For δabc, ∠a = 4x + 7, ∠b = 2x + 3, and ∠c = 6x - 10. if δabc undergoes a dilation by a scale factor of 2 to create δa'b'c' with ∠a' = 5x - 8, ∠b' = 3x - 12, and ∠c' = 7x - 25, which confirms that δabc∼δa'b'c by the aa criterion?

Bizzare problem.  What are the choices?

The triangle angles add to 180.

4x + 7 + 2x +3 + 6x - 10 = 180

12x = 180

x = 15

Dilation doesn't change the angles so we must have

4x + 7 = 5x - 8  and 2x + 3 = 3x - 12 and 6x - 10 = 7x - 25

Fortunately x=15 is the solution to all of these so it's consistent.  

The value of x is 15.

Step-by-step explanation:

It is given that in ΔABC, ∠A = 4x + 7, ∠B = 2x + 3, and ∠C = 6x - 10.

According to angle sum property of triangle, the sum of all interior angles of a triangle is 180°.

In ΔABC,

Combine like terms.

Divide both sides by 12.

The value of x is 15.

We get

Two corresponding angles of triangles are congruent. By AA criterion

Hence proved.

do you have the measurements?