Question

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

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.

Since, CD is the height to AB, therefore, CD is perpendicular to AB.

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

Also, $m\angle CAD$ is same as $m\angle A$.

Therefore, $m\angle CAD=\alpha$°

Now, triangles ΔCBD and ΔCAD are right angled triangles.

So, for the right angled triangle ΔCAD,

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

Now, from the figure,

$m\angle C=m\angle ACD+m\angle BCD\\90=90- \alpha + m\angle BCD\\\therefore m\angle BCD=\alpha$

From $\DeltaCBD$,

$m\angle BCD+m\angle CBD=90\\\alpha + m\angle CBD=90\\m\angle CBD=90- \alpha$

Hence, all the angles of the triangles ΔCBD and ΔCAD are:

$m\angle CDB=90$°

$m\angle CBD=90- \alpha$°

$m\angle BCD=\alpha$°

$m\angle CDA=90$°

$m\angle CAD=\alpha$°

$m\angle ACD=90- \alpha$°