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For ΔABC, ∠A = 3x - 8, ∠B = 5x - 6, and ∠C = 4x + 2. If ΔABC undergoes a dilation by a scale factor of 1/2 to create ΔA'B'C' with ∠A' = 2x + 8, ∠B' = 90 - x, and ∠C' = 5x - 14, which confirms that ΔABC∼ΔA'B'C by the AA criterion?


A) ∠A = ∠A' = 37° and ∠B = ∠B' = 69°
B) ∠A = ∠A' = 22° and ∠C = ∠C' = 42°
C) ∠B = ∠B' = 37° and ∠C = ∠C' = 33°
D) ∠B = ∠B' = 74° and ∠C = ∠C' = 66°

Respuesta :

Answer:

D. ∠B = ∠B' = 74° and ∠C = ∠C' = 66°

Step-by-step explanation:

Given,

[tex]\angle A=3x-8\\\angle B=5x-6\\\angle C=4x+2[/tex]

We know,sum of all 3 angles of a triangle is [tex]180[/tex]°

Doing so ,

[tex]\angle A+\angle B+\angle C=180\\3x-8+5x-6+4x+2=180[/tex]

Simplifying all like terms,

[tex]12x-12=180\\12x-12+12=180+12\\12x=192[/tex]

Dividing both side by 12

[tex]x=\frac{192}{12}[/tex]

[tex]x=16[/tex]

Now we plug this in each angles and find their exact values.

[tex]\angle A=3x-8=(3\times16)-8=48-8=40[/tex]°

[tex]\angle B=5x-6=(5\times16)-6=80-6 =74[/tex]°

[tex]\angle C=4x+2 = (4\times16)+2=64+2=66[/tex]°

Now substituting the [tex]x=16[/tex] in angles of dilated triangle.

[tex]\angle A'=(2\times16)+8=32+8= 40[/tex]°

[tex]\angle B'=90-16=74 [/tex]°

[tex]\angle C'=(5\times16)-14=80-14=66[/tex]°

We see that all the corresponding angles are equal and as far as the options are concerned only option (D) matches.

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