Trigonometry - Primary Trigonometric Ratios - All Level2 All PDF ... - Free Printable

Let’s solve each triangle one by one. We’ll use the sine rule because we’re given two sides and a non-included angle (or sometimes two angles and a side — but here it’s mostly SSA or SAS with an angle not between the sides). The sine rule says:

> In any triangle,
> $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

We can also use the fact that the sum of angles in a triangle is 180°.

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Problem 1: Triangle with sides 6 ft, 5.2 ft, and angle opposite 5.2 ft is unknown? Wait — let’s label properly.

Looking at diagram 1:
- Side opposite angle F is 5.2 ft
- Side opposite angle G is ? → actually, let’s read carefully.

Actually, from the diagram:

Triangle 1:
- Angle at F is unknown → call it ∠F
- Side opposite ∠F is 5.2 ft
- Side opposite ∠G is 6 ft? Wait — no.

Wait — better to assign labels based on standard notation.

In triangle 1:
Vertices: F, G, H
Side FH = 6 ft
Side GH = 5.2 ft
Angle at H is marked as right angle? No — wait, looking again...

Actually, in diagram 1, there’s a small square at vertex H — that means it’s a right angle! So triangle 1 is a right triangle with right angle at H.

So:
- Right angle at H → so angles at F and G are acute and add to 90°.
- Side FH = 6 ft → this is adjacent to angle F
- Side GH = 5.2 ft → this is opposite to angle F
- Hypotenuse FG = ?

Since it’s a right triangle, we can use basic trig ratios.

Let’s find angle F first.

In right triangle FGH, right-angled at H:

→ tan(F) = opposite / adjacent = GH / FH = 5.2 / 6

Calculate that:

5.2 ÷ 6 = 0.8667

Now, tan⁻¹(0.8667) ≈ ?

Use calculator: tan⁻¹(0.8667) ≈ 40.9° → round to nearest tenth: 40.9°

Then angle G = 90° - 40.9° = 49.1°

Now, hypotenuse FG = ?

Use Pythagoras: FG² = FH² + GH² = 6² + 5.2² = 36 + 27.04 = 63.04

FG = √63.04 ≈ 7.94 ft → round to nearest tenth: 7.9 ft

✔ So for triangle 1:
∠F ≈ 40.9°, ∠G ≈ 49.1°, side FG ≈ 7.9 ft

But the question asks for “primary trigonometric ratios for the mentioned angle” — which angle? Looking back at the worksheet, for each triangle, one angle is labeled with a letter and we need to find its sine, cosine, tangent? Or maybe find missing sides/angles?

Wait — re-read the instruction: “Find primary trigonometric ratios for the mentioned angle.”

And in each triangle, one angle is named: e.g., in #1, angle F is mentioned? Actually, looking at the diagrams:

In #1: angle F is shown with arc — probably the angle we need to find ratios for.

Similarly, #2: angle F is shown.

#3: angle X

#4: angle B

So for each, we need to compute sin, cos, tan of that specific angle.

But to do that, we need all three sides relative to that angle.

Let’s go one by one.

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## 🔢 Problem 1: Triangle FGH, right-angled at H

Given:
- FH = 6 ft (adjacent to angle F)
- GH = 5.2 ft (opposite to angle F)
- Right angle at H → so FG is hypotenuse

We already calculated FG ≈ 7.94 ft → use 7.9 ft for rounding, but for accuracy, keep more digits during calculation.

Actually, let’s use exact values.

Compute FG exactly:

FG = √(6² + 5.2²) = √(36 + 27.04) = √63.04

√63.04 = ? Let's calculate:

7.94² = 63.0436 → very close → so FG ≈ 7.94 ft

Now, for angle F:

- Opposite = GH = 5.2
- Adjacent = FH = 6
- Hypotenuse = FG ≈ 7.94

So:

sin(F) = opposite/hypotenuse = 5.2 / 7.94 ≈ 0.655

cos(F) = adjacent/hypotenuse = 6 / 7.94 ≈ 0.756

tan(F) = opposite/adjacent = 5.2 / 6 ≈ 0.867

Round to 3 decimal places as typical for trig ratios.

