Answer:
1. 90°
2. 90°
3. 40°
4. 45°
5. 45°
Step-by-step explanation:
Given: ΔABC,
CD⊥AB,
m∠A=50°,
m∠ACB=85°
Solution:
1. ∠ADC is a right ange, because CD⊥AB, so
[tex]m\angle ADC=90^{\circ}[/tex]
2. ∠CDB is a right ange, because CD⊥AB, so
[tex]m\angle CDB=90^{\circ}[/tex]
3. Consider triangle ACD. The sum of the measures of all interior angles is always 180°, so
[tex]m\angle A+m\angle ADC+m\angle ACD=180^{\circ}\\ \\50^{\circ}+90^{\circ}+m\angle ACD=180^{\circ}\\ \\m\angle ACD=180^{\circ}-50^{\circ}-90^{\circ}=40^{\circ}[/tex]
4. By Angle Addition Postulate,
[tex]m\angle ACD+m\angle BCD=m\angle ACB\\ \\40^{\circ}+m\angle BCD=85^{\circ}\\ \\m\angle BCD=85^{\circ}-40^{\circ}=45^{\circ}[/tex]
5. Consider triangle ABC. The sum of the measures of all interior angles is always 180°, so
[tex]m\angle A+m\angle ACB+m\angle B=180^{\circ}\\ \\50^{\circ}+85^{\circ}+m\angle B=180^{\circ}\\ \\m\angle B=180^{\circ}-50^{\circ}-85^{\circ}=45^{\circ}[/tex]
