Right Triangle Trigonometry Notes and Worksheets - Lindsay Bowden - Free Printable

Let's solve each problem step by step using trigonometric ratios for right triangles. The three basic trigonometric ratios are:

- Sine (sin) = Opposite / Hypotenuse
- Cosine (cos) = Adjacent / Hypotenuse
- Tangent (tan) = Opposite / Adjacent

We'll use these definitions for each triangle, identifying the correct sides relative to the given angle.

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1. Triangle ABC

Right angle at B.
AB = 9 (vertical leg), BC = 12 (horizontal leg), AC = 15 (hypotenuse)

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

$$
\sin(A) = \frac{12}{15} = \frac{4}{5}
$$
$$
\cos(A) = \frac{9}{15} = \frac{3}{5}
$$
$$
\tan(A) = \frac{12}{9} = \frac{4}{3}
$$

✔ Answer:
- sin(A) = $ \frac{4}{5} $
- cos(A) = $ \frac{3}{5} $
- tan(A) = $ \frac{4}{3} $

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2. Triangle DEF

Right angle at D.
DE = 16, DF = 12, EF = 20 (hypotenuse)

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

$$
\sin(F) = \frac{16}{20} = \frac{4}{5}
$$
$$
\cos(F) = \frac{12}{20} = \frac{3}{5}
$$
$$
\tan(F) = \frac{16}{12} = \frac{4}{3}
$$

✔ Answer:
- sin(F) = $ \frac{4}{5} $
- cos(F) = $ \frac{3}{5} $
- tan(F) = $ \frac{4}{3} $

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3. Triangle QRS

Right angle at R.
QR = 5, RS = 12, QS = 13 (hypotenuse)

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

$$
\sin(Q) = \frac{12}{13}
$$
$$
\cos(Q) = \frac{5}{13}
$$
$$
\tan(Q) = \frac{12}{5}
$$

✔ Answer:
- sin(Q) = $ \frac{12}{13} $
- cos(Q) = $ \frac{5}{13} $
- tan(Q) = $ \frac{12}{5} $

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4. Triangle MNO

Right angle at N.
MN = 24, NO = 10, MO = 26 (hypotenuse)

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

$$
\sin(M) = \frac{10}{26} = \frac{5}{13}
$$
$$
\cos(M) = \frac{24}{26} = \frac{12}{13}
$$
$$
\tan(M) = \frac{10}{24} = \frac{5}{12}
$$

✔ Answer:
- sin(M) = $ \frac{5}{13} $
- cos(M) = $ \frac{12}{13} $
- tan(M) = $ \frac{5}{12} $

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5. Triangle TUV

Right angle at U.
TU = 3, UV = 4, TV = 5 (hypotenuse)

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

$$
\sin(V) = \frac{3}{5}, \quad \cos(V) = \frac{4}{5}, \quad \tan(V) = \frac{3}{4}
$$

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

$$
\sin(T) = \frac{4}{5}, \quad \cos(T) = \frac{3}{5}, \quad \tan(T) = \frac{4}{3}
$$

✔ Answer:
- sin(V) = $ \frac{3}{5} $, cos(V) = $ \frac{4}{5} $, tan(V) = $ \frac{3}{4} $
- sin(T) = $ \frac{4}{5} $, cos(T) = $ \frac{3}{5} $, tan(T) = $ \frac{4}{3} $

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6. Triangle GHI

Right angle at H.
GH = 9, HI = 40, GI = 41 (hypotenuse)

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

$$
\sin(G) = \frac{40}{41}, \quad \cos(G) = \frac{9}{41}, \quad \tan(G) = \frac{40}{9}
$$

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

$$
\sin(I) = \frac{9}{41}, \quad \cos(I) = \frac{40}{41}, \quad \tan(I) = \frac{9}{40}
$$

✔ Answer:
- sin(G) = $ \frac{40}{41} $, cos(G) = $ \frac{9}{41} $, tan(G) = $ \frac{40}{9} $
- sin(I) = $ \frac{9}{41} $, cos(I) = $ \frac{40}{41} $, tan(I) = $ \frac{9}{40} $

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7. Triangle FGH

Right angle at G.
FG = 8, GH = 15, FH = 17 (hypotenuse)

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

$$
\sin(F) = \frac{15}{17}, \quad \cos(F) = \frac{8}{17}, \quad \tan(F) = \frac{15}{8}
$$

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

$$
\sin(H) = \frac{8}{17}, \quad \cos(H) = \frac{15}{17}, \quad \tan(H) = \frac{8}{15}
$$

✔ Answer:
- sin(F) = $ \frac{15}{17} $, cos(F) = $ \frac{8}{17} $, tan(F) = $ \frac{15}{8} $
- sin(H) = $ \frac{8}{17} $, cos(H) = $ \frac{15}{17} $, tan(H) = $ \frac{8}{15} $

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8. Triangle RST

Right angle at S.
RS = 12, ST = 35, RT = 37 (hypotenuse)

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

$$
\sin(R) = \frac{35}{37}, \quad \cos(R) = \frac{12}{37}, \quad \tan(R) = \frac{35}{12}
$$

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

$$
\sin(T) = \frac{12}{37}, \quad \cos(T) = \frac{35}{37}, \quad \tan(T) = \frac{12}{35}
$$

✔ Answer:
- sin(R) = $ \frac{35}{37} $, cos(R) = $ \frac{12}{37} $, tan(R) = $ \frac{35}{12} $
- sin(T) = $ \frac{12}{37} $, cos(T) = $ \frac{35}{37} $, tan(T) = $ \frac{12}{35} $

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✔ Final Answers Summary:

| Problem | Angle | sin | cos | tan |
|--------|-------|-----|-----|-----|
| 1 | A | $ \frac{4}{5} $ | $ \frac{3}{5} $ | $ \frac{4}{3} $ |
| 2 | F | $ \frac{4}{5} $ | $ \frac{3}{5} $ | $ \frac{4}{3} $ |
| 3 | Q | $ \frac{12}{13} $ | $ \frac{5}{13} $ | $ \frac{12}{5} $ |
| 4 | M | $ \frac{5}{13} $ | $ \frac{12}{13} $ | $ \frac{5}{12} $ |
| 5 | V | $ \frac{3}{5} $ | $ \frac{4}{5} $ | $ \frac{3}{4} $<br>T: $ \frac{4}{5}, \frac{3}{5}, \frac{4}{3} $ |
| 6 | G | $ \frac{40}{41} $ | $ \frac{9}{41} $ | $ \frac{40}{9} $<br>I: $ \frac{9}{41}, \frac{40}{41}, \frac{9}{40} $ |
| 7 | F | $ \frac{15}{17} $ | $ \frac{8}{17} $ | $ \frac{15}{8} $<br>H: $ \frac{8}{17}, \frac{15}{17}, \frac{8}{15} $ |
| 8 | R | $ \frac{35}{37} $ | $ \frac{12}{37} $ | $ \frac{35}{12} $<br>T: $ \frac{12}{37}, \frac{35}{37}, \frac{12}{35} $ |

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📌 Key Tips:

- Always identify the right angle and the angle of interest.
- Use SOHCAHTOA:
- Sin = Opposite / Hypotenuse
- Cos = Adjacent / Hypotenuse
- Tan = Opposite / Adjacent
- Reduce fractions whenever possible.
- Check that your answers make sense: sine and cosine should be between 0 and 1 (for acute angles), tangent can be any positive number.

Let me know if you'd like this formatted as a printable answer sheet!