Answer:
Explanation:
It seems easiest to relate the angles if we can take advantage of the fact that alternate interior angles where a transversal crosses parallel lines are congruent. We can use this fact a couple of ways:
1. draw line CF to the right from point C parallel to AB and DE. Then angle BCF is 35°, matching angle CBA.
Angles FCD and CDE are supplementary, being same-side angles where transversal CD crosses parallel lines CF and DE. Hence angle FCD is 180° -120° = 60°.
Angle C is the sum of angles BCF and FCD, so is 35° + 60° = 95°. In short, ...
a° = 95°
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2. We can extend lines BC and ED so they meet at point G, forming triangle CGD. The angle at G is an alternate interior angle with angle B where transversal BG crosses parallel lines AB and GE. Hence angle G is 35°.
Angle CDG is the supplement to angle CDE, so is 180° -120° = 60°. And angle a° is the sum of opposite interior angles CDG and CGD, so is ...
a° = ∠CDG + ∠CGD = 60° +35°
a° = 95°