Question
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
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
Answers
GPT-4o mini
To show that \(\tan \angle D = \tan \angle A\) based on the dilation of triangles BAC and BDE with a scale factor of 2, we need to compare the ratios of their side lengths corresponding to the angles.
Since triangle BAC is a dilation of triangle BDE with a scale factor of 2, we know the following:
1. The sides of triangle BAC are double the lengths of the corresponding sides of triangle BDE.
2. For right triangle BDE, we can find \(\tan \angle D\) and for triangle BAC, \(\tan \angle A\).
Given:
- \( BD = 2 \) (adjacent side for \(\angle D\)),
- \( BE = 3 \) (opposite side for \(\angle D\)),
- \( DE \) can be computed using the Pythagorean theorem:
\[
DE = \sqrt{BE^2 + BD^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}.
\]
Now we can find \(\tan \angle D\):
\[
\tan \angle D = \frac{BE}{BD} = \frac{3}{2}.
\]
For triangle BAC, since it is a dilation of triangle BDE with a scale factor of 2, the corresponding sides will be:
- Side corresponding to \( BD \): \( AB = 2 \times BD = 2 \times 2 = 4 \),
- Side corresponding to \( BE \): \( AC = 2 \times BE = 2 \times 3 = 6 \).
Now we can find \(\tan \angle A\):
\[
\tan \angle A = \frac{AC}{AB} = \frac{6}{4} = \frac{3}{2}.
\]
Since both \(\tan \angle D\) and \(\tan \angle A\) are equal:
\[
\tan \angle D = \frac{3}{2} \quad \text{and} \quad \tan \angle A = \frac{3}{2},
\]
Thus, we have that:
\[
\tan \angle D = \tan \angle A.
\]
From the choices you provided, the proportion that proves this is **3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths**, where the left side approximates the same ratio as the right side. So, in context, the proportions that correspond with the tangent ratios being identical can also be related to the scale factor of the lengths that sustain the angles' tangents.
Since triangle BAC is a dilation of triangle BDE with a scale factor of 2, we know the following:
1. The sides of triangle BAC are double the lengths of the corresponding sides of triangle BDE.
2. For right triangle BDE, we can find \(\tan \angle D\) and for triangle BAC, \(\tan \angle A\).
Given:
- \( BD = 2 \) (adjacent side for \(\angle D\)),
- \( BE = 3 \) (opposite side for \(\angle D\)),
- \( DE \) can be computed using the Pythagorean theorem:
\[
DE = \sqrt{BE^2 + BD^2} = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}.
\]
Now we can find \(\tan \angle D\):
\[
\tan \angle D = \frac{BE}{BD} = \frac{3}{2}.
\]
For triangle BAC, since it is a dilation of triangle BDE with a scale factor of 2, the corresponding sides will be:
- Side corresponding to \( BD \): \( AB = 2 \times BD = 2 \times 2 = 4 \),
- Side corresponding to \( BE \): \( AC = 2 \times BE = 2 \times 3 = 6 \).
Now we can find \(\tan \angle A\):
\[
\tan \angle A = \frac{AC}{AB} = \frac{6}{4} = \frac{3}{2}.
\]
Since both \(\tan \angle D\) and \(\tan \angle A\) are equal:
\[
\tan \angle D = \frac{3}{2} \quad \text{and} \quad \tan \angle A = \frac{3}{2},
\]
Thus, we have that:
\[
\tan \angle D = \tan \angle A.
\]
From the choices you provided, the proportion that proves this is **3 over 3 and 61 hundredths = 6 over 7 and 22 hundredths**, where the left side approximates the same ratio as the right side. So, in context, the proportions that correspond with the tangent ratios being identical can also be related to the scale factor of the lengths that sustain the angles' tangents.