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

To solve the trigonometric ratios for each right triangle, we need to identify the sides opposite and adjacent to the given angle, as well as the hypotenuse. The trigonometric ratios are defined as follows:

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

Let's solve each problem step by step.

---

Problem 1:

Triangle \( \triangle ABC \)
- \( AB = 9 \) (opposite side to \( \angle A \))
- \( BC = 12 \) (adjacent side to \( \angle A \))
- \( AC = 15 \) (hypotenuse)

#### Trigonometric Ratios for \( \angle A \):
1. \(\sin(A)\):
\[
\sin(A) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{9}{15} = \frac{3}{5}
\]

2. \(\cos(A)\):
\[
\cos(A) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{12}{15} = \frac{4}{5}
\]

3. \(\tan(A)\):
\[
\tan(A) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \frac{9}{12} = \frac{3}{4}
\]

Answer for Problem 1:
\[
\sin(A) = \frac{3}{5}, \quad \cos(A) = \frac{4}{5}, \quad \tan(A) = \frac{3}{4}
\]

---

Problem 2:

Triangle \( \triangle DEF \)
- \( DE = 16 \) (adjacent side to \( \angle F \))
- \( DF = 12 \) (opposite side to \( \angle F \))
- \( EF = 20 \) (hypotenuse)

#### Trigonometric Ratios for \( \angle F \):
1. \(\sin(F)\):
\[
\sin(F) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{DF}{EF} = \frac{12}{20} = \frac{3}{5}
\]

2. \(\cos(F)\):
\[
\cos(F) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{DE}{EF} = \frac{16}{20} = \frac{4}{5}
\]

3. \(\tan(F)\):
\[
\tan(F) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{DF}{DE} = \frac{12}{16} = \frac{3}{4}
\]

Answer for Problem 2:
\[
\sin(F) = \frac{3}{5}, \quad \cos(F) = \frac{4}{5}, \quad \tan(F) = \frac{3}{4}
\]

---

Problem 3:

Triangle \( \triangleQRS \)
- \( QR = 5 \) (adjacent side to \( \angle Q \))
- \( RS = 12 \) (opposite side to \( \angle Q \))
- \( QS = 13 \) (hypotenuse)

#### Trigonometric Ratios for \( \angle Q \):
1. \(\sin(Q)\):
\[
\sin(Q) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{RS}{QS} = \frac{12}{13}
\]

2. \(\cos(Q)\):
\[
\cos(Q) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{QR}{QS} = \frac{5}{13}
\]

3. \(\tan(Q)\):
\[
\tan(Q) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{RS}{QR} = \frac{12}{5}
\]

Answer for Problem 3:
\[
\sin(Q) = \frac{12}{13}, \quad \cos(Q) = \frac{5}{13}, \quad \tan(Q) = \frac{12}{5}
\]

---

Problem 4:

Triangle \( \triangle MNO \)
- \( MN = 24 \) (opposite side to \( \angle M \))
- \( MO = 10 \) (adjacent side to \( \angle M \))
- \( NO = 26 \) (hypotenuse)

#### Trigonometric Ratios for \( \angle M \):
1. \(\sin(M)\):
\[
\sin(M) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{MN}{NO} = \frac{24}{26} = \frac{12}{13}
\]

2. \(\cos(M)\):
\[
\cos(M) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{MO}{NO} = \frac{10}{26} = \frac{5}{13}
\]

3. \(\tan(M)\):
\[
\tan(M) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{MN}{MO} = \frac{24}{10} = \frac{12}{5}
\]

Answer for Problem 4:
\[
\sin(M) = \frac{12}{13}, \quad \cos(M) = \frac{5}{13}, \quad \tan(M) = \frac{12}{5}
\]

---

Problem 5:

Triangle \( \triangle TUV \)
- \( TU = 3 \) (opposite side to \( \angle V \))
- \( UV = 4 \) (adjacent side to \( \angle V \))
- \( TV = 5 \) (hypotenuse)

#### Trigonometric Ratios for \( \angle V \):
1. \(\sin(V)\):
\[
\sin(V) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{TU}{TV} = \frac{3}{5}
\]

2. \(\cos(V)\):
\[
\cos(V) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{UV}{TV} = \frac{4}{5}
\]

3. \(\tan(V)\):
\[
\tan(V) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{TU}{UV} = \frac{3}{4}
\]

#### Trigonometric Ratios for \( \angle T \):
1. \(\sin(T)\):
\[
\sin(T) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{UV}{TV} = \frac{4}{5}
\]

2. \(\cos(T)\):
\[
\cos(T) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{TU}{TV} = \frac{3}{5}
\]

3. \(\tan(T)\):
\[
\tan(T) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{UV}{TU} = \frac{4}{3}
\]

Answer for Problem 5:
\[
\sin(V) = \frac{3}{5}, \quad \cos(V) = \frac{4}{5}, \quad \tan(V) = \frac{3}{4}
\]
\[
\sin(T) = \frac{4}{5}, \quad \cos(T) = \frac{3}{5}, \quad \tan(T) = \frac{4}{3}
\]

---

Problem 6:

Triangle \( \triangle GHI \)
- \( GH = 6 \) (adjacent side to \( \angle G \))
- \( HI = 41 \) (opposite side to \( \angle G \))
- \( GI = 40 \) (hypotenuse)

#### Trigonometric Ratios for \( \angle G \):
1. \(\sin(G)\):
\[
\sin(G) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{HI}{GI} = \frac{41}{40}
\]

2. \(\cos(G)\):
\[
\cos(G) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{GH}{GI} = \frac{6}{40} = \frac{3}{20}
\]

