In parallelogram abcd, the angle abc is 150 °. The bisector of the bad angle intersects side bc at point L. Find the angle ALC.

Let’s find the values that are needed to solve the problem. Determine the value of the angle BAD:
Angle CBA + Angle BAD = 180;
150 + angle BAD = 180;
angle BAD = 180 – 150 = 30 degrees.
The bisector divides the angle in half, so the angle BAL = angle LAD = 30: 2 = 15.
Consider triangle ABL:
angle ABL + angle BAL + angle BLA = 180;
150 + 15 + angle BLA = 180;
angle BLA = 180 – 150 – 15;
BLA angle = 15 degrees.
Let’s find the required angle:
angle BLA + angle ALC = 180;
15 + angle ALC = 180;
angle ALC = 180 – 15;
angle ALC = 165.
Answer: 165 degrees.