Check:

5.2 / 7.94 = 520 / 794 = simplify? Divide numerator and denominator by 2: 260 / 397 ≈ 0.6549 → 0.655

6 / 7.94 = 600 / 794 = 300 / 397 ≈ 0.7557 → 0.756

5.2 / 6 = 52/60 = 13/15 ≈ 0.8667 → 0.867

✔ So for angle F in triangle 1:

sin F ≈ 0.655
cos F ≈ 0.756
tan F ≈ 0.867

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## 🔢 Problem 2: Triangle FGH, with angle F = 119°, side FG = 6 m, side FH = 8.9 m

Wait — diagram shows:

Vertices F, G, H

Angle at F is 119° — obtuse angle.

Side FG = 6 m → this is side from F to G

Side FH = 8.9 m → from F to H

So we have two sides and the included angle? No — angle at F is between sides FG and FH? Yes!

Actually, in triangle FGH, if angle at F is 119°, and sides FG and FH are given, then yes — it’s SAS: two sides and included angle.

But the problem says “find primary trig ratios for the mentioned angle” — which is angle F.

But angle F is given as 119° — so why find ratios? Maybe they want us to find other angles or sides?

Wait — perhaps I misread. Let me check the diagram description.

User said: “Trigonometric Ratios” and “for the mentioned angle”.

In diagram 2: angle F is labeled 119°, and sides FG=6m, FH=8.9m — so likely, we need to find the other sides or angles? But the instruction is to find trig ratios for the mentioned angle — which is already given.

That doesn’t make sense. Perhaps “mentioned angle” means the angle we are to focus on, and we need to find sin, cos, tan of that angle — but if it’s given, we can just compute them directly.

For example, for angle F = 119°, we can compute sin(119°), cos(119°), tan(119°) using calculator.

But that seems too straightforward — and the other problems involve finding missing parts.

Perhaps in some cases, the angle is not given, and we need to find it first.

Let’s look at diagram 2 again.

Actually, in many such worksheets, when an angle is given and two sides, you might need to find the third side or other angles, but the task specifically says “trigonometric ratios for the mentioned angle”.

Maybe for angle F, even though it’s given, we still report its sin, cos, tan.

But let’s see problem 3 and 4.

Problem 3: triangle XYZ, with XY = 5.6 cm, YZ = 4.03 cm, angle at Y is not given, but angle at X is mentioned? Diagram shows angle X with arc.

Sides: XY = 5.6 cm, YZ = 4.03 cm — but these are not both adjacent to angle X.

Let’s define:

In triangle XYZ:

- Vertex X, Y, Z
- Side XY = 5.6 cm
- Side YZ = 4.03 cm
- Angle at X is the one we care about.

To find trig ratios for angle X, we need the sides relative to angle X.

But we don't have enough info yet — we need either another side or another angle.

This suggests that for each triangle, we may need to first find missing sides or angles using laws of sines or cosines, then compute the trig ratios for the specified angle.

Let’s reinterpret the task.

The worksheet says: “Find primary trigonometric ratios for the mentioned angle.”

And for each triangle, one angle is highlighted or named (F, F, X, B), and we need to find sin, cos, tan of that angle.

To do that, we need the lengths of the sides forming that angle and the opposite side, or we need to know the angle measure.

In problem 1, it was a right triangle, so easy.

In problem 2, angle F is given as 119°, so we can directly compute its trig ratios.

In problem 3, angle X is not given, so we need to find it first.

In problem 4, angle B is not given, so find it first.

So let's proceed accordingly.

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## ✔ Problem 1: Right triangle, angle F

As above:

sin F = 5.2 / 7.94 ≈ 0.655
cos F = 6 / 7.94 ≈ 0.756
tan F = 5.2 / 6 ≈ 0.867

We'll box these later.

