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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 =

In triangle ΔABC C is a right angle and CD is the height to AB Find the angles in ΔCBD and ΔCAD if mA αmCDB mCBD mBCD mCDA m CAD mACD class=

Respuesta :

Answer:

[tex]m\angle CDB=90[/tex]°

[tex]m\angle CBD=90- \alpha[/tex]°

[tex]m\angle BCD=\alpha[/tex]°

[tex]m\angle CDA=90[/tex]°

[tex]m\angle CAD=\alpha[/tex]°

[tex]m\angle ACD=90- \alpha[/tex]°

Step-by-step explanation:

The triangles are drawn below.

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

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

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

Therefore, [tex]m\angle CAD=\alpha[/tex]°

Now, triangles ΔCBD and ΔCAD are right angled triangles.

So, for the right angled triangle ΔCAD,

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

Now, from the figure,

[tex]m\angle C=m\angle ACD+m\angle BCD\\90=90- \alpha + m\angle BCD\\\therefore m\angle BCD=\alpha[/tex]

From [tex]\DeltaCBD[/tex],

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

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

[tex]m\angle CDB=90[/tex]°

[tex]m\angle CBD=90- \alpha[/tex]°

[tex]m\angle BCD=\alpha[/tex]°

[tex]m\angle CDA=90[/tex]°

[tex]m\angle CAD=\alpha[/tex]°

[tex]m\angle ACD=90- \alpha[/tex]°

Ver imagen DarcySea

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