For δabc, ∠a = 3x, ∠b = 2x - 3, and ∠c = x + 3. if δabc undergoes a dilation by a scale factor of 2 to create δa'b'c' with ∠a' = 2x + 30, ∠b' = x + 27, and ∠c' = 1/2x + 18, which confirms that δabc∼δa'b'c by the aa criterion? a) ∠a = ∠a' = 75° and ∠b = ∠b' = 47° b) ∠a = ∠a' = 60° and ∠b = ∠b' = 37° c) ∠b = ∠b' = 57° and ∠c = ∠c' = 33° d) ∠a = ∠a' = 105° and ∠c = ∠c' = 38°

C)∠B = ∠B' = 57° and ∠C = ∠C' = 33°

Step-by-step explanation:

GIven:- In ΔABC, ∠A = 3x, ∠B = 2x - 3, and ∠C = x + 3.

In ΔA'B'C' ∠A' = 2x + 30, ∠B' = x + 27, and ∠C' = 1/2x + 18

By angle sum property in  ΔABC

⇒ ∠A =3(30)=90°

∠B=2(30)-3=57°

∠C=30+3=33°

By angle sum property in  ΔA'B'C'

∠A'=2(30)+30=90°

∠B'=30+27=57°

∠C'=

therefore, C) ∠B = ∠B' = 57° and ∠C = ∠C' = 33° gives ΔABC∼ΔA'B'C by the AA criterion.

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.