Here are the steps to solve for $\cos \alpha$ and $\sin \beta$ based on the given triangle:

Step 1: Label the sides

• In $\triangle PQR$:
  • $PQ = 16$ (opposite side of $\alpha$)
  • $PR = 20$ (hypotenuse of $\triangle PQR$)
  • $QR$ is unknown, but you can use the Pythagorean theorem to find it.

Step 2: Use the Pythagorean theorem for $\triangle PQR$

The Pythagorean theorem is: $PR^2 = PQ^2 + QR^2$ Substitute the known values: $20^2 = 16^2 + QR^2$ $400 = 256 + QR^2$ $QR^2 = 400 - 256 = 144$ $QR = \sqrt{144} = 12$ Now, we know $QR = 12$.

Step 3: Find $\cos \alpha$

In $\triangle PQR$, $\cos \alpha$ is the ratio of the adjacent side ($QR$) to the hypotenuse ($PR$): $\cos \alpha = \frac{QR}{PR} = \frac{12}{20} = 0.6$

Step 4: Use the Pythagorean theorem for $\triangle QRS$

In $\triangle QRS$, use the Pythagorean theorem to find $RS$: $RS^2 = QR^2 + QS^2$ Substitute the known values: $RS^2 = 12^2 + 21^2$ $RS^2 = 144 + 441 = 585$ $RS = \sqrt{585} \approx 24.19$

Step 5: Find $\sin \beta$

In $\triangle QRS$, $\sin \beta$ is the ratio of the opposite side ($QR$) to the hypotenuse ($RS$): $\sin \beta = \frac{QR}{RS} = \frac{12}{24.19} \approx 0.496$

Final Answers:

• $\cos \alpha = 0.6$
• $\sin \beta \approx 0.496$

Let me know if you want more details or have questions!

5 Related Questions:

1. How do you find the sine of an angle in a right triangle?
2. What is the difference between sine and cosine?
3. How do you apply the Pythagorean theorem to right triangles?
4. Why is the hypotenuse always the longest side in a right triangle?
5. How can you find angles using sine and cosine values?

Tip:

Always make sure to use the correct ratio of sides when solving for trigonometric functions (opposite, adjacent, hypotenuse).