For δabc, ∠a = 2x - 2, ∠b = 2x + 2, and ∠c = 5x. if δabc undergoes a dilation by a scale factor of 1 2 to create δa'b'c' with ∠a' = 58 - x, ∠b' = 3x - 18, and ∠c' = 120 - x, which confirms that δabc∼δa'b'c by the aa criterion? a) ∠a = ∠a' = 38° and ∠b = ∠b' = 42° b) ∠a = ∠a' = 19° and ∠c = ∠c' = 21° c) ∠b = ∠b' = 18° and ∠c = ∠c' = 50° d) ∠b = ∠b' = 32° and ∠c = ∠c' = 75°

A)∠A = ∠A' = 38° and ∠B = ∠B' = 42°

The sum of angles in ∆ABC is 180°, so ...

... (2x -2) + (2x +2) + (5x) = 180

... 9x = 180

... x = 20

and the angles of ∆ABC are ∠A = 38°, ∠B = 42°, ∠C = 100°.

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The sum of angles of ∆A'B'C' is 180°, so ...

... (58 -x) +(3x -18) +(120 -x) = 180

... x +160 = 180

... x = 20

and ∠A' = 38°, ∠B' = 42°, ∠C' = 100°.

The values of angle measures of ∆ABC match those of ∆A'B'C', so we can conclude ...

... A)∠A = ∠A' = 38° and ∠B = ∠B' = 42°

29 is the correct answer

step-by-step explanation:

ou would have 12 of each