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

Absolutely! Let's solve each problem step by step, using the basic trigonometric ratios for right triangles:

> SOH-CAH-TOA:
> - sin(θ) = Opposite / Hypotenuse
> - cos(θ) = Adjacent / Hypotenuse
> - tan(θ) = Opposite / Adjacent

We'll go one problem at a time.

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Problem 1: Triangle ABC (∠B = 90°)

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

We need sin(A), cos(A), tan(A).

For angle A:
- Opposite side = BC = 12
- Adjacent side = AB = 9
- Hypotenuse = AC = 15

✔ sin(A) = Opposite/Hypotenuse = 12/15 = 4/5
✔ cos(A) = Adjacent/Hypotenuse = 9/15 = 3/5
✔ tan(A) = Opposite/Adjacent = 12/9 = 4/3

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Problem 2: Triangle DEF (∠D = 90°)

Sides:
- DE = 16
- DF = 12
- EF = 20 (hypotenuse)

Angle F is at vertex F.

For angle F:
- Opposite side = DE = 16
- Adjacent side = DF = 12
- Hypotenuse = EF = 20

✔ sin(F) = 16/20 = 4/5
✔ cos(F) = 12/20 = 3/5
✔ tan(F) = 16/12 = 4/3

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Problem 3: Triangle QRS (∠R = 90°)

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

Angle Q is at vertex Q.

For angle Q:
- Opposite side = RS = 12
- Adjacent side = QR = 5
- Hypotenuse = QS = 13

✔ sin(Q) = 12/13
✔ cos(Q) = 5/13
✔ tan(Q) = 12/5

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Problem 4: Triangle MNO (∠N = 90°)

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

Angle M is at vertex M.

For angle M:
- Opposite side = NO = 10
- Adjacent side = MN = 24
- Hypotenuse = MO = 26

✔ sin(M) = 10/26 = 5/13
✔ cos(M) = 24/26 = 12/13
✔ tan(M) = 10/24 = 5/12

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Problem 5: Triangle TUV (∠U = 90°)

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

We need trig ratios for both angles V and T.

#### For angle V:
- Opposite = TU = 3
- Adjacent = UV = 4
- Hypotenuse = TV = 5

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

#### For angle T:
- Opposite = UV = 4
- Adjacent = TU = 3
- Hypotenuse = TV = 5

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

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Problem 6: Triangle GHI (∠H = 90°)

Sides:
- GH = 9
- HI = 40
- GI = 41 (hypotenuse — 9-40-41 is a Pythagorean triple!)

We need trig ratios for both angles G and I.

#### For angle G:
- Opposite = HI = 40
- Adjacent = GH = 9
- Hypotenuse = GI = 41

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

#### For angle I:
- Opposite = GH = 9
- Adjacent = HI = 40
- Hypotenuse = GI = 41

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

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Problem 7: Triangle FGH (∠G = 90°)

Sides:
- FG = 8
- GH = 15
- FH = 17 (hypotenuse — 8-15-17 is a Pythagorean triple!)

We need trig ratios for both angles F and H.

#### For angle F:
- Opposite = GH = 15
- Adjacent = FG = 8
- Hypotenuse = FH = 17

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

#### For angle H:
- Opposite = FG = 8
- Adjacent = GH = 15
- Hypotenuse = FH = 17

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

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Problem 8: Triangle RST (∠S = 90°)

Sides:
- RS = 12
- ST = 35
- RT = 37 (hypotenuse — 12-35-37 is a Pythagorean triple!)

We need trig ratios for both angles R and T.

#### For angle R:
- Opposite = ST = 35
- Adjacent = RS = 12
- Hypotenuse = RT = 37

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

#### For angle T:
- Opposite = RS = 12
- Adjacent = ST = 35
- Hypotenuse = RT = 37

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

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## ✔ Final Answers Summary:

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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

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All fractions are reduced as required. Great job practicing SOH-CAH-TOA! 🎯