Trigonometric Ratios Worksheets - Math Monks - Free Printable

To solve the problems involving trigonometric ratios in right triangles, we will use the basic trigonometric functions: sine (sin), cosine (cos), and tangent (tan). The relationships are as follows:

- $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
- $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
- $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$

Part 1: Finding Missing Sides

#### Problem 1:
Given:
- $\angle A = 70^\circ$
- $BC = 7.2$
- Find $x$ (hypotenuse) and $y$ (opposite side).

Using $\cos 70^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7.2}{x}$:
\[
x = \frac{7.2}{\cos 70^\circ} = \frac{7.2}{0.342} \approx 21.05
\]

Using $\sin 70^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{x}$:
\[
y = x \cdot \sin 70^\circ = 21.05 \cdot 0.940 \approx 19.86
\]

So, $x \approx 21.05$ and $y \approx 19.86$.

#### Problem 2:
Given:
- $\angle Q = 16^\circ$
- $PR = 20$
- Find $x$ (adjacent side) and $y$ (opposite side).

Using $\cos 16^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{20}$:
\[
x = 20 \cdot \cos 16^\circ = 20 \cdot 0.961 \approx 19.22
\]

Using $\sin 16^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{20}$:
\[
y = 20 \cdot \sin 16^\circ = 20 \cdot 0.276 \approx 5.52
\]

So, $x \approx 19.22$ and $y \approx 5.52$.

#### Problem 3:
Given:
- $\angle A = 50^\circ$
- $AB = 5$
- Find $x$ (hypotenuse) and $y$ (opposite side).

Using $\sin 50^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{x}$:
\[
y = x \cdot \sin 50^\circ
\]

Using $\cos 50^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{x}$:
\[
x = \frac{5}{\cos 50^\circ} = \frac{5}{0.643} \approx 7.78
\]

Using $y = x \cdot \sin 50^\circ$:
\[
y = 7.78 \cdot 0.766 \approx 5.96
\]

So, $x \approx 7.78$ and $y \approx 5.96$.

#### Problem 4:
Given:
- $\angle B = 64^\circ$
- $AC = 10$
- Find $x$ (opposite side) and $y$ (adjacent side).

Using $\sin 64^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{10}$:
\[
x = 10 \cdot \sin 64^\circ = 10 \cdot 0.899 \approx 8.99
\]

Using $\cos 64^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{10}$:
\[
y = 10 \cdot \cos 64^\circ = 10 \cdot 0.438 \approx 4.38
\]

So, $x \approx 8.99$ and $y \approx 4.38$.

#### Problem 5:
Given:
- $\angle A = 45^\circ$
- $\angle C = 45^\circ$
- $AB = x$
- $BC = y$
- Since it's a 45-45-90 triangle, $x = y$.

Using the properties of a 45-45-90 triangle:
\[
x = y = \frac{AC}{\sqrt{2}}
\]

Since $AC = x\sqrt{2}$, we can use the Pythagorean theorem:
\[
x^2 + y^2 = AC^2 \implies x^2 + x^2 = (x\sqrt{2})^2 \implies 2x^2 = 2x^2
\]

So, $x = y = 7.07$ (if $AC = 10$).

#### Problem 6:
Given:
- $\angle A = 30^\circ$
- $\angle C = 60^\circ$
- $AB = 1$
- Find $x$ (hypotenuse) and $y$ (adjacent side).

Using the properties of a 30-60-90 triangle:
\[
x = 2 \cdot AB = 2 \cdot 1 = 2
\]
\[
y = AB \cdot \sqrt{3} = 1 \cdot \sqrt{3} = \sqrt{3} \approx 1.73
\]

So, $x = 2$ and $y \approx 1.73$.

#### Problem 7:
Given:
- $\angle A = 60^\circ$
- $\angle C = 30^\circ$
- $BC = 27$
- Find $x$ (adjacent side) and $y$ (opposite side).

Using the properties of a 30-60-90 triangle:
\[
x = BC \cdot \sqrt{3} = 27 \cdot \sqrt{3} \approx 46.77
\]
\[
y = 2 \cdot BC = 2 \cdot 27 = 54
\]

So, $x \approx 46.77$ and $y = 54$.

#### Problem 8:
Given:
- $\angle A = 30^\circ$
- $\angle C = 60^\circ$
- $AC = 6$
- Find $x$ (adjacent side) and $y$ (opposite side).

Using the properties of a 30-60-90 triangle:
\[
x = AC \cdot \sqrt{3} = 6 \cdot \sqrt{3} \approx 10.39
\]
\[
y = 2 \cdot AC = 2 \cdot 6 = 12
\]

So, $x \approx 10.39$ and $y = 12$.

Part 2: Finding Unknown Angles

#### Problem 9:
Given:
- $AB = 6$
- $BC = 4$
- Find $x$ (angle opposite to $AB$) and $y$ (angle opposite to $BC$).

Using $\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{4} = 1.5$:
\[
x = \tan^{-1}(1.5) \approx 56^\circ
\]
\[
y = 90^\circ - x = 90^\circ - 56^\circ = 34^\circ
\]

So, $x \approx 56^\circ$ and $y \approx 34^\circ$.

#### Problem 10:
Given:
- $AB = 14$
- $BC = 7$
- Find $x$ (angle opposite to $AB$) and $y$ (angle opposite to $BC$).

Using $\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{14}{7} = 2$:
\[
x = \tan^{-1}(2) \approx 63^\circ
\]
\[
y = 90^\circ - x = 90^\circ - 63^\circ = 27^\circ
\]

So, $x \approx 63^\circ$ and $y \approx 27^\circ$.

Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & x \approx 21.05, y \approx 19.86 \\
2. & x \approx 19.22, y \approx 5.52 \\
3. & x \approx 7.78, y \approx 5.96 \\
4. & x \approx 8.99, y \approx 4.38 \\
5. & x = 7.07, y = 7.07 \\
6. & x = 2, y \approx 1.73 \\
7. & x \approx 46.77, y = 54 \\
8. & x \approx 10.39, y = 12 \\
9. & x \approx 56^\circ, y \approx 34^\circ \\
10. & x \approx 63^\circ, y \approx 27^\circ \\
\end{array}
}
\]