PLEASE HELP! (05.01 MC) Triangle BAC was dilated from triangle BDE at a scale factor of 2. What proportion proves that cos∠D = cos∠A?

Triangles BDE and BAC, in which angle B is a right angle, point D is between points B and A, and point E is between points B and C; BD measures 2 units, BE measures 3 units, and DE measures 3 and 61 hundredths units.

A. 2 over 3 and 61 hundredths = 4 over 7 and 22 hundredths
B. three halves = six fourths
C. 3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths
D. two thirds = four sixths

3) Option C is correct.

3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths

Step-by-step explanation:

Triangle BDE is dilated by a scale factor of 2 to obtain triangle BAC.

Dilation by a scale factor of 2 means that all the sides of triangle BAC are twice as much as the corresponding sides of triangle BDE.

And triangle BDE is similar to triangle BAC.

It also means that the corresponding angles are necessarily equal.

BD = 2

BE = 3

ED = 3.61

BA = 2 × BD = 2 × 2 = 4

BC = 2 × BE = 2 × 3 = 6

CA = 2 × ED = 2 × 3.61 = 7.22

Cos ∠D according to trigonometric relations is given as (adj/hyp)

Adj = Adjacent side = BE = 3

Hyp = hypotenuse side = ED = 3.61

Cos ∠D = (3/3.61)

Cos ∠A can also be similarly obtained from trigonometric relations as (adj/hyp)

Adj = Adjacent side = BC = 6

Hyp = hypotenuse side = CA = 7.22

Cos ∠A = (6/7.22)

Since the two angles are corresponding angles of two similar triangles,

We can easily see that

Cos ∠D = Cos ∠A

(3/3.61) = (6/7.22) = 0.8310

Which is necessarily equal to each other

Hence, the proportion that proves that Cos ∠D = Cos ∠A is 3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths.

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