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

m∠CDB =

m∠CBD =

m∠BCD =

m∠CDA =

m∠CAD=

m∠ACD =

Answer 1

`m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha`

Explanation:

The triangles are drawn below.

CD is perpendicular to AB as CD is height to AB.

Therefore, angles
`m\angle CDB=m\angle CDA=90`°

So, triangles ΔCBD and ΔCAD are right angled triangles.

Now, from the right angled triangle ΔABC,

`m\angle A+m\angle B =90\\\alpha+m\angle B=90\\m\angle B=90-\alpha`

From ΔCBD,

`m\angle CBD` is same as
`m\angle B`.

So,
`m\angle CBD=90-\alpha`

`m\angle BCD+m\angle BDC =90\\m\angle BCD+90-\alpha=90\\m\angle BCD=\alpha`

Now, from ΔCAD,

`m\angle CAD` is same as
`m\angle A`

So,
`m\angle CAD=\alpha`

`m\angle CAD+m\angle ACD =90\\\alpha+m\angle ACD=90\\m\angle ACD=90-\alpha`

Hence, the unknown angles of both the triangles are:

`m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha`