1 Answer

RedJandal · Answer 1 · 2020-03-10T08:05:45+0000

Answer:

$m\angle CDB=90$ °

$m\angle CBD=90- \alpha$ °

$m\angle BCD=\alpha$ °

$m\angle CDA=90$ °

$m\angle CAD=\alpha$ °

$m\angle ACD=90- \alpha$ °

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
$\DeltaCBD$ ,

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