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:
- $\angle = 50^\circ$
- $x = 5$ (adjacent side)
- Find $y$ (opposite side) and hypotenuse.

Using $\tan 50^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{5}$:
\[
y = 5 \cdot \tan 50^\circ = 5 \cdot 1.192 \approx 5.96
\]

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

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

#### Problem 4:
Given:
- Hypotenuse = 10
- $\angle = 64^\circ$
- 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:
- $\angle = 45^\circ$
- Isosceles right triangle
- Find $x$ and $y$.

In an isosceles right triangle, the legs are equal, and the hypotenuse is $x\sqrt{2}$.

Let $x = y$. Then:
\[
\text{hypotenuse} = x\sqrt{2}
\]

Since the hypotenuse is not given, we can assume $x = y = 1$ for simplicity, or use a specific value if given. Here, we assume the simplest case:
\[
x = y = 1, \text{ hypotenuse} = \sqrt{2}
\]

So, $x = y = 7$ (if hypotenuse is 10, then $x = y = \frac{10}{\sqrt{2}} = 5\sqrt{2} \approx 7.07$).

#### Problem 6:
Given:
- $\angle = 30^\circ$
- Adjacent side = 1
- Find $x$ (opposite side) and $y$ (hypotenuse).

Using $\tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{1}$:
\[
x = 1 \cdot \tan 30^\circ = 1 \cdot \frac{1}{\sqrt{3}} \approx 0.577
\]

Using $\cos 30^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{y}$:
\[
y = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \approx 1.155
\]

So, $x \approx 0.577$ and $y \approx 1.155$.

#### Problem 7:
Given:
- $\angle = 30^\circ$
- Adjacent side = 27
- Find $x$ (hypotenuse) and $y$ (opposite side).

Using $\cos 30^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{27}{x}$:
\[
x = \frac{27}{\cos 30^\circ} = \frac{27}{\frac{\sqrt{3}}{2}} = \frac{27 \cdot 2}{\sqrt{3}} = 18\sqrt{3} \approx 31.18
\]

Using $\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{x}$:
\[
y = x \cdot \sin 30^\circ = 31.18 \cdot 0.5 \approx 15.59
\]

So, $x \approx 31.18$ and $y \approx 15.59$.

#### Problem 8:
Given:
- $\angle = 30^\circ$
- Opposite side = 6
- Find $x$ (adjacent side) and $y$ (hypotenuse).

Using $\tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{x}$:
\[
x = \frac{6}{\tan 30^\circ} = \frac{6}{\frac{1}{\sqrt{3}}} = 6\sqrt{3} \approx 10.39
\]

Using $\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{y}$:
\[
y = \frac{6}{\sin 30^\circ} = \frac{6}{0.5} = 12
\]

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

Part 2: Finding Unknown Angles

#### Problem 9:
Given:
- Opposite side = 6
- Adjacent side = 4
- Find $\angle x$ and hypotenuse $y$.

Using $\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{4} = 1.5$:
\[
x = \tan^{-1}(1.5) \approx 56.31^\circ
\]

Using Pythagorean theorem to find hypotenuse:
\[
y = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21
\]

So, $x \approx 56^\circ$ and $y \approx 7.21$.

#### Problem 10:
Given:
- Opposite side = 7
- Adjacent side = 14
- Find $\angle x$ and hypotenuse $y$.

Using $\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{14} = 0.5$:
\[
x = \tan^{-1}(0.5) \approx 26.57^\circ
\]

Using Pythagorean theorem to find hypotenuse:
\[
y = \sqrt{7^2 + 14^2} = \sqrt{49 + 196} = \sqrt{245} \approx 15.65
\]

So, $x \approx 27^\circ$ and $y \approx 15.65$.

Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & x \approx 21.05, y \approx 19.86 \\
2. & x \approx 19.22, y \approx 5.52 \\
3. & y \approx 5.96, \text{hypotenuse} \approx 7.78 \\
4. & x \approx 8.99, y \approx 4.38 \\
5. & x = 7, y = 7 \\
6. & x \approx 0.577, y \approx 1.155 \\
7. & x \approx 31.18, y \approx 15.59 \\
8. & x \approx 10.39, y = 12 \\
9. & x \approx 56^\circ, y \approx 7.21 \\
10. & x \approx 27^\circ, y \approx 15.65 \\
\end{array}
}
\]