Free Printable Trigonometric Ratios Worksheet Answers - Free Printable

Let’s solve each problem step by step. We’re finding the six trigonometric ratios for a given angle in a right triangle: sine, cosine, tangent, cosecant, secant, and cotangent.

Remember:
- sin = opposite / hypotenuse
- cos = adjacent / hypotenuse
- tan = opposite / adjacent
- csc = 1/sin = hypotenuse / opposite
- sec = 1/cos = hypotenuse / adjacent
- cot = 1/tan = adjacent / opposite

We’ll go one problem at a time.

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Problem 1: ∠X

Triangle XYZ is right-angled at Z.

Sides:
- XZ = 8 (adjacent to ∠X)
- YZ = ? → Wait, we don’t have it yet. But we have XY = 10 (hypotenuse), and XZ = 8.

Wait — actually, looking again: The side labeled “8” is from X to Z, and “10” is from X to Y. Since it’s right-angled at Z, then:

- Side opposite ∠X is YZ
- Side adjacent to ∠X is XZ = 8
- Hypotenuse is XY = 10

But we need YZ. Use Pythagoras:

YZ² + XZ² = XY²
→ YZ² + 8² = 10²
→ YZ² + 64 = 100
→ YZ² = 36
→ YZ = 6

So now:

For ∠X:
- Opposite = YZ = 6
- Adjacent = XZ = 8
- Hypotenuse = XY = 10

Now compute:

sin X = opp/hyp = 6/10 = 3/5 ✔
cos X = adj/hyp = 8/10 = 4/5 ✔
tan X = opp/adj = 6/8 = 3/4 ✔
csc X = 1/sin = 5/3 ✔
sec X = 1/cos = 5/4 ✔
cot X = 1/tan = 4/3 ✔

All match the answer key.

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Problem 2: ∠Q

Triangle PQR, right-angled at P.

Given:
- PR = 16√2 (this is adjacent to ∠Q? Let’s see…)

Actually, let’s label carefully.

Right angle at P → so sides are:

- PQ and PR are legs
- QR is hypotenuse = 24

∠Q is at vertex Q.

So for ∠Q:
- Opposite side = PR = 16√2
- Adjacent side = PQ = ?
- Hypotenuse = QR = 24

We need PQ. Use Pythagoras:

PQ² + PR² = QR²
→ PQ² + (16√2)² = 24²
→ PQ² + 256 * 2 = 576
→ PQ² + 512 = 576
→ PQ² = 64
→ PQ = 8

So now:

For ∠Q:
- Opposite = PR = 16√2
- Adjacent = PQ = 8
- Hypotenuse = 24

Compute:

sin Q = opp/hyp = 16√2 / 24 = simplify: divide numerator and denominator by 8 → 2√2 / 3 ✔
cos Q = adj/hyp = 8 / 24 = 1/3 ✔
tan Q = opp/adj = 16√2 / 8 = 2√2 ✔
csc Q = 1/sin = 3/(2√2) → rationalize: multiply numerator and denominator by √2 → (3√2)/(2*2) = 3√2/4 ✔
sec Q = 1/cos = 3/1 = 3 ✔
cot Q = 1/tan = 1/(2√2) → rationalize: √2/(2*2) = √2/4 ✔

All match.

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Problem 3: ∠K

Triangle KLM, right-angled at L.

Given:
- KL = 20 (adjacent to ∠K)
- LM = 21 (opposite to ∠K)
- KM = ? → hypotenuse

Use Pythagoras:

KM² = KL² + LM² = 20² + 21² = 400 + 441 = 841
→ KM = √841 = 29

So for ∠K:
- Opposite = LM = 21
- Adjacent = KL = 20
- Hypotenuse = KM = 29

Compute:

sin K = 21/29 ✔
cos K = 20/29 ✔
tan K = 21/20 ✔
csc K = 29/21 ✔
sec K = 29/20 ✔
cot K = 20/21 ✔

Perfect.

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Final Answer:

1) For ∠X:
sin X = 3/5, cos X = 4/5, tan X = 3/4,
csc X = 5/3, sec X = 5/4, cot X = 4/3

2) For ∠Q:
sin Q = 2√2/3, cos Q = 1/3, tan Q = 2√2,
csc Q = 3√2/4, sec Q = 3, cot Q = √2/4

3) For ∠K:
sin K = 21/29, cos K = 20/29, tan K = 21/20,
csc K = 29/21, sec K = 29/20, cot K = 20/21