∠BCD = 52°
∠CBD = 76°
m∠DAC = 34°
m∠DBA = 142°
m∠ADC = 73°
m∠BAD = 17°
Step-by-step explanation:
Given:
m∠BAC = 17°
m∠BDC = 52°
Considering triangle DBC:
two sides of the triange are equal. Hence, the base angles will be equal.
The base angls are ∠BDC and ∠BCD
∠BDC = ∠BCD
∠BCD = 52°
∠BDC + ∠BCD + ∠CBD = 180° (angles in a triangle)
52 + 52 + ∠CBD = 180°
104 + ∠CBD = 180°
∠CBD = 180 - 104
∠CBD = 76°
B bisects angle ∠DAC. This divides it into two equal halves.
m∠BAC = m∠BAD, m∠BAC = 17
m∠BAD = 17°
m∠DAC = m∠BAC + m∠BAD
m∠DAC = 17 + 17 = 34°
m∠CDB + m∠DBA + m∠CBA = 360° (angles at a point)
m∠DBA = m∠CBA (side AD = ide AC)
Angles opposite the sides will also be equal
76 + 2(m∠DBA) = 360
2(m∠DBA) = 360 - 76
2(m∠DBA) = 284
m∠DBA = 284/2
m∠DBA = 142°
m∠ADC = m∠ADB + m∠BDC
m∠ADB + m∠DBA + m∠BAD = 180° (angles in a triangle)
m∠ADB + 142 + 17 = 180
m∠ADB + 159 = 180
m∠ADB 180 - 159
m∠ADB = 21°
m∠ADC = 21 + 52
m∠ADC = 73°