https://file.aiquickdraw.com/mathbot/file/1727282494728z72hy3ug.jpg

The image shows several right-angled triangles with missing trigonometric values. Let's solve each triangle step by step using trigonometric relationships.

### 1. Triangle ABC: 
- Given: \( AB = 8 \), \( BC = 10 \), \( \angle B = 90^\circ \)
- To find: \( sinA, cosA, tanA, cscA, secA, cotA \)

Using the Pythagorean theorem:
\[
AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 10^2} = \sqrt{64 + 100} = \sqrt{164} \approx 12.81
\]

Now, we can find the trigonometric values:
- \( sinA = \frac{BC}{AC} = \frac{10}{12.81} \approx 0.780 \)
- \( cosA = \frac{AB}{AC} = \frac{8}{12.81} \approx 0.625 \)
- \( tanA = \frac{BC}{AB} = \frac{10}{8} = 1.25 \)
- \( cscA = \frac{1}{sinA} = \frac{1}{0.780} \approx 1.282 \)
- \( secA = \frac{1}{cosA} = \frac{1}{0.625} = 1.6 \)
- \( cotA = \frac{1}{tanA} = \frac{1}{1.25} = 0.8 \)

### 2. Triangle PQR: 
- Given: \( PQ = 15 \), \( PR = 8 \), \( \angle R = 90^\circ \)
- To find: \( sinQ, cosQ, tanQ, cscQ, secQ, cotQ \)

Using the Pythagorean theorem:
\[
QR = \sqrt{PQ^2 - PR^2} = \sqrt{15^2 - 8^2} = \sqrt{225 - 64} = \sqrt{161} \approx 12.69
\]

Now, we can find the trigonometric values:
- \( sinQ = \frac{PR}{PQ} = \frac{8}{15} = 0.533 \)
- \( cosQ = \frac{QR}{PQ} = \frac{12.69}{15} \approx 0.846 \)
- \( tanQ = \frac{PR}{QR} = \frac{8}{12.69} \approx 0.631 \)
- \( cscQ = \frac{1}{sinQ} = \frac{1}{0.533} \approx 1.876 \)
- \( secQ = \frac{1}{cosQ} = \frac{1}{0.846} \approx 1.182 \)
- \( cotQ = \frac{1}{tanQ} = \frac{1}{0.631} \approx 1.585 \)

### 3. Triangle XYZ: 
- Given: \( XY = 4 \), \( YZ = 1 \), \( \angle Y = 90^\circ \)
- To find: \( sinX, cosX, tanX, cscX, secX, cotX \)

Using the Pythagorean theorem:
\[
XZ = \sqrt{XY^2 + YZ^2} = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17} \approx 4.123
\]

Now, we can find the trigonometric values:
- \( sinX = \frac{YZ}{XZ} = \frac{1}{4.123} \approx 0.242 \)
- \( cosX = \frac{XY}{XZ} = \frac{4}{4.123} \approx 0.970 \)
- \( tanX = \frac{YZ}{XY} = \frac{1}{4} = 0.25 \)
- \( cscX = \frac{1}{sinX} = \frac{1}{0.242} \approx 4.132 \)
- \( secX = \frac{1}{cosX} = \frac{1}{0.970} \approx 1.031 \)
- \( cotX = \frac{1}{tanX} = \frac{1}{0.25} = 4 \)

### 4. Triangle RST: 
- Given: \( RS = \frac{\sqrt{2}}{3} \), \( \angle S = 90^\circ \)
- To find: \( sinR, cosR, tanR, cscR, secR, cotR \)

Without the second side, it is hard to directly calculate trigonometric values for this triangle. If additional information is provided, I can proceed.

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### Related Questions:
1. How is the Pythagorean theorem applied in right triangles?
2. What are the reciprocal trigonometric functions, and how are they calculated?
3. How do we determine angles using trigonometric functions if the sides are known?
4. Why is the tangent of an angle the ratio of the opposite and adjacent sides?
5. How do we approximate square roots in trigonometric problems?

### Tip:
For quick calculations of trigonometric functions, remember that calculators typically provide sine, cosine, and tangent, while their reciprocals need manual computation.

Right-angled triangles with missing trigonometric values. Solve for sin, cos, tan, and their reciprocal functions for each triangle.

Learn how to find the sine, cosine, and tangent values along with their reciprocals for different right-angled triangles. We use the Pythagorean theorem and basic trigonometric formulas to solve each triangle step-by-step. Suitable for Grades 9-12.

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