
    Refer to the figure shown below.

    Given:

    m∠C = 90°, because ∠C is a right angle.

    m∠D = 90°, because CD is the height to AB.

    m∠A = α

    Because the sum of angles in a triangle is 180°, therefore

    m∠DBC + 90° + α = 180°

    m∠DBC = 90° - α

    Again, for the same reason,

    m∠DCB + m∠DBC + 90° = 180°

    m∠DCB + 90° - α + 90° = 180°

    m∠DCB = α

    For the same reason,

    m∠ACD + 90° + α = 180°

    m∠ACD = 90° - α

    m∠ADC = 90° (by definition)

    m∠CDB = 90° (by definition)

    Answer:

    m∠DBC = 90° - α

    m∠DCB = α

    m∠CDB = 90°

    m∠ACD = 90° - α

    m∠ADC = 90°
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