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## ✔ Problem 2: Angle F = 119°, find sin, cos, tan of angle F

Even though it's given, we compute the ratios.

Use calculator:

sin(119°) = sin(180° - 61°) = sin(61°) ≈ 0.8746 → 0.875

cos(119°) = -cos(61°) ≈ -0.4848 → -0.485

tan(119°) = -tan(61°) ≈ -1.8040 → -1.804

So:

sin F ≈ 0.875
cos F ≈ -0.485
tan F ≈ -1.804

Note: since 119° is in second quadrant, sin positive, cos negative, tan negative.

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## ✔ Problem 3: Triangle XYZ, with XY = 5.6 cm, YZ = 4.03 cm, and we need trig ratios for angle X.

First, identify the sides relative to angle X.

In triangle XYZ:

- Angle at X
- Side opposite to angle X is YZ = 4.03 cm
- Side adjacent to angle X could be XY, but we need another side.

Actually, to define trig ratios for angle X, we need the three sides of the triangle, or at least the sides forming the angle and the opposite side.

But we only have two sides: XY and YZ.

XY is from X to Y, YZ is from Y to Z.

So side XZ is missing.

Also, we don't have any angles given.

This is insufficient unless we assume something.

Perhaps the diagram shows that angle at Y is right angle? But user didn't say.

Looking back at user's description: in diagram 3, it's triangle XYZ with XY=5.6cm, YZ=4.03cm, and angle X is mentioned.

But without more info, we can't proceed.

Unless... perhaps it's a right triangle? But no indication.

Another possibility: maybe "mentioned angle" means we need to find the angle first using law of cosines or sines, but we need three pieces of information.

We have two sides, but no included angle or other side.

Perhaps I missed something.

Let's think differently.

In some diagrams, the side lengths are given, and the angle is to be found, then its trig ratios computed.

But for that, we need three elements.

For problem 3: sides XY=5.6, YZ=4.03, and perhaps XZ is given? User didn't specify.

User said: "3) ∠X" and "5.6 cm", "4.03 cm" — probably those are two sides, but which ones?

Assume that in triangle XYZ:

- Side XY = c = 5.6 cm (between X and Y)
- Side YZ = a = 4.03 cm (between Y and Z)
- Side XZ = b = ?

But we need another piece.

Perhaps the diagram shows that angle at Y is the one between the two given sides? But not stated.

Another idea: perhaps for angle X, the two sides forming it are XY and XZ, but we only have XY, not XZ.

This is ambiguous.

Let's look at problem 4 for clue.

Problem 4: triangle ABC, with AB = 64 m, BC = 72 m, angle B mentioned.

Sides AB and BC are given, and angle at B is between them? If so, then we have SAS.

In triangle ABC:

- AB = 64 m
- BC = 72 m
- Angle at B is the included angle? But it's not given; we need to find trig ratios for angle B, so probably we need to find the angle first or the sides.

If we have two sides and the included angle, we can find the third side, but here angle B is not given.

Perhaps in diagram 4, angle B is not given, but we have two sides and need to find the angle.

But with only two sides, we can't determine the triangle uniquely.

Unless it's a right triangle.

Let's check if any of the triangles are right-angled.

In problem 1, it was right-angled at H.

In problem 2, angle F is 119°, so not right.

In problem 3, no indication.

In problem 4, no indication.

Perhaps for problem 3 and 4, we are to assume that the triangle is configured such that we can use the given sides to find the angle.

Another thought: in problem 3, sides XY=5.6 cm, YZ=4.03 cm, and perhaps XZ is the base, but not given.

Perhaps the "mentioned angle" is at X, and the two sides are XY and XZ, but we only have XY.

I think there might be a mistake in my approach.

Let's search for similar problems online or recall standard format.

Perhaps "primary trigonometric ratios" means for the angle, we need to express sin, cos, tan in terms of the sides, but since sides are given, we can compute if we know which is which.

For problem 3: let's assume that in triangle XYZ, with vertices X,Y,Z, side XY = 5.6 cm, side YZ = 4.03 cm, and side XZ is unknown, but perhaps angle at Y is given or something.

User didn't provide full details.

Perhaps in the diagram, for problem 3, it's shown that angle at Y is 90 degrees or something.

To resolve this, let's assume that for each triangle, the given sides are sufficient to find the required angle's trig ratios, and for problem 3 and 4, we may need to use the law of cosines or sines after finding missing parts.

But for problem 3, with only two sides, it's impossible.

Unless the third side is implied or can be found.

Another idea: perhaps in problem 3, the side XZ is not given, but we can find angle X if we had more, but we don't.

Let's look at the numbers.

In problem 3: XY = 5.6 cm, YZ = 4.03 cm.

If we assume that angle at Y is the angle between them, then we can find side XZ using law of cosines, but we don't have angle at Y.

Perhaps the "mentioned angle" is angle X, and we need to find it, but we need three elements.

I think there might be a typo or missing information.

Perhaps for problem 3, the side XZ is given as well, but user didn't mention.

User said: "3) ∠X" and "5.6 cm", "4.03 cm" — perhaps those are the two sides enclosing angle X? But 5.6 and 4.03 are XY and YZ, which are not both incident to X.

XY is incident to X, YZ is not; YZ is between Y and Z.

So for angle at X, the two sides are XY and XZ.

So if only XY is given, we need XZ.

Perhaps in the diagram, XZ is given, but user omitted.

To make progress, let's assume that for problem 3, the triangle has sides XY = 5.6 cm, XZ = ? , YZ = 4.03 cm, and we need angle at X.

But still missing.

Another possibility: perhaps "4.03 cm" is the length of XZ, not YZ.

Let's check the user's input: "3) ∠X" and "5.6 cm", "4.03 cm" — and in the text, it's "5.6 cm" next to XY, "4.03 cm" next to YZ, but perhaps it's a different labeling.

Perhaps in triangle XYZ, side opposite X is YZ = 4.03 cm, side adjacent is XY = 5.6 cm, but for trig ratios, we need the hypotenuse or other side.

I think I need to guess that for problem 3, it's a right triangle with right angle at Y or something.

Let's calculate the ratio.

Suppose we assume that angle at Y is 90 degrees. Then in triangle XYZ, right-angled at Y, then:

- XY = 5.6 cm (one leg)
- YZ = 4.03 cm (other leg)
- Then XZ = hypotenuse = sqrt(5.6^2 + 4.03^2) = sqrt(31.36 + 16.2409) = sqrt(47.6009) ≈ 6.899 cm

Then for angle X:

- Opposite side is YZ = 4.03 cm
- Adjacent side is XY = 5.6 cm
- Hypotenuse is XZ ≈ 6.899 cm

So sin X = opposite/hyp = 4.03 / 6.899 ≈ 0.584

cos X = adjacent/hyp = 5.6 / 6.899 ≈ 0.812

tan X = opposite/adjacent = 4.03 / 5.6 ≈ 0.720

This makes sense, and the numbers work.

Similarly, for problem 4, if we assume right angle at C or something.

In problem 4: triangle ABC, AB = 64 m, BC = 72 m, angle B mentioned.

If we assume right angle at C, then:

- BC = 72 m (leg)
- AC = ?
- AB = 64 m — but AB would be hypotenuse, but 64 < 72, impossible.

If right angle at A, then AB and AC are legs, BC hypotenuse.

But we have AB = 64, BC = 72, so if right-angled at A, then BC should be hypotenuse, so BC > AB, 72 > 64, ok.

Then AC = sqrt(BC^2 - AB^2) = sqrt(72^2 - 64^2) = sqrt(5184 - 4096) = sqrt(1088) = sqrt(64*17) = 8sqrt(17) ≈ 8*4.123 = 32.984 m

Then for angle B:

- Opposite side is AC ≈ 32.984 m
- Adjacent side is AB = 64 m
- Hypotenuse is BC = 72 m

So sin B = opposite/hyp = 32.984 / 72 ≈ 0.458

cos B = adjacent/hyp = 64 / 72 ≈ 0.889

tan B = opposite/adjacent = 32.984 / 64 ≈ 0.515

This also works.

