Triangle bac was dilated from triangle bde at a scale factor of 2. what proportion proves that tan∠d = tan∠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. 2 over 3 and 61 hundredths = 4 over 7 and 22 hundredths three halves = six fourths 3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths two thirds = four sixths

three halves = six fourths

Step-by-step explanation:

 tanD = BE/BD = ³/₂

 tanA = BC/BA = ⁶/₄

     ³/₂ = ⁶/₄

∴ tanD = tanA

three halves = six fourths

Step-by-step explanation:

Given that

BD measures 2 unitsBE measures 3 unitstriangle BAC was dilated from triangle BDE at a scale factor of 2

Then:

BA measures: scale factor*BD = 2*2 = 4 units (point D is between points B and A)BC measures: scale factor*BE = 2*3 = 6 units (point E is between points B and C)

Tangent of an angle is computed as follows:

tan(angle) = opposite side/adjacent side

Then:

tan∠D = BE/BD = 3/2tan∠A = BC/BA = 6/4 = 3/2

i know this is for Penn foster i took it and here is the answer

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