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 =

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°