Question
Right triangle ABC has 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. What is the ratio of the side opposite angle A to the hypotenuse of triangle ABC. Write your answer is a simplified fraction. (10 points)
Answers
Answered by
GPT-4o mini
In triangle ABC, we know that:
- \( AB = 6 \)
- \( BC = 8 \)
- \( CA = 10 \)
Here, angle A is opposite side BC, which has a length of \( 8 \), and the hypotenuse, which is side CA, has a length of \( 10 \).
Next, we need to find the ratio of the side opposite angle A (BC) to the hypotenuse (CA):
\[
\text{Ratio} = \frac{\text{BC}}{\text{CA}} = \frac{8}{10}
\]
To simplify this fraction, we divide both the numerator and the denominator by their greatest common divisor, which is \( 2 \):
\[
\frac{8 \div 2}{10 \div 2} = \frac{4}{5}
\]
Thus, the ratio of the side opposite angle A to the hypotenuse of triangle ABC is
\[
\boxed{\frac{4}{5}}.
\]
- \( AB = 6 \)
- \( BC = 8 \)
- \( CA = 10 \)
Here, angle A is opposite side BC, which has a length of \( 8 \), and the hypotenuse, which is side CA, has a length of \( 10 \).
Next, we need to find the ratio of the side opposite angle A (BC) to the hypotenuse (CA):
\[
\text{Ratio} = \frac{\text{BC}}{\text{CA}} = \frac{8}{10}
\]
To simplify this fraction, we divide both the numerator and the denominator by their greatest common divisor, which is \( 2 \):
\[
\frac{8 \div 2}{10 \div 2} = \frac{4}{5}
\]
Thus, the ratio of the side opposite angle A to the hypotenuse of triangle ABC is
\[
\boxed{\frac{4}{5}}.
\]