Let’s solve each problem step by step. We’re given right triangles and asked to find trigonometric ratios using the side lengths shown.
Remember: In a right triangle, for any acute angle θ:
- sin θ = opposite / hypotenuse
- cos θ = adjacent / hypotenuse
- tan θ = opposite / adjacent
- sec θ = 1 / cos θ = hypotenuse / adjacent
- csc θ = 1 / sin θ = hypotenuse / opposite
- cot θ = 1 / tan θ = adjacent / opposite
We’ll go one by one.
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Problem 1: cos θ
Triangle has sides:
- Adjacent to θ: 5 (horizontal leg)
- Opposite to θ: 3 (vertical leg)
- Hypotenuse: ? → Use Pythagoras: √(3² + 5²) = √(9 + 25) = √34
So, cos θ = adjacent / hypotenuse =
5 / √34
But we usually rationalize the denominator:
Multiply numerator and denominator by √34:
→ (5√34) / 34
✔ Final Answer for #1:
5√34 / 34
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Problem 2: cot θ
Triangle has sides:
- Adjacent to θ: 8 (horizontal leg)
- Opposite to θ: 6 (vertical leg)
cot θ = adjacent / opposite =
8 / 6 = simplify →
4 / 3
✔ Final Answer for #2:
4/3
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Problem 3: sec θ
Triangle has sides:
- Adjacent to θ: 12 (side next to θ, not hypotenuse)
- Opposite to θ: 5
- Hypotenuse: ? → √(5² + 12²) = √(25 + 144) = √169 = 13
sec θ = hypotenuse / adjacent =
13 / 12
✔ Final Answer for #3:
13/12
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Problem 4: cosec θ
Triangle has sides:
- Opposite to θ: 7 (side across from θ)
- Adjacent to θ: 24
- Hypotenuse: ? → √(7² + 24²) = √(49 + 576) = √625 = 25
cosec θ = hypotenuse / opposite =
25 / 7
✔ Final Answer for #4:
25/7
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Problem 5: cot θ
Triangle has sides:
- Adjacent to θ: 8 (side next to θ)
- Opposite to θ: 15 (side across from θ)
cot θ = adjacent / opposite =
8 / 15
✔ Final Answer for #5:
8/15
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Problem 6: sec θ
Triangle has sides:
- Adjacent to θ: 7 (side next to θ)
- Opposite to θ: 24
- Hypotenuse: ? → √(7² + 24²) = √(49 + 576) = √625 = 25
sec θ = hypotenuse / adjacent =
25 / 7
✔ Final Answer for #6:
25/7
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Final Answer:
1) 5√34 / 34
2) 4/3
3) 13/12
4) 25/7
5) 8/15
6) 25/7
Parent Tip: Review the logic above to help your child master the concept of geometry trigonometric ratios worksheet.
