The ABC triangle is isosceles, the MC is parallel to the AC, the KMC angle is 30 °, find the BMC angle.

After the triangle ABC has been divided into two triangles by the line MK, we have formed a trapezoid AMKC.
Since the triangle was originally isosceles, the trapezoid AMKC should also have the same lateral faces MA and KC.
It also has the same paired angles AMK and MKC and MAC and KCA.
The angles МКС and КМС are equal to 30 °. This means that the remaining angles will be the difference between the sum of all angles and the sum of known angles.
Let us find the sum of the angles MAC and KCA:
360 ° – 30 ° – 30 ° = 300 °.
Find the angle BMK:
180 ° – 30 ° = 150 °.
ANSWER: ∠ВМK = 150 °.