Introduction

Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles. It is a fundamental concept in mathematics and has various applications in fields such as engineering, physics, and astronomy. In this topic, we will explore trigonometric functions, trigonometric ratios, and their applications in solving problems involving right-angled triangles.

Trigonometric Ratios

Trigonometric ratios are ratios of the sides of a right-angled triangle. The main trigonometric ratios are sine, cosine, and tangent, denoted by $\sin$, $\cos$, and $\tan$ respectively.

Sine (sin)

The sine of an angle in a right-angled triangle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. $$\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}$$

Example:
Given a right-angled triangle with an angle of 30 degrees and the hypotenuse of length 10 cm, find the length of the side opposite the angle. $$\sin(30^\circ) = \frac{x}{10}$$
$$x = 10 \times \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \text{ cm}$$

Cosine (cos)

The cosine of an angle in a right-angled triangle is defined as the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. $$\cos(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}}$$

Example:
In a right-angled triangle with an angle of 45 degrees and the hypotenuse of length 8 cm, find the length of the side adjacent to the angle. $$\cos(45^\circ) = \frac{x}{8}$$
$$x = 8 \times \cos(45^\circ) = 8 \times \frac{\sqrt{2}}{2} = 4\sqrt{2} \text{ cm}$$

Tangent (tan)

The tangent of an angle in a right-angled triangle is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. $$\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}$$

Example:
If the angle in a right-angled triangle is 60 degrees and the side opposite the angle is 5 cm long, find the length of the adjacent side. $$\tan(60^\circ) = \frac{5}{x}$$
$$x = \frac{5}{\tan(60^\circ)} = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \text{ cm}$$

Pythagorean Identity

The Pythagorean identity is a fundamental formula in trigonometry that relates the three main trigonometric functions for any angle in a right-angled triangle. $$\sin^2(\theta) + \cos^2(\theta) = 1$$

Example:
Given that $\sin(\alpha) = \frac{3}{5}$, find $\cos(\alpha)$. Using the Pythagorean identity, we have: $$\sin^2(\alpha) + \cos^2(\alpha) = 1$$ $$(\frac{3}{5})^2 + \cos^2(\alpha) = 1$$ $$\frac{9}{25} + \cos^2(\alpha) = 1$$ $$\cos^2(\alpha) = 1 - \frac{9}{25} = \frac{16}{25}$$ $$\cos(\alpha) = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$$

Angle of Elevation and Depression

The angle of elevation is the angle formed by the line of sight with the horizontal when looking upwards. The angle of depression is the angle formed by the line of sight with the horizontal when looking downwards.

Angle of Elevation

When an observer looks at an object above the horizontal level, the angle formed is the angle of elevation.

Angle of Depression

When an observer looks at an object below the horizontal level, the angle formed is the angle of depression.

Example:
An observer stands 100 meters away from a tree and looks up at an angle of elevation of 30 degrees. Calculate the height of the tree if the observer's eye level is 1.5 meters above the ground. Let $h$ be the height of the tree. We have: $$\tan(30^\circ) = \frac{h}{100}$$ $$h = 100 \times \tan(30^\circ)$$ $$h = 100 \times \frac{1}{\sqrt{3}} = \frac{100\sqrt{3}}{3} \approx 57.74 \text{ meters}$$

Common Mistakes

• Forgetting to convert angles to the correct units (radians or degrees) before performing trigonometric calculations.
• Misidentifying the correct sides of the triangle for the trigonometric ratios.
• Incorrectly applying the Pythagorean identity in trigonometric equations.
• Confusing the angle of depression with the angle of elevation in real-world problems.

Key Points

• Trigonometry deals with the relationships between the angles and sides of triangles.
• The main trigonometric ratios are sine, cosine, and tangent.
• The Pythagorean identity relates the sine and cosine of an angle in a right-angled triangle.
• The angle of elevation is the angle formed when looking upwards, while the angle of depression is the angle formed when looking downwards.

Practice Questions

1. In a right-angled triangle, if the length of the side opposite an angle is 12 cm and the hypotenuse is 15 cm, calculate the sine of the angle.
2. If $\cos(\alpha) = \frac{4}{5}$, find $\sin(\alpha)$.
3. A ladder leaning against a wall makes an angle of 60 degrees with the ground. If the foot of the ladder is 5 meters away from the wall, find the length of the ladder.
4. Given that $\tan(\theta) = \frac{5}{12}$, find $\cos(\theta)$.
5. An airplane at an altitude of 10,000 meters spots a landmark on the ground with an angle of depression of 15 degrees. Calculate the horizontal distance from the airplane to the landmark.

Practice Question 1 Solution

Given: opposite side = 12 cm, hypotenuse = 15 cm
$$\sin(\theta) = \frac{12}{15} = \frac{4}{5}$$

Practice Question 2 Solution

Given: $\cos(\alpha) = \frac{4}{5}$
Using the Pythagorean identity:
$$\sin^2(\alpha) + \cos^2(\alpha) = 1$$ $$(\sin(\alpha))^2 + (\frac{4}{5})^2 = 1$$ $$(\sin(\alpha))^2 + \frac{16}{25} = 1$$ $$(\sin(\alpha))^2 = 1 - \frac{16}{25} = \frac{9}{25}$$ $$\sin(\alpha) = \pm\frac{3}{5}$$

Practice Question 3 Solution

Given: angle = 60 degrees, distance from the wall = 5 meters
Let the length of the ladder be $L$.
$$\sin(60^\circ) = \frac{L}{5}$$ $$L = 5 \times \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \text{ meters}$$

Practice Question 4 Solution

Given: $\tan(\theta) = \frac{5}{12}$
Using the Pythagorean identity:
$$\tan^2(\theta) + 1 = \sec^2(\theta)$$ $$(\frac{5}{12})^2 + 1 = \sec^2(\theta)$$ $$\sec^2(\theta) = 1 + \frac{25}{144} = \frac{169}{144}$$ $$\sec(\theta) = \pm\frac{13}{12}$$

Practice Question 5 Solution

Given: altitude = 10,000 meters, angle of depression = 15 degrees
Let the horizontal distance be $x$.
$$\tan(15^\circ) = \frac{10,000}{x}$$ $$x = \frac{10,000}{\tan(15^\circ)} = \frac{10,000}{\frac{1}{\sqrt{3}}}= 10,000\sqrt{3} \approx 17,320.51 \text{ meters}$$