Question #0de9f

I won’t provide the entire solution, but rather the process that you can use to complete the exercise.

Given the parallel lines and the angle at E, the angle at K must be the same (#69^o#).
This leads to angle 12 being (180 – 69 – 90) = #21^o#.
From that, angle 8 must be 21 + 90 = #111^o#, and then the angle on the other side of it must be (180 - 111) = #69^o#.
EFG is an isosceles triangle, so angle 7 is also #69^o#, leading to angle 4 = (180 – 69 – 69) = #42^o#. Then angle 3 is (180 – 42 – 69) = #69^o# also.

Further, the angle at M is (180 – 90 – 21) = #69^o#. This is the same angle as that at G, so angle 8 is #111^o# also, and confirms our previous determination of angles 4 and 7.
Angle 9 is #116^o#, so angle 10 is #64^o#.

So far: Angle 3: #69^o# Angle 4: #42^o# Angle 5: ?? Angle 6: ?? Angle 7: #69^o# Angle 8: #111^o# Angle 9: #116^o# Angle 10: #64^o# Angle 11: ?? Angle 12: #21^o#

Continue on, using the angles you calculate with the properties of the parallel lines and triangles to calculate the remaining angles 5, 6 and 11