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

To solve the trigonometric ratios for each right triangle, we will use the definitions of sine (sin), cosine (cos), and tangent (tan):

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

Let's solve each problem step by step.

---

Problem 1:

Triangle ABC with sides:
- AB = 9 (opposite to ∠A)
- BC = 12 (adjacent to ∠A)
- AC = 15 (hypotenuse)

#### Trigonometric Ratios for ∠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 DEF with sides:
- DE = 16 (adjacent to ∠F)
- DF = 12 (opposite to ∠F)
- EF = 20 (hypotenuse)

#### Trigonometric Ratios for ∠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:

TriangleQRS with sides:
- QR = 5 (adjacent to ∠Q)
- RS = 12 (opposite to ∠Q)
- QS = 13 (hypotenuse)

#### Trigonometric Ratios for ∠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 MNO with sides:
- MN = 24 (opposite to ∠M)
- MO = 10 (adjacent to ∠M)
- NO = 26 (hypotenuse)

#### Trigonometric Ratios for ∠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 TUV with sides:
- TU = 3 (opposite to ∠V)
- UV = 4 (adjacent to ∠V)
- TV = 5 (hypotenuse)

#### Trigonometric Ratios for ∠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 ∠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 GHI with sides:
- GH = 6 (adjacent to ∠G)
- HI = 41 (opposite to ∠G)
- GI = 40 (hypotenuse)

#### Trigonometric Ratios for ∠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 ∠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 FGH with sides:
- FG = 8 (adjacent to ∠F)
- GH = 15 (opposite to ∠F)
- FH = 17 (hypotenuse)

#### Trigonometric Ratios for ∠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 ∠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 STR with sides:
- ST = 35 (adjacent to ∠R)
- TR = 12 (opposite to ∠R)
- SR = 37 (hypotenuse)

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

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

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

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

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

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

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:

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