Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB and cos ∠X = 2 and 5 tenths over 5 and 59 hundredths. Part A: Use complete sentences to explain the special relationship between the trigonometric ratios of triangles XYZ and ABC. You must show all work and calculations to receive full credit. (5 points) Part B: Explain how to find the measures of segments AC and AB. You must show all work and calculations to receive full credit. (5 points) Triangles XYZ and ACB; angles Y and C both measure 90 degrees, angles A and X are congruent.

Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB and cos ∠X = 2 and 5 tenths over 5 and 59 hundredths. Part A: Use complete sentences to explain the special relationship between the trigonometric ratios of triangles XYZ and ABC. You must show all work and calculations to receive full credit. (5 points) Part B: Explain how to find the measures of segments AC and AB. You must show all work and calculations to receive full credit. (5 points) Triangles XYZ and ACB; angles Y and C both measure 90 degrees, angles A and X are congruent.

Step-by-step explanation:

step-by-step explanation:

equation

you are asked to write the equation

  expression = 0

using the given expression. when you do that, you have

  2x^2-2x-12 = 0

factoring

you can see that each of the coefficients is even, so a factor of 2 can be removed immediately.

  2(x^2 -x -6) = 0

if you multiply the binomials (x+a)(x+b), you get

  (x +a)(x +b) = x^2 +(a+b)x +ab

this tells you the coefficient of x is the sum of two factors of the constant term. here, that means -1 is the sum of two factors of -6. we want to know those factors so we can factor the trinomial.

  -6 = -1×6 = -2×3 = -3×2 = -6×1

the factor pair that sums to -1 is {-3, 2}, so our factorization is

  2(x -3)(x +2) = 0

solutions

the solutions to this equation are the values of x that make the product zero. a product is zero only when only of the factors of the product is zero. here, that means

  x - 3 = 0

  x = 3 . . . add 3 to both sides of the equation

or

  x +2 = 0

  x = -2 . . . add -2 to both sides of the equation

the solutions to the equation are x=3 or x=-2. both of these values of x satisfy the equation—they make it be true.