Question

In triangle ΔABC, ∠C is a right angle and CD is the height to

Answer 1

Answer:

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

Step-by-step 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:

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