Kari drew two parallel lines PQ and RS intersected by a transversal KL, as shown below:
2 answers:
It is given in the question that
Two parallel lines PQ and RS are drawn with KL as a transversal intersecting PQ at point M and RS at point N. Angle QMN is shown congruent to angle LNS.
In parallel lines, alternate interior angles and alternate exterior angles are congruent .
And sum of angles of same side interior angles is 180 degree.
Therefore sum of measurements of angles QML and SNK is 180 degree. So they are supplementary angles.
Correct option is C.
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By the law of cosines, the angle 
satisfies
which reduces to
which converts to approximately 77 degrees (this can be done my multiplying by 180 and dividing by pi).
which reduces to
which converts to approximately 77 degrees (this can be done my multiplying by 180 and dividing by pi).
Since ABCD is a parallelogram, the opposite sides will be parallel and equal,
Consider that AC acts as a transversal to the parallel lines AB and CD, so we can write,
So by the ASA criteria, the triangle AED is congruent to the triangle CEB,
Then the corresponding parts of the triangles will be equal,
Hence Proved.