Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB, YZ = 5, and XZ = 5.59.

Triangles XYZ and ACB; angles Y and C both measure 90 degrees, angles A and X are congruent.

Part A: Find sin ∠X. Explain how sin ∠X compares to sin ∠A. (5 points)

Part B: Find CB and BA. You must show all work and calculations to receive full credit. (5 points)

To find sin ∠X, we can use the formula sin(∠X) = opposite/hypotenuse. In triangle XYZ, since angle Y measures 90 degrees, angle X is the right angle. Therefore, the side opposite to angle X is side YZ, which is 5. Since the hypotenuse is XZ, which is 5.59, we have:

sin(∠X) = 5/5.59 ≈ 0.8951

To compare sin ∠X to sin ∠A, note that angles A and X are congruent. Since sine is a trigonometric function that depends only on the angle, sin ∠X = sin ∠A. Therefore, sin ∠X is equal to sin ∠A.

Part B:

Since triangle ACB was dilated from triangle XYZ by a scale factor of 2, the corresponding sides are in a ratio of 1:2. Therefore, CB = 2*YZ = 2*5 = 10.

To find BA, we can use the Pythagorean theorem in triangle ACB.

Using AC as the hypotenuse, we have:

AC^2 = AB^2 + BC^2

AC^2 = BA^2 + 10^2

AC^2 = BA^2 + 100

Since AC is twice the length of XZ (5.59), we have:

AC = 2*XZ = 2*5.59 = 11.18

Substitute AC and solve for BA:

(11.18)^2 = BA^2 + 100

124.9924 = BA^2 + 100

BA^2 = 24.9924

BA ≈ √24.9924

BA ≈ 5

Therefore, CB = 10 units and BA ≈ 5 units.