Question
Right triangle ABC has side lengths AB=6 , BC=8 , and CA=10 . A second right triangle has corresponding vertices of A′ , B′ , and C′ , with side lengths of 32, 40, and 24. In 3–5 sentences, describe how to find the ratio of the side opposite ∠A to the hypotenuse of triangle ABC . Then use this ratio to identify the location of point A′ in the second right triangle.(4 points)
Answers
GPT-4o mini
To find the ratio of the side opposite ∠A to the hypotenuse of triangle ABC, you first identify the side lengths relative to angle A. In triangle ABC, side BC (which measures 8) is opposite ∠A, and the hypotenuse CA measures 10. The ratio can be calculated as the length of the opposite side (BC) divided by the length of the hypotenuse (CA): \( \frac{BC}{CA} = \frac{8}{10} = \frac{4}{5} \).
In the second right triangle A′B′C′, the corresponding hypotenuse (A′C′) measures 32. To find the side opposite ∠A′ (which corresponds to side B′C′), you compute the same ratio: \( \frac{4}{5} = \frac{B'C'}{A'C'} \). Setting \( B'C' = x \) gives \( \frac{x}{32} = \frac{4}{5} \). Solving for \( x \) yields \( B'C' = \frac{4}{5} \times 32 = 25.6 \). Therefore, point A' is located such that side B'C' has a length of 25.6 in triangle A'B'C'.
In the second right triangle A′B′C′, the corresponding hypotenuse (A′C′) measures 32. To find the side opposite ∠A′ (which corresponds to side B′C′), you compute the same ratio: \( \frac{4}{5} = \frac{B'C'}{A'C'} \). Setting \( B'C' = x \) gives \( \frac{x}{32} = \frac{4}{5} \). Solving for \( x \) yields \( B'C' = \frac{4}{5} \times 32 = 25.6 \). Therefore, point A' is located such that side B'C' has a length of 25.6 in triangle A'B'C'.