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### Problem 2: Let \( \triangle ABC \) be a right triangle, where \( \angle C = 90^\circ \).

**(a) Given that \( a = 7 \) and \( b = 4 \), find \( \sin \alpha \), \( \cos \alpha \), \( \sin \beta \), and \( \cos \beta \).**

- Here, \( a \) is the adjacent side to angle \( \beta \), and \( b \) is the opposite side to angle \( \beta \). To solve this, we also need the hypotenuse \( c \), which can be found using the Pythagorean theorem:
  \[
  c = \sqrt{a^2 + b^2} = \sqrt{7^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65}
  \]
  So, \( c = \sqrt{65} \).

Now, let's compute the trigonometric ratios:

- \( \sin \alpha = \frac{\text{opposite to } \alpha}{\text{hypotenuse}} = \frac{4}{\sqrt{65}} \)
- \( \cos \alpha = \frac{\text{adjacent to } \alpha}{\text{hypotenuse}} = \frac{7}{\sqrt{65}} \)
- \( \sin \beta = \frac{\text{opposite to } \beta}{\text{hypotenuse}} = \frac{7}{\sqrt{65}} \)
- \( \cos \beta = \frac{\text{adjacent to } \beta}{\text{hypotenuse}} = \frac{4}{\sqrt{65}} \)

**(b) Given that \( \cos \beta = \frac{12}{13} \) and \( c = 13 \), find \( a \), \( b \), \( \sin \beta \), and \( \tan \beta \).**

We know \( c = 13 \), and \( \cos \beta = \frac{a}{c} \), so:
\[
\frac{a}{13} = \frac{12}{13} \implies a = 12
\]
Using the Pythagorean theorem to find \( b \):
\[
b = \sqrt{c^2 - a^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5
\]

Now we can find the trigonometric ratios:
- \( \sin \beta = \frac{b}{c} = \frac{5}{13} \)
- \( \tan \beta = \frac{b}{a} = \frac{5}{12} \)

**(c) Given that \( c = 1 \) and \( a = 0.6 \), find all six trigonometric ratios for angle \( \beta \).**

We already know \( a = 0.6 \) and \( c = 1 \). Using the Pythagorean theorem to find \( b \):
\[
b = \sqrt{c^2 - a^2} = \sqrt{1^2 - 0.6^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8
\]

Now, we can compute all six trigonometric ratios:
- \( \sin \beta = \frac{b}{c} = \frac{0.8}{1} = 0.8 \)
- \( \cos \beta = \frac{a}{c} = \frac{0.6}{1} = 0.6 \)
- \( \tan \beta = \frac{b}{a} = \frac{0.8}{0.6} = \frac{4}{3} \)
- \( \csc \beta = \frac{1}{\sin \beta} = \frac{1}{0.8} = 1.25 \)
- \( \sec \beta = \frac{1}{\cos \beta} = \frac{1}{0.6} \approx 1.6667 \)
- \( \cot \beta = \frac{1}{\tan \beta} = \frac{1}{\frac{4}{3}} = \frac{3}{4} = 0.75 \)

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Here are 5 related questions you can explore:
1. How do you derive the Pythagorean theorem for any right triangle?
2. What is the significance of the six trigonometric ratios, and how are they interrelated?
3. How can you use trigonometry to solve real-world problems involving right triangles?
4. Why are the sine and cosine functions bounded between -1 and 1?
5. What is the geometric meaning of tangent, secant, cosecant, and cotangent?

**Tip:** Always remember that the hypotenuse is the longest side in any right triangle!

Let ABC be a right triangle where angle C = 90 degrees. (a) Given that a = 7 and b = 4, find sin(α), cos(α), sin(β), and cos(β). (b) Given that cos(β) = 12/13 and c = 13, find a, b, sin(β), and tan(β). (c) Given that c = 1 and a = 0.6, find all six trigonometric ratios for angle β.

Solve right triangle trigonometric ratio problems, including finding sine, cosine, and tangent using the Pythagorean Theorem. This problem involves calculating values like sin(α), cos(β), and more, based on different triangle configurations. Ideal for Grades 9-11.

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