Question

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°

Answer 1

I am not 100% sure but I think the answer is C I hope I helped you and Good luck

Answer 2

Answer: 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

$\Rightarrow3x+2x-3+x+3=180^{\circ}\\\Rightarrow6x=180^{\circ}\\\Rightarrow\ x=30^{\circ}$

⇒ ∠A =3(30)=90°

∠B=2(30)-3=57°

∠C=30+3=33°

By angle sum property in  ΔA'B'C'

$\Rightarrow2x+x+27+\frac{1}{2}x+18=180^{\circ}\\\Rightarrow\frac{7}{2}x+75=180^{\circ}\\\Rightarrow7x+150=180^{\circ}\\\Rightarrow7x=210^{\circ}\\\Rightarrow\ x=30^{\circ}$

∠A'=2(30)+30=90°

∠B'=30+27=57°

∠C'=$\frac{1}{2}(30)+18=33^{\circ}$

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