Question

[Trần Bá Hưng]

Cho tam giác ABC cân tại A , góc A = 80 độ . Điểm M nằm trong tam giác sao cho góc MAC= góc MCA = 10 độ . Tính số đo của góc AMB

Answer 1

[Trần Quốc Anh]

# Angles in triangle ABC
angle_A = 80
angle_B = 50
angle_C = 50

# Angles related to point M
angle_MAC = 10
angle_MCA = 10

# Calculate angle AMC
angle_AMC = 180 - angle_MAC - angle_MCA
print(f"Angle AMC = {angle_AMC}")

# Calculate other angles
angle_MAB = angle_A - angle_MAC
print(f"Angle MAB = {angle_MAB}")
angle_MCB = angle_C - angle_MCA
print(f"Angle MCB = {angle_MCB}")

# Consider triangle BCM. Construct a point N such that triangle BCN is isosceles
# with angle NBC = angle NCB = 40. Then angle BNC = 100.
# Angle NBA = angle ABC - angle NBC = 50 - 40 = 10.
# Angle NCA = angle ACB - angle NCB = 50 - 40 = 10.
# So N coincides with M.
# In triangle BCM, angle MBC = 40, angle MCB = 40, angle BMC = 180 - 40 - 40 = 100.

# Now consider triangle ABM.
# Angle MAB = 70.
# Angle ABM = angle ABC - angle MBC = 50 - 40 = 10.
# Angle AMB = 180 - angle_MAB - angle_ABM
angle_AMB = 180 - 70 - 10
print(f"Angle AMB = {angle_AMB}")

Python

import math

def degrees_to_radians(degrees):
    return degrees * math.pi / 180

def radians_to_degrees(radians):
    return radians * 180 / math.pi

# Using Law of Sines

# Assume BC = 1 for simplicity
angle_A_rad = degrees_to_radians(80)
angle_B_rad = degrees_to_radians(50)
angle_C_rad = degrees_to_radians(50)

AC = math.sin(angle_B_rad) / math.sin(angle_A_rad)
AB = math.sin(angle_C_rad) / math.sin(angle_A_rad)

angle_MAC_rad = degrees_to_radians(10)
angle_MCA_rad = degrees_to_radians(10)
angle_AMC_rad = degrees_to_radians(160)

AM = AC * math.sin(angle_MCA_rad) / math.sin(angle_AMC_rad)
CM = AC * math.sin(angle_MAC_rad) / math.sin(angle_AMC_rad)

angle_MAB_rad = degrees_to_radians(70)
angle_MCB_rad = degrees_to_radians(40)

# Using Law of Sines in triangle BCM
# BM / sin(angle_MCB_rad) = BC / sin(angle_BMC_rad)
# BM / sin(degrees_to_radians(40)) = 1 / sin(angle_BMC_rad)

# Using Law of Sines in triangle ABM
# BM / sin(angle_MAB_rad) = AB / sin(angle_AMB_rad)
# BM / sin(degrees_to_radians(70)) = AB / sin(angle_AMB_rad)

# This approach is still complicated.

# Geometric approach revisited:
# Construct point N on AB such that angle BCN = 40.
# Triangle BNC is isosceles with BN = CN and angle BNC = 100.
# Angle ACN = angle ACB - angle BCN = 50 - 40 = 10.
# So N lies on CM.

# Construct point K on BC such that angle BAK = 50.
# Triangle ABK is isosceles with BK = AK and angle BKA = 80.
# Angle KAC = 80 - 50 = 30.

# Consider a point O such that triangle OBC is isosceles with OB = OC and angle BOC = 80.
# Then angle OBC = angle OCB = 50. So O is the center of the circumcircle of ABC.

# Let's use the fact that angle MBC = 40 (derived earlier).
# In triangle ABM, angle MAB = 70, angle ABM = 10.
# Angle AMB = 180 - 70 - 10 = 100.

Final Answer: The final answer is boxed100

Answer 2

[Trần Bá Hưng]

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