Question

Right triangle ABC has side lengths AB=8, BC=15, and AC=17. A second right triangle, A'B'C', has side lengths of 34, 30, and 16. First, find the ratio of the side opposite <C to to the hypotenuse of triangle ABC. Then, use this ratio to identify the side lengths opposite point C' on the second right triangle.

The length of the side opposite C' is ____ units.

Answer 1

To find the ratio of the side opposite angle C to the hypotenuse in triangle ABC, we first identify the relevant side lengths. In triangle ABC, it is given that:

- The side opposite angle C (side AB) = 8 units.
- The hypotenuse (side AC) = 17 units.

The ratio of the side opposite angle C to the hypotenuse is calculated as follows:

\[
\text{Ratio} = \frac{\text{side opposite C}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{8}{17}.
\]

Next, we apply this ratio to triangle A'B'C' to find the side lengths opposite point C'. In triangle A'B'C', the side opposite angle C' corresponds to A'B', which has a length of 30 units (as A'B' = BC in triangle ABC). The hypotenuse A'C' corresponds to AC in triangle ABC, which has a length of 34 units.

To find the length of the side opposite angle C' in triangle A'B'C', we use the ratio established:

Let \( x \) be the length of the side opposite angle C' in triangle A'B'C'. We set up the equation according to the ratio:

\[
\frac{x}{34} = \frac{8}{17}.
\]

To isolate \( x \), we can cross-multiply:

\[
17x = 8 \times 34.
\]

Calculating \( 8 \times 34 \):

\[
8 \times 34 = 272.
\]

So we have:

\[
17x = 272.
\]

Now we divide both sides by 17 to solve for \( x \):

\[
x = \frac{272}{17} = 16.
\]

Thus, the length of the side opposite point C' on the second right triangle is:

\[
\boxed{16} \text{ units.}
\]