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Right triangle \( A B C \) has side lengths \( A B=6, B C=6 \sqrt{3} \), and \( A C=12 \). A second right triangle, \( A^{\prime} B^{\prime} C^{\prime} \), has side lengths of \( 8 \sqrt{3}, 8 \), and 16 . Find the ratio of the side opposite \( \angle C \) ) the hypotenuse of triangle \( A B C \). Then use this ratio to identify the location of point \( C^{\prime} \) in the second right triangle. (1 point)

Ask by Wang Frazier. in the United States
Jan 23,2025

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The ratio of the side opposite \( \angle C \) to the hypotenuse in triangle \( ABC \) is \( \frac{1}{2} \). Applying this ratio to triangle \( A'B'C' \), point \( C' \) is located opposite the side of length 8.

Solución

In triangle \( ABC \): - The hypotenuse is \( AC = 12 \). - The side opposite \( \angle C \) is \( AB = 6 \). The ratio is: \[ \frac{\text{Opposite } \angle C}{\text{Hypotenuse}} = \frac{6}{12} = \frac{1}{2} \] Applying this ratio to triangle \( A'B'C' \) where the hypotenuse is 16: \[ \frac{\text{Opposite } \angle C'}{16} = \frac{1}{2} \implies \text{Opposite } \angle C' = 8 \] Thus, point \( C' \) is located opposite the side of length 8 in the second triangle. **Answer:** After simplifying, the ratio is ½. Thus C′ lies opposite the side of length 8 in triangle A′B′C′.

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To find the ratio of the side opposite \( \angle C \) to the hypotenuse of triangle \( ABC \), we identify the components. In triangle \( ABC \), the side opposite \( \angle C \) is \( BC = 6\sqrt{3} \) and the hypotenuse is \( AC = 12 \). Thus, the ratio is: \[ \text{Ratio} = \frac{BC}{AC} = \frac{6\sqrt{3}}{12} = \frac{\sqrt{3}}{2} \] In triangle \( A' B' C' \), we need to find where \( C' \) would be located based on this ratio. Here, the hypotenuse \( A'C' = 16 \) and we can apply the same ratio to identify the side opposite \( \angle C' \): \[ \text{Side opposite } \angle C' = \text{Ratio} \times A'C' = \frac{\sqrt{3}}{2} \times 16 = 8\sqrt{3} \] So, point \( C' \) is located such that it is opposite \( 8\sqrt{3} \) in triangle \( A' B' C' \).

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