Given:
ΔABC, m∠C = 90º, m∠B = 30º
CM - angle bisector
Find: m∠AMC
![Given ΔABC mC 90º mB 30º CM angle bisector Find mAMC class=](https://us-static.z-dn.net/files/d16/f401300541b5168570636ede576567e9.png)
Answer:
75°
Step-by-step explanation:
1. Consider right triangle ABC. In this triangle
m∠C = 90º and CM - angle bisector.
If CM is angle C bisector, then
m∠ACM = m∠MCB = 45º
2. Consider triangle CMB. The sum of the measures of all interior angles of the triangle is always 180°, then
m∠MCB + m∠CBM + m∠BMC = 180°,
m∠BMC = 180° - 45° - 30°
m∠BMC = 105°
3. Angles AMC and BMC are supplementary angles, so
m∠AMC + m∠BMC = 180°
m∠AMC = 180° - 105°
m∠AMC = 75°