Trigonometric Ratios in Right Triangles worksheet with problems to find missing sides and angles.

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 $x\sqrt{2}$.

Using $\sin 45^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{x\sqrt{2}} = \frac{1}{\sqrt{2}}$:
\[
x = y
\]
\[
\text{hypotenuse} = x\sqrt{2}
\]

So, $x = y$ and hypotenuse $= x\sqrt{2}$.

#### 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 $x$ (opposite side) and $y$ (hypotenuse).

Using $\tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{27}$:
\[
x = 27 \cdot \tan 30^\circ = 27 \cdot \frac{1}{\sqrt{3}} \approx 15.59
\]

Using $\cos 30^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{27}{y}$:
\[
y = \frac{27}{\cos 30^\circ} = \frac{27}{\frac{\sqrt{3}}{2}} = \frac{54}{\sqrt{3}} \approx 31.18
\]

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

#### 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.

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

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

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

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

Using $\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{14} = 0.5$:
\[
x = \tan^{-1}(0.5) \approx 27^\circ
\]

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

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

Final Answers:

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