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. Describe how to find the ratio of the side opposite angle A to the hypotenuse of triangle ABC.

Answer 1

To find the ratio of the side opposite angle A to the hypotenuse of triangle ABC, you can follow these steps:

1. **Identify the sides of triangle ABC**:
- Side opposite angle A (let's denote this as \( a \)): This is side BC, which has a length of \( 8 \).
- The hypotenuse (side AC): This is \( 10 \).

2. **Set up the ratio**:
The ratio of the side opposite angle A to the hypotenuse can be represented as:

\[
\text{Ratio} = \frac{\text{side opposite angle A}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{8}{10}
\]

3. **Simplify the ratio**:
You can simplify \( \frac{8}{10} \) by dividing both the numerator and the denominator by their greatest common divisor, which is \( 2 \):

\[
\text{Ratio} = \frac{8 \div 2}{10 \div 2} = \frac{4}{5}
\]

So the ratio of the side opposite angle A to the hypotenuse of triangle ABC is \( \frac{4}{5} \).