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 = αm∠CDB = m∠CBD = m∠BCD = m∠CDA = m∠CAD= m∠ACD =
Accepted Solution
A:
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]°