To solve for the trigonometric ratios given in the image, we’ll use the right triangle properties. Let’s identify the lengths and calculate each trigonometric function.

Step 1: Find the Hypotenuse

Given:

• Opposite side to $\alpha$ (adjacent to $\beta$) = 8
• Adjacent side to $\alpha$ (opposite to $\beta$) = 14

Using the Pythagorean theorem to find the hypotenuse $c$: $c = \sqrt{8^2 + 14^2} = \sqrt{64 + 196} = \sqrt{260} = 2\sqrt{65} \approx 16.12$

Step 2: Calculate the Trigonometric Ratios

1. $\sin(\alpha)$ and $\cos(\beta)$:

   • $\sin(\alpha) = \frac{\text{opposite to } \alpha}{\text{hypotenuse}} = \frac{8}{16.12} \approx 0.496$
   • $\cos(\beta) = \sin(\alpha)$, so $\cos(\beta) \approx 0.496$

2. $\tan(\alpha)$ and $\cot(\beta)$:

   • $\tan(\alpha) = \frac{\text{opposite to } \alpha}{\text{adjacent to } \alpha} = \frac{8}{14} = \frac{4}{7} \approx 0.571$
   • $\cot(\beta) = \tan(\alpha)$, so $\cot(\beta) \approx 0.571$

3. $\sec(\alpha)$ and $\csc(\beta)$:

   • $\sec(\alpha) = \frac{\text{hypotenuse}}{\text{adjacent to } \alpha} = \frac{16.12}{14} \approx 1.15$
   • $\csc(\beta) = \sec(\alpha)$, so $\csc(\beta) \approx 1.15$

Final Answers

(a) $\sin(\alpha) \approx 0.496$ and $\cos(\beta) \approx 0.496$
(b) $\tan(\alpha) \approx 0.571$ and $\cot(\beta) \approx 0.571$
(c) $\sec(\alpha) \approx 1.15$ and $\csc(\beta) \approx 1.15$

Would you like a more detailed explanation or have any questions?

Here are five related questions to expand your understanding:

1. What is the Pythagorean identity, and how is it applied in trigonometry?
2. How are trigonometric functions related for complementary angles in a right triangle?
3. What are the exact values for trigonometric functions at common angles (like 30°, 45°, and 60°)?
4. How can we use the unit circle to find trigonometric values of angles beyond 90°?
5. What are the reciprocal trigonometric functions, and how are they derived?

Tip: Memorizing the basic trigonometric ratios for special triangles, like 45-45-90 and 30-60-90, can speed up your calculations.