Intro to Trigonometry practice worksheet featuring eight right triangles with labeled sides and angles for calculating trigonometric ratios.

Let’s solve each problem step by step. We’ll use the definitions:

For any right triangle and angle θ (not the right angle):

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

We must reduce all fractions to simplest form.

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Problem 1: Triangle ABC, right angle at B

Sides:
- AB = 9 (adjacent to A)
- BC = 12 (opposite to A)
- AC = 15 (hypotenuse)

Angle A:

- sin(A) = opposite/hypotenuse = BC/AC = 12/15 → reduce by 3 → 4/5
- cos(A) = adjacent/hypotenuse = AB/AC = 9/15 → reduce by 3 → 3/5
- tan(A) = opposite/adjacent = BC/AB = 12/9 → reduce by 3 → 4/3

✔ Check: 9-12-15 is a multiple of 3-4-5 triangle → correct ratios.

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Problem 2: Triangle DEF, right angle at D

Sides:
- DF = 12 (adjacent to F? Wait — let’s label carefully)

Right angle at D → so legs are DF and DE, hypotenuse is EF = 20.

Angle F:

- Opposite side to F is DE = 16
- Adjacent side to F is DF = 12
- Hypotenuse = EF = 20

So:

- sin(F) = opposite/hypotenuse = 16/20 → reduce by 4 → 4/5
- cos(F) = adjacent/hypotenuse = 12/20 → reduce by 4 → 3/5
- tan(F) = opposite/adjacent = 16/12 → reduce by 4 → 4/3

✔ Same as Problem 1 — makes sense, same ratio triangle.

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Problem 3: Triangle QRS, right angle at R

Sides:
- QR = 5
- RS = 12
- QS = 13 (hypotenuse)

Angle Q:

- Opposite to Q is RS = 12
- Adjacent to Q is QR = 5
- Hypotenuse = QS = 13

So:

- sin(Q) = 12/13 → already reduced → 12/13
- cos(Q) = 5/13 → already reduced → 5/13
- tan(Q) = 12/5 → already reduced → 12/5

✔ 5-12-13 is a Pythagorean triple → correct.

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Problem 4: Triangle MNO, right angle at N

Sides:
- MN = 24
- NO = 10
- MO = 26 (hypotenuse)

Angle M:

- Opposite to M is NO = 10
- Adjacent to M is MN = 24
- Hypotenuse = MO = 26

So:

- sin(M) = 10/26 → reduce by 2 → 5/13
- cos(M) = 24/26 → reduce by 2 → 12/13
- tan(M) = 10/24 → reduce by 2 → 5/12

✔ 10-24-26 is double 5-12-13 → correct.

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Problem 5: Triangle TUV, right angle at U

Sides:
- TU = 3
- UV = 4
- TV = 5 (hypotenuse)

First, find for angle V:

- Opposite to V is TU = 3
- Adjacent to V is UV = 4
- Hypotenuse = TV = 5

→ sin(V) = 3/5
→ cos(V) = 4/5
→ tan(V) = 3/4

Now for angle T:

- Opposite to T is UV = 4
- Adjacent to T is TU = 3
- Hypotenuse = TV = 5

→ sin(T) = 4/5
→ cos(T) = 3/5
→ tan(T) = 4/3

✔ Classic 3-4-5 triangle.

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Problem 6: Triangle GHI, right angle at H

Sides:
- GH = 9
- HI = 40
- GI = 41 (hypotenuse)

Check: 9² + 40² = 81 + 1600 = 1681 = 41² → yes.

Angle G:

- Opposite to G is HI = 40
- Adjacent to G is GH = 9
- Hypotenuse = GI = 41

→ sin(G) = 40/41
→ cos(G) = 9/41
→ tan(G) = 40/9

Angle I:

- Opposite to I is GH = 9
- Adjacent to I is HI = 40
- Hypotenuse = GI = 41

→ sin(I) = 9/41
→ cos(I) = 40/41
→ tan(I) = 9/40

✔ All fractions already reduced.

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Problem 7: Triangle FGH, right angle at G

Sides:
- FG = 17
- GH = 8
- FH = 15? Wait — check labels.

Wait — diagram shows:

Points: F, G, H. Right angle at G.

Side opposite G is FH — but labeled 17? Let’s read carefully.

Actually, from diagram:

- Side FG = 17? No — wait, looking again:

In problem 7:

Triangle with points F, G, H. Right angle at G.

Side between F and H is labeled 17 → that’s hypotenuse.

Side between F and G is labeled 8? Or 15?

Wait — user image says:

“F” at top, “G” bottom right (right angle), “H” bottom left.

Side FG = ? Not labeled directly.

Wait — actually, in the diagram:

- Side FH = 17 (hypotenuse)
- Side GH = 15
- Side FG = 8

Because it says: next to FG is 8, next to GH is 15, and FH is 17.

Check: 8² + 15² = 64 + 225 = 289 = 17² → yes.

So:

Angle F:

- Opposite to F is GH = 15
- Adjacent to F is FG = 8
- Hypotenuse = FH = 17

→ sin(F) = 15/17
→ cos(F) = 8/17
→ tan(F) = 15/8

Angle H:

- Opposite to H is FG = 8
- Adjacent to H is GH = 15
- Hypotenuse = FH = 17

→ sin(H) = 8/17
→ cos(H) = 15/17
→ tan(H) = 8/15

✔ Correct.

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Problem 8: Triangle RST, right angle at S

Sides:
- RS = 12
- ST = 35
- RT = 37 (hypotenuse)

Check: 12² + 35² = 144 + 1225 = 1369 = 37² → yes.

Angle R:

- Opposite to R is ST = 35
- Adjacent to R is RS = 12
- Hypotenuse = RT = 37

→ sin(R) = 35/37
→ cos(R) = 12/37
→ tan(R) = 35/12

Angle T:

- Opposite to T is RS = 12
- Adjacent to T is ST = 35
- Hypotenuse = RT = 37

→ sin(T) = 12/37
→ cos(T) = 35/37
→ tan(T) = 12/35

✔ All good.

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Final Answer:

1.
sin(A) = 4/5
cos(A) = 3/5
tan(A) = 4/3

2.
sin(F) = 4/5
cos(F) = 3/5
tan(F) = 4/3

3.
sin(Q) = 12/13
cos(Q) = 5/13
tan(Q) = 12/5

4.
sin(M) = 5/13
cos(M) = 12/13
tan(M) = 5/12

5.
sin(V) = 3/5, sin(T) = 4/5
cos(V) = 4/5, cos(T) = 3/5
tan(V) = 3/4, tan(T) = 4/3

6.
sin(G) = 40/41, sin(I) = 9/41
cos(G) = 9/41, cos(I) = 40/41
tan(G) = 40/9, tan(I) = 9/40

7.
sin(F) = 15/17, sin(H) = 8/17
cos(F) = 8/17, cos(H) = 15/17
tan(F) = 15/8, tan(H) = 8/15

8.
sin(R) = 35/37, sin(T) = 12/37
cos(R) = 12/37, cos(T) = 35/37
tan(R) = 35/12, tan(T) = 12/35