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:
1. \( \sin(A) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{9}{15} = \frac{3}{5} \)
2. \( \cos(A) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{12}{15} = \frac{4}{5} \)
3. \( \tan(A) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \frac{9}{12} = \frac{3}{4} \)

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

---

Problem 2:

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

#### Trigonometric Ratios:
1. \( \sin(F) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{DF}{EF} = \frac{12}{20} = \frac{3}{5} \)
2. \( \cos(F) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{DE}{EF} = \frac{16}{20} = \frac{4}{5} \)
3. \( \tan(F) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{DF}{DE} = \frac{12}{16} = \frac{3}{4} \)

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

---

Problem 3:

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

#### Trigonometric Ratios:
1. \( \sin(Q) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{RS}{QS} = \frac{12}{13} \)
2. \( \cos(Q) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{QR}{QS} = \frac{5}{13} \)
3. \( \tan(Q) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{RS}{QR} = \frac{12}{5} \)

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

---

Problem 4:

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

#### Trigonometric Ratios:
1. \( \sin(M) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{MN}{NO} = \frac{24}{26} = \frac{12}{13} \)
2. \( \cos(M) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{MO}{NO} = \frac{10}{26} = \frac{5}{13} \)
3. \( \tan(M) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{MN}{MO} = \frac{24}{10} = \frac{12}{5} \)

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

---

Problem 5:

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

#### Trigonometric Ratios:
1. \( \sin(V) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{TU}{TV} = \frac{3}{5} \)
2. \( \cos(V) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{UV}{TV} = \frac{4}{5} \)
3. \( \tan(V) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{TU}{UV} = \frac{3}{4} \)

For \( \angle T \):
1. \( \sin(T) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{UV}{TV} = \frac{4}{5} \)
2. \( \cos(T) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{TU}{TV} = \frac{3}{5} \)
3. \( \tan(T) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{UV}{TU} = \frac{4}{3} \)

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

---

Problem 6:

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

#### Trigonometric Ratios:
1. \( \sin(G) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{HI}{GI} = \frac{41}{40} \)
2. \( \cos(G) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{GH}{GI} = \frac{6}{40} = \frac{3}{20} \)
3. \( \tan(G) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{HI}{GH} = \frac{41}{6} \)

For \( \angle I \):
1. \( \sin(I) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{GH}{GI} = \frac{6}{40} = \frac{3}{20} \)
2. \( \cos(I) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{HI}{GI} = \frac{41}{40} \)
3. \( \tan(I) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{GH}{HI} = \frac{6}{41} \)

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

---

Problem 7:

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

#### Trigonometric Ratios:
1. \( \sin(F) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{GH}{FH} = \frac{15}{17} \)
2. \( \cos(F) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{FG}{FH} = \frac{8}{17} \)
3. \( \tan(F) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{GH}{FG} = \frac{15}{8} \)

For \( \angle H \):
1. \( \sin(H) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{FG}{FH} = \frac{8}{17} \)
2. \( \cos(H) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{GH}{FH} = \frac{15}{17} \)
3. \( \tan(H) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{FG}{GH} = \frac{8}{15} \)

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

---

Problem 8:

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

#### Trigonometric Ratios:
1. \( \sin(R) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{ST}{SR} = \frac{12}{37} \)
2. \( \cos(R) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{TR}{SR} = \frac{35}{37} \)
3. \( \tan(R) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{ST}{TR} = \frac{12}{35} \)

For \( \angle T \):
1. \( \sin(T) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{TR}{SR} = \frac{35}{37} \)
2. \( \cos(T) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{ST}{SR} = \frac{12}{37} \)
3. \( \tan(T) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{TR}{ST} = \frac{35}{12} \)

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

---

Final Answer:

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