Hello from MrBillDoesMath!
Answer:
x = 67
Discussion:
This is a tough one and it's late. But it looks good to me at 8:30pm );
Study attachment where I added angle labels "a", "b", and c"". I also drew auxiliary line LS parallel to QM (and hence parallel to PN)
1. As PN = NL, the bases angles ("b") of the isosceles triangle PNL are equal
2. As LQ bisects angles PQM, let the equal angles be labeled "a".
3. Angle PLS = b as alternate interior angles of parallel lines (PN, SL) are equal.
4. Angle SLQ= a as alternate interior angles of parallel lines (QM, SL) are equal.
5. In triangle QLM, a + angle QLM + 54 = 180, so angle QLM = 180 - 54 - a = 126- a
6. Line NLM has 180 degrees so
180 = (126-a) + a + b + b => 180 = 126 + 2b => b =27
7. Angle PLQ = 70 = a + b. Using 6) this gives 70 = a + 27 or
a = 70 - 27 = 43.
8. From triangle QPL,
a + x + 70 = 180 =>
43 + x + 70 = 180 =>
113 + x = 180 =>
x = 180 - 113 = 67
Good luck. Please tell me if you see any errors.
Thank you,
MrB