3. \(\tan(G)\):
\[
\tan(G) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{HI}{GH} = \frac{41}{6}
\]

#### Trigonometric Ratios for \( \angle I \):
1. \(\sin(I)\):
\[
\sin(I) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{GH}{GI} = \frac{6}{40} = \frac{3}{20}
\]

2. \(\cos(I)\):
\[
\cos(I) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{HI}{GI} = \frac{41}{40}
\]

3. \(\tan(I)\):
\[
\tan(I) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{GH}{HI} = \frac{6}{41}
\]

Answer for Problem 6:
\[
\sin(G) = \frac{41}{40}, \quad \cos(G) = \frac{3}{20}, \quad \tan(G) = \frac{41}{6}
\]
\[
\sin(I) = \frac{3}{20}, \quad \cos(I) = \frac{41}{40}, \quad \tan(I) = \frac{6}{41}
\]

---

Problem 7:

Triangle \( \triangle FGH \)
- \( FG = 8 \) (adjacent side to \( \angle F \))
- \( GH = 15 \) (opposite side to \( \angle F \))
- \( FH = 17 \) (hypotenuse)

#### Trigonometric Ratios for \( \angle F \):
1. \(\sin(F)\):
\[
\sin(F) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{GH}{FH} = \frac{15}{17}
\]

2. \(\cos(F)\):
\[
\cos(F) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{FG}{FH} = \frac{8}{17}
\]

3. \(\tan(F)\):
\[
\tan(F) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{GH}{FG} = \frac{15}{8}
\]

#### Trigonometric Ratios for \( \angle H \):
1. \(\sin(H)\):
\[
\sin(H) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{FG}{FH} = \frac{8}{17}
\]

2. \(\cos(H)\):
\[
\cos(H) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{GH}{FH} = \frac{15}{17}
\]

3. \(\tan(H)\):
\[
\tan(H) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{FG}{GH} = \frac{8}{15}
\]

Answer for Problem 7:
\[
\sin(F) = \frac{15}{17}, \quad \cos(F) = \frac{8}{17}, \quad \tan(F) = \frac{15}{8}
\]
\[
\sin(H) = \frac{8}{17}, \quad \cos(H) = \frac{15}{17}, \quad \tan(H) = \frac{8}{15}
\]

---

Problem 8:

Triangle \( \triangle STR \)
- \( ST = 12 \) (opposite side to \( \angle R \))
- \( TR = 35 \) (adjacent side to \( \angle R \))
- \( SR = 37 \) (hypotenuse)

#### Trigonometric Ratios for \( \angle R \):
1. \(\sin(R)\):
\[
\sin(R) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{ST}{SR} = \frac{12}{37}
\]

2. \(\cos(R)\):
\[
\cos(R) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{TR}{SR} = \frac{35}{37}
\]

3. \(\tan(R)\):
\[
\tan(R) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{ST}{TR} = \frac{12}{35}
\]

#### Trigonometric Ratios for \( \angle T \):
1. \(\sin(T)\):
\[
\sin(T) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{TR}{SR} = \frac{35}{37}
\]

2. \(\cos(T)\):
\[
\cos(T) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{ST}{SR} = \frac{12}{37}
\]

3. \(\tan(T)\):
\[
\tan(T) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{TR}{ST} = \frac{35}{12}
\]

Answer for Problem 8:
\[
\sin(R) = \frac{12}{37}, \quad \cos(R) = \frac{35}{37}, \quad \tan(R) = \frac{12}{35}
\]
\[
\sin(T) = \frac{35}{37}, \quad \cos(T) = \frac{12}{37}, \quad \tan(T) = \frac{35}{12}
\]

---

Final Answers:

\[
\boxed{
\begin{array}{l}
\text{Problem 1: } \sin(A) = \frac{3}{5}, \cos(A) = \frac{4}{5}, \tan(A) = \frac{3}{4} \\
\text{Problem 2: } \sin(F) = \frac{3}{5}, \cos(F) = \frac{4}{5}, \tan(F) = \frac{3}{4} \\
\text{Problem 3: } \sin(Q) = \frac{12}{13}, \cos(Q) = \frac{5}{13}, \tan(Q) = \frac{12}{5} \\
\text{Problem 4: } \sin(M) = \frac{12}{13}, \cos(M) = \frac{5}{13}, \tan(M) = \frac{12}{5} \\
\text{Problem 5: } \sin(V) = \frac{3}{5}, \cos(V) = \frac{4}{5}, \tan(V) = \frac{3}{4}; \sin(T) = \frac{4}{5}, \cos(T) = \frac{3}{5}, \tan(T) = \frac{4}{3} \\
\text{Problem 6: } \sin(G) = \frac{41}{40}, \cos(G) = \frac{3}{20}, \tan(G) = \frac{41}{6}; \sin(I) = \frac{3}{20}, \cos(I) = \frac{41}{40}, \tan(I) = \frac{6}{41} \\
\text{Problem 7: } \sin(F) = \frac{15}{17}, \cos(F) = \frac{8}{17}, \tan(F) = \frac{15}{8}; \sin(H) = \frac{8}{17}, \cos(H) = \frac{15}{17}, \tan(H) = \frac{8}{15} \\
\text{Problem 8: } \sin(R) = \frac{12}{37}, \cos(R) = \frac{35}{37}, \tan(R) = \frac{12}{35}; \sin(T) = \frac{35}{37}, \cos(T) = \frac{12}{37}, \tan(T) = \frac{35}{12}
\end{array}
}
\]