Trigonometry - Primary Trigonometric Ratios - Segments All PDF ... - Free Printable

Let’s solve each problem one by one. We’ll use the basic trigonometric ratios:

- sin θ = opposite / hypotenuse
- cos θ = adjacent / hypotenuse
- tan θ = opposite / adjacent

We need to identify which side is which for angle θ in each triangle.

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Problem 1: sin θ

Triangle PQR, right-angled at Q. Angle θ is at P.

- Opposite to θ (at P) → QR
- Hypotenuse → PR (always across from right angle)

So,
sin θ = QR / PR

But we don’t have numbers — so we leave it as a ratio of sides.

Wait — looking again: The triangle has points P, Q, R. Right angle at Q. So:

- Side opposite θ (angle at P) is QR.
- Hypotenuse is PR.

✔ Answer: QR / PR

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Problem 2: cos θ

Triangle EFG, right-angled at G. Angle θ is at F.

- Adjacent to θ (at F) → FG (side next to θ that’s not hypotenuse)
- Hypotenuse → EF

So,
cos θ = FG / EF

✔ Answer: FG / EF

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Problem 3: tan θ

Triangle XYZ, right-angled at Z. Angle θ is at Y.

- Opposite to θ (at Y) → XZ
- Adjacent to θ → YZ

So,
tan θ = XZ / YZ

✔ Answer: XZ / YZ

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Problem 4: sin θ

Triangle ABC, right-angled at B. Angle θ is at C.

- Opposite to θ (at C) → AB
- Hypotenuse → AC

So,
sin θ = AB / AC

✔ Answer: AB / AC

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Problem 5: cos θ

Triangle STU, right-angled at U. Angle θ is at S.

- Adjacent to θ (at S) → SU
- Hypotenuse → ST

So,
cos θ = SU / ST

✔ Answer: SU / ST

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Problem 6: tan θ

Triangle LMN, right-angled at M. Angle θ is at N.

- Opposite to θ (at N) → LM
- Adjacent to θ → MN

So,
tan θ = LM / MN

✔ Answer: LM / MN

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Final Answer:
1) QR / PR
2) FG / EF
3) XZ / YZ
4) AB / AC
5) SU / ST
6) LM / MN