AABC has vertices A(-3,-4), B(1,5), and C(4.1). AABC is translated so that B'is located at point (-3, 2). State the
1 answer:
You might be interested in
Angles formed when two parallel are intersected by a common transversal include, corresponding, alternate interior and exterior, same-side interior, and vertically opposite angles
The correct option is Same–Side Int. ∠s Thm. m∠1 = 30°, m∠5 = 150°
The reason the selection is correct is as follows:
The given angles are;
m∠1 = (30·x - 30)°. m∠5 = (40·x + 70)°
∠1, and ∠5, are located on the same side of the transversal and are both formed on the interior side of the parallel lines
Therefore, they are same–side interior angles
Same side interior angles theorem states that same side interior angles formed between parallel lines are supplementary
Therefore, we have;
m∠1 + m∠5 = 180°
Which gives;
(30·x - 30)° + (40·x + 70)° = 180°
(70·x + 40)° = 180°
70·x = 180° - 40° = 140°
m∠1 = (30·x - 30)° = (30 × 2 - 30)° = 30°
m∠1 = 30°
m∠5 = (40·x + 70)° = (40 × 2 + 70)° = 150°
m∠5 = 150°
The correct option is therefore;
Same-Side Int. ∠s Thm. m∠1 = 30°, m∠5 = 150°
Learn more about parallel lines cut by a transversal here: