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 =

Question

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Respuesta :

DarcySea DarcySea · Answer 1 · 2019-11-14T09:03:12+01:00

Answer:

[tex]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[/tex]

Step-by-step explanation:

The triangles are drawn below.

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

Therefore, angles [tex]m\angle CDB=m\angle CDA=90[/tex]°

So, triangles ΔCBD and ΔCAD are right angled triangles.

Now, from the right angled triangle ΔABC,

[tex]m\angle A+m\angle B =90\\\alpha+m\angle B=90\\m\angle B=90-\alpha[/tex]

From ΔCBD,

[tex]m\angle CBD[/tex] is same as [tex]m\angle B[/tex].

So, [tex]m\angle CBD=90-\alpha[/tex]

[tex]m\angle BCD+m\angle BDC =90\\m\angle BCD+90-\alpha=90\\m\angle BCD=\alpha[/tex]

Now, from ΔCAD,

[tex]m\angle CAD[/tex] is same as [tex]m\angle A[/tex]

So, [tex]m\angle CAD=\alpha[/tex]

[tex]m\angle CAD+m\angle ACD =90\\\alpha+m\angle ACD=90\\m\angle ACD=90-\alpha[/tex]

Hence, the unknown angles of both the triangles are:

[tex]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[/tex]