Step-by-step explanation:
This is your answer
Given that PQ and RS are drawn with KL as tranversal intersecting PQ at M and RS at point N. Angle QMN is congruent to angle LNS because they are alternate to each other. The theorem that Kari can use to show that the meansure of QML is supplementary to the measure of angle SNK is Alternate Exterior Angles Theorem.
This is because angle KNR is equal to QML by alternate exterior angles theorem so is angle MLP and SNK
Answer:
sin A =cos B
Step-by-step explanation:
sin A =opp/hyp =BC/AB
cos B=adj/hyp = BC/AB
