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°

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

idk

step-by-step explanation: