Question

Right triangle ABC has side lengths AB =6, BC =8, and CA=10. A second right triangle has corresponding vertices of A', B', C', with side lengths of 32. 40 and 24. In 3-5 sentences, describe how to find the ratio of the side opposites<A to the hypotenuse of triangle ABC. Then use this ratio to identify the location of point A' in the second right triangle.

Answer 1

To find the ratio of the side opposite angle \( A \) to the hypotenuse of triangle \( ABC \), we first identify the lengths of the relevant sides: the side opposite angle \( A \) is \( BC \) (which is \( 8 \)), and the hypotenuse \( CA \) is \( 10 \). The ratio can be calculated as \( \frac{BC}{CA} = \frac{8}{10} = \frac{4}{5} \).

For the second right triangle with side lengths 32, 40, and 24, we can identify the hypotenuse as 40 (which corresponds to \( CA \) in the first triangle) and the side opposite \( A' \) as 24 (which corresponds to \( BC \)). To find the location of point \( A' \), we need to maintain the same ratio; thus, we calculate the length of the side opposite \( A' \) using the hypotenuse: \( \frac{4}{5} \) of \( 40 \), which equals \( 32 \). Since the sides are in proportion, we may place point \( A' \) at the point that is \( 24 \) units from \( C' \) (the right-angle vertex) alongside the ratio, while ensuring the triangle properties are maintained.