FREE Trigonometry ratio review worksheet. | Trigonometry ... - Free Printable

To solve the problems in the image, we will use the trigonometric ratios: sine (sin), cosine (cos), and tangent (tan). The steps are as follows:

General Approach:

1. Identify the given sides of the triangle.
2. Determine which trigonometric ratio to use based on the given sides.
3. Set up the equation using the appropriate trigonometric ratio.
4. Solve for the angle \( x \) using a calculator.

Trigonometric Ratios Recap:

- Sine (\(\sin\)): \(\sin x = \frac{\text{opposite}}{\text{hypotenuse}}\)
- Cosine (\(\cos\)): \(\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}\)
- Tangent (\(\tan\)): \(\tan x = \frac{\text{opposite}}{\text{adjacent}}\)

Solving Each Problem:

#### Problem 1:
- Given: Opposite = 8, Hypotenuse = 17
- Use: \(\sin x = \frac{\text{opposite}}{\text{hypotenuse}}\)
- Equation: \(\sin x = \frac{8}{17}\)
- Solve: \( x = \sin^{-1}\left(\frac{8}{17}\right) \approx 28^\circ \)

#### Problem 2:
- Given: Adjacent = 55, Hypotenuse = 60
- Use: \(\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}\)
- Equation: \(\cos x = \frac{55}{60}\)
- Solve: \( x = \cos^{-1}\left(\frac{55}{60}\right) \approx 19^\circ \)

#### Problem 3:
- Given: Opposite = 30, Hypotenuse = 10
- Use: \(\sin x = \frac{\text{opposite}}{\text{hypotenuse}}\)
- Equation: \(\sin x = \frac{30}{10} = 3\) (This is not possible since \(\sin x \leq 1\). There might be a typo in the problem.)

#### Problem 4:
- Given: Adjacent = 27, Hypotenuse = 42
- Use: \(\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}\)
- Equation: \(\cos x = \frac{27}{42}\)
- Solve: \( x = \cos^{-1}\left(\frac{27}{42}\right) \approx 50^\circ \)

#### Problem 5:
- Given: Opposite = 9, Hypotenuse = 15
- Use: \(\sin x = \frac{\text{opposite}}{\text{hypotenuse}}\)
- Equation: \(\sin x = \frac{9}{15} = \frac{3}{5}\)
- Solve: \( x = \sin^{-1}\left(\frac{3}{5}\right) \approx 37^\circ \)

#### Problem 6:
- Given: Opposite = 25, Hypotenuse = 85
- Use: \(\sin x = \frac{\text{opposite}}{\text{hypotenuse}}\)
- Equation: \(\sin x = \frac{25}{85} = \frac{5}{17}\)
- Solve: \( x = \sin^{-1}\left(\frac{5}{17}\right) \approx 17^\circ \)

#### Problem 7:
- Given: Opposite = 36, Adjacent = 48
- Use: \(\tan x = \frac{\text{opposite}}{\text{adjacent}}\)
- Equation: \(\tan x = \frac{36}{48} = \frac{3}{4}\)
- Solve: \( x = \tan^{-1}\left(\frac{3}{4}\right) \approx 37^\circ \)

#### Problem 8:
- Given: Opposite = 20, Adjacent = 4
- Use: \(\tan x = \frac{\text{opposite}}{\text{adjacent}}\)
- Equation: \(\tan x = \frac{20}{4} = 5\)
- Solve: \( x = \tan^{-1}(5) \approx 79^\circ \)

#### Problem 9:
- Given: Opposite = 58, Hypotenuse = 65
- Use: \(\sin x = \frac{\text{opposite}}{\text{hypotenuse}}\)
- Equation: \(\sin x = \frac{58}{65}\)
- Solve: \( x = \sin^{-1}\left(\frac{58}{65}\right) \approx 60^\circ \)

Final Answers:

\[
\boxed{
\begin{array}{ccc}
1 & 2 & 3 \\
28^\circ & 19^\circ & \text{Not possible} \\
4 & 5 & 6 \\
50^\circ & 37^\circ & 17^\circ \\
7 & 8 & 9 \\
37^\circ & 79^\circ & 60^\circ \\
\end{array}
}
\]