For ΔABC, ∠A = x + 30, ∠B = 2x - 4, and ∠C = 4x. If ΔABC undergoes a dilation by a scale factor of 1 2 to create ΔA'B'C' with ∠A' = 2x + 8, ∠B' = x + 18, and ∠C' = 5x - 22, which confirms that ΔABC∼ΔA'B'C by the AA criterion? A) ∠A = ∠A' = 50° and ∠B = ∠B' = 36° B) ∠A = ∠A' = 26° and ∠C = ∠C' = 44° C) ∠A = ∠A' = 52° and ∠C = ∠C' = 88° D) ∠B = ∠B' = 46° and ∠C = ∠C' = 100°

Respuesta :

i took one for the team, the answer is C, i got it wrong on usa test prep

Answer:

The option C). ∠A = ∠A' = 52° and ∠C = ∠C' = 88° is correct

Step-by-step explanation:

In ΔABC ,

Given : ∠A = x + 30, ∠B = 2x - 4, and ∠C = 4x

Now, using angle sum property of a triangle

∠A + ∠B + ∠C = 180°

⇒ x + 30 + 2x - 4 + 4x = 180°

⇒ 7x + 26 = 180°

⇒ 7x = 154°

⇒ x = 22

Now, ∠A = 22 + 30 = 52°,

∠B = 2 × 22 - 4 = 40°

∠C = 4 × 22 = 88°

Now, ΔABC ~ ΔA'B'C'

So, by using property of similar triangles

∠A = ∠A' , ∠B = ∠B' , ∠C = ∠C'

⇒ ∠A = ∠A' = 52° and ∠C = ∠C' = 88°

Hence, the option C). ∠A = ∠A' = 52° and ∠C = ∠C' = 88° is correct