And for problem 2, we have angle given, so direct computation.

For problem 1, right triangle, done.

So probably, in the diagrams, triangles 3 and 4 are right-angled, but the right angle is not explicitly marked in the user's description, but in the actual image, it might be.

Since the user said "printable math worksheets", and often in such sheets, right angles are marked with squares, like in problem 1.

In problem 1, it was marked, in others not, but perhaps for 3 and 4, it's assumed or marked.

To confirm, in problem 3, if right-angled at Y, then angle at X is acute, and we can compute.

Similarly for 4.

So I'll proceed with that assumption.

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## ✔ Problem 3: Assume right-angled at Y

Triangle XYZ, right-angled at Y.

Given:
- XY = 5.6 cm (adjacent to angle X)
- YZ = 4.03 cm (opposite to angle X)
- Hypotenuse XZ = sqrt(5.6^2 + 4.03^2) = sqrt(31.36 + 16.2409) = sqrt(47.6009) = 6.899 cm (as before)

So for angle X:

sin X = opposite/hyp = YZ / XZ = 4.03 / 6.899 ≈ 0.5841 → 0.584

cos X = adjacent/hyp = XY / XZ = 5.6 / 6.899 ≈ 0.8117 → 0.812

tan X = opposite/adjacent = YZ / XY = 4.03 / 5.6 = 0.7196 → 0.720

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## ✔ Problem 4: Assume right-angled at A

Triangle ABC, right-angled at A.

Given:
- AB = 64 m (adjacent to angle B)
- BC = 72 m (hypotenuse)
- AC = opposite to angle B = sqrt(BC^2 - AB^2) = sqrt(72^2 - 64^2) = sqrt(5184 - 4096) = sqrt(1088)

Simplify sqrt(1088): 1088 ÷ 16 = 68, so sqrt(16*68) = 4*sqrt(68) = 4*sqrt(4*17) = 8*sqrt(17)

sqrt(17) ≈ 4.12310562562, so 8*4.12310562562 = 32.98484500496 m

So AC ≈ 32.985 m

For angle B:

- Opposite = AC ≈ 32.985 m
- Adjacent = AB = 64 m
- Hypotenuse = BC = 72 m

sin B = 32.985 / 72 ≈ 0.458125 → 0.458

cos B = 64 / 72 = 8/9 ≈ 0.888888... → 0.889

tan B = 32.985 / 64 ≈ 0.515390625 → 0.515

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Now, summarize all answers.

For each problem, the "mentioned angle" is:

1. Angle F in triangle FGH (right-angled at H)
2. Angle F in triangle FGH (given as 119°)
3. Angle X in triangle XYZ (assumed right-angled at Y)
4. Angle B in triangle ABC (assumed right-angled at A)

And we need sin, cos, tan for each.

So final answers:

Problem 1:
sin F ≈ 0.655
cos F ≈ 0.756
tan F ≈ 0.867

Problem 2:
sin F ≈ 0.875
cos F ≈ -0.485
tan F ≈ -1.804

Problem 3:
sin X ≈ 0.584
cos X ≈ 0.812
tan X ≈ 0.720

Problem 4:
sin B ≈ 0.458
cos B ≈ 0.889
tan B ≈ 0.515

All rounded to three decimal places.

Now, for the Final Answer section, since the user asked for the solution, and there are four parts, I'll list them clearly.

Final Answer:
For triangle 1 (angle F): sin F = 0.655, cos F = 0.756, tan F = 0.867
For triangle 2 (angle F): sin F = 0.875, cos F = -0.485, tan F = -1.804
For triangle 3 (angle X): sin X = 0.584, cos X = 0.812, tan X = 0.720
For triangle 4 (angle B): sin B = 0.458, cos B = 0.889, tan B = 0.515