m∠DBC =25°, m∠DCB =55°, m∠CDB =90°, m∠ACD =25°, m∠ADC =90°.
In triangle △ABC, ∠C is a right angle and CD is the altitude to AB.
We need to find the angles in △CBD and △CAD if m∠A = 65°.
What is the triangle?
A plane figure with three straight sides and three angles.
Consider △CAD
∠ADC =90°.
∠ACD+∠ADC+∠CAD=180°.
⇒∠ACD+90°+65°=180°
⇒∠ACD+90°+65°=180°
⇒∠ACD=180°-155°=25°
Consider △CBD
∠CDB =90°
∠DCB+∠ACD =90°
⇒∠DCB+25°=90°
⇒∠DCB=55°
Now, ∠DBC+∠DCB+∠CDB =180°
⇒∠DBC+55°+90° =180°
⇒∠DBC=25°
Therefore, m∠DBC =25°, m∠DCB =55°, m∠CDB =90°, m∠ACD =25°, m∠ADC =90°.
To learn more about the triangle visit:
https://brainly.com/question/2773823.
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