The ABC triangle is isosceles with the BC base. The straight line MK is parallel to the base M belongs to BC, K belongs to AB. Find the angles of the CME triangle if the angle B = 66 degrees, the angle C = 48 degrees.
Given: isosceles triangle ABC:
base of the aircraft;
MK is parallel to the BC:
M belongs to the AC;
K belongs to AB;
angle B = 66 degrees;
angle C = 48 degrees.
Find the angles of the triangle KAM -?
1) Consider an isosceles triangle ABC. Angle B = angle C = 66 degrees;
2) If MK is parallel to BC, then the angle KBC = AKM = 66 degrees, as these are the corresponding angles for these parallel sides and secant AB;
3) If MK is parallel to BC, then the angle BCM = KMA = 66 degrees as these are the corresponding angles for these parallel sides and secant AC;
4) Angle A is common. Then the angle KAM = 48 degrees.
Answer: 66 degrees; 66 degrees; 48 degrees.
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