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Let's go through each part of this assignment.

### Part A: Find the Six Trigonometric Ratios for Angle \( \theta \)

Given triangle has:
- Opposite side = 7
- Adjacent side = 5
- Hypotenuse (using the Pythagorean theorem) = \( \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} \)

Now, we can find the six trigonometric ratios:

1. **\(\sin \theta\)**: \(\frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{\sqrt{74}}\)
2. **\(\cos \theta\)**: \(\frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{\sqrt{74}}\)
3. **\(\tan \theta\)**: \(\frac{\text{opposite}}{\text{adjacent}} = \frac{7}{5}\)
4. **\(\csc \theta\)**: \(\frac{\text{hypotenuse}}{\text{opposite}} = \frac{\sqrt{74}}{7}\)
5. **\(\sec \theta\)**: \(\frac{\text{hypotenuse}}{\text{adjacent}} = \frac{\sqrt{74}}{5}\)
6. **\(\cot \theta\)**: \(\frac{\text{adjacent}}{\text{opposite}} = \frac{5}{7}\)

### Part B: Solve the Indicated Sides and Angles

The triangle \( \triangle ABC \) has:
- Angle \( C = 30^\circ \)
- Side \( AB = 100 \)

We need to find:
1. **\(\angle CAB\)** (let's call it \( \angle A \))
2. **Side \( AB \) (already given as 100)**
3. **Side \( AC \)**
4. **Side \( BC \)**

Since we know angle \( C \) and side \( AB \) (the hypotenuse), we can use trigonometric ratios to find \( AC \) (adjacent to \( \angle C \)) and \( BC \) (opposite to \( \angle C \)):

1. **\(\angle CAB\)**: This is the remaining angle, so \( \angle A = 90^\circ - 30^\circ = 60^\circ \).
   
2. **\( AC \)**: \( AC = AB \cdot \cos 30^\circ = 100 \cdot \frac{\sqrt{3}}{2} = 50\sqrt{3} \approx 86.6 \)

3. **\( BC \)**: \( BC = AB \cdot \sin 30^\circ = 100 \cdot \frac{1}{2} = 50 \)

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Would you like further details or have any questions?

### Additional Related Questions:
1. How would the trigonometric ratios change if we altered the lengths of the sides in Part A?
2. What would happen to the angles in Part B if the hypotenuse \( AB \) was reduced?
3. Can you solve for the six trigonometric ratios if only the hypotenuse and one side are known?
4. How would the calculations change if angle \( C \) was \(45^\circ\) in Part B?
5. What if the triangle in Part A was not a right triangle? How would we find the ratios then?

### Tip:
Always double-check your trigonometric values (like \( \sin 30^\circ \) or \( \cos 30^\circ \)) as they are common sources of error in calculations.

Find the six trigonometric ratios with respect to angle θ in a right triangle with sides 5 and 7, and solve the indicated sides and angles in a triangle with AB = 100 and ∠C = 30°.

Learn how to find the six trigonometric ratios for a right triangle with sides 5 and 7, and solve a triangle with a 100-unit hypotenuse and a 30° angle. This detailed solution covers trigonometric ratios and the Pythagorean Theorem, suitable for Grades 9-10.

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