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The given triangle is a right triangle, and we are tasked with finding trigonometric ratios using the provided side lengths and angles. Here’s how to proceed:

### Given information:
- \( AB = 8 \)
- \( BC = AC = 4\sqrt{2} \) (as given in part a)

### b) \( \sin(A) \):
\( \sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} \)

- \( BC = 4\sqrt{2} \)
- \( AB = 8 \)

So,
\[
\sin(A) = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2} \approx 0.707
\]

- Decimal: \( 0.707 \)
- Fraction: \( \frac{\sqrt{2}}{2} \)

### c) \( \cos(A) \):
\( \cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AC}{AB} \)

Since \( AC = 4\sqrt{2} \) and \( AB = 8 \), this is the same as the sine function:
\[
\cos(A) = \frac{\sqrt{2}}{2} \approx 0.707
\]

- Decimal: \( 0.707 \)
- Fraction: \( \frac{\sqrt{2}}{2} \)

### d) \( \tan(B) \):
\( \tan(B) = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{AC} \)

Since \( AB = 8 \) and \( AC = 4\sqrt{2} \):
\[
\tan(B) = \frac{8}{4\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414
\]

- Decimal: \( 1.414 \)
- Fraction: \( \sqrt{2} \)

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Would you like any further explanation on these steps or any other details?

### 5 related questions:
1. How do we use the Pythagorean theorem to verify side lengths?
2. What is the process of simplifying radical expressions like \( \frac{4\sqrt{2}}{8} \)?
3. How can we find the tangent of angle \( A \) instead of angle \( B \)?
4. How do we convert a trigonometric ratio to its corresponding angle?
5. What happens to the trigonometric values if the side lengths change?

**Tip**: Always double-check radical simplifications to ensure proper trigonometric ratios!

Use the Pythagorean theorem to find the missing length, and then write the desired trigonometric ratios (sin(A), cos(A), tan(B)) in decimals and simplified fractions for the given right triangle with side lengths: AB = 8, BC = AC = 4√2.

Learn how to use the Pythagorean theorem to find side lengths in a right triangle and calculate trigonometric ratios (sin, cos, tan) with step-by-step solutions. This problem is suitable for students in Grades 9-10.

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