Question

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

Answer 1

Answer:

Given:

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

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

m∠A = α

Since 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°

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

In triangle ΔABC, ∠C is a right angle

m∠C = 90

m∠A = α

so m∠B = 180-90-α = 90 - α

In triangle ΔCBD, CD is the height to AB

so ∠CDB is a right angle

m∠CDB = 90

∠CBD = ∠CBA

m∠CBD = m∠B = 90 - α

m∠DCB = 180 - m∠CDB - m∠CBD

= 180-90-(90-α)

= α

In triangle ΔCAD, CD is the height to AB

so ∠CDA is a right angle

m∠CDA = 90

∠CAD = ∠CAB

m∠CAD = m∠A = α

m∠DCA = 180 - m∠CDA - m∠CAD

= 180-90-α

= 90 - α