Question

ABC a right triangle has side lenghts AB=4, BC=4 square root of 3, and CA=8. A second right triangle has side lenghts of 8 square root of 3, 16 square root of 3 and 24. This second right triangle is uspposed to ahve corresponding vertices of A' B' and C', but these are not labeled. Find the ratio of the side opposite < A to the hypotenuse of ABC. use this ratio to identify the location of 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 sides based on the vertices provided.

In triangle \( ABC \):

• \( AB = 4 \) (one leg)
• \( BC = 4 \sqrt{3} \) (the other leg)
• \( CA = 8 \) (the hypotenuse)

Step 1: Find the sine of angle \( A \)

The side opposite angle \( A \) is \( BC \) which has a length of \( 4\sqrt{3} \). The hypotenuse is \( CA \) which has a length of \( 8 \).

The sine of angle \( A \) is given by the ratio:

\[ \sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{CA} = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2} \]

Step 2: Identify the corresponding triangle and location of \( A' \)

Now, we consider the second triangle which has the following side lengths:

• \( 8\sqrt{3} \)
• \( 16\sqrt{3} \)
• \( 24 \)

We need to check which side corresponds to the hypotenuse. The largest side, which is \( 24 \), will be the hypotenuse.

Now, we compute the sine of angle \( A' \) in the second triangle using the same ratio. The side opposite \( A' \) in the second triangle is \( 16\sqrt{3} \), and the hypotenuse is \( 24 \).

Calculating the sine:

\[ \sin A' = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{16\sqrt{3}}{24} = \frac{2\sqrt{3}}{3} \]

Step 3: Matching the sine values

From our previous calculation, we found that:

\[ \sin A = \frac{\sqrt{3}}{2} \]

Now we need to verify if \( \sin A' = \frac{\sqrt{3}}{2} \) holds for either \( 8\sqrt{3} \) or \( 16\sqrt{3} \) in the second triangle.

Verify whether \( A' \) corresponds to the calculated values:

For both triangles' angles:

• The angle \( A \) corresponds to \( 60^\circ \) since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).
• The angle \( A' \) is \( 60^\circ \) corresponding to the larger side:

Using our hypotenuse and side relationships:

\[ \text{Identifying}\ A'\text{ } \text{correspondence to the side opposite}\ 60^\circ: \]

Conclusion:

The location of \( A' \) in the second triangle is opposite the side of length \( 16\sqrt{3} \).

Thus, in conclusion:

The ratio of the side opposite \( <A \) to the hypotenuse of triangle \( ABC \) is \( \frac{\sqrt{3}}{2} \), and the location of \( A' \) corresponds to \( 16\sqrt{3} \) in the second triangle.