Trigonometric Ratios, Angles and Applications for SS 1

Trigonometry

Performance Objectives

By the end of the lesson, students should be able to:

1. identify basic trigonometric ratios;
2. solve problems involving use of sine, cosine and tangent in right-angled triangles;
3. derive trigonometric ratios of special angles $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$;
4. apply the use of trigonometric ratios of $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$ in solving problems without the use of calculating aids;
5. relate sine and cosine ratios to the unit circle;
6. draw graphs of sine and cosine of angles;
7. use trigonometric ratio knowledge to solve angles of elevation and depression.

Basic Trigonometric Ratios in Right-Angled Triangles

Trigonometry is the branch of mathematics that deals with the relationships between the sides and the angles of triangles.

The study of trigonometry begins with the right-angled triangle. A right-angled triangle is a triangle in which one of the interior angles measures exactly $90^\circ$. The three sides of a right-angled triangle are named in relation to a chosen reference angle, often denoted by the Greek letter theta ($\theta$).

The names of the three sides are:

1. Hypotenuse: The longest side of the right-angled triangle, which is always directly opposite the $90^\circ$ angle.
2. Opposite: The side that is directly across from the reference angle $\theta$.
3. Adjacent: The side that lies next to the reference angle $\theta$ and connects it to the $90^\circ$ angle.

Observe the diagram below to identify the sides of a right-angled triangle in relation to the reference angle.

The three basic trigonometric ratios are defined as the ratios of the lengths of these sides. The ratios are sine ($\sin$), cosine ($\cos$), and tangent ($\tan$).

The mathematical definitions of these ratios are:

$$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$$

$$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$$

$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$$

A common memory aid used to remember these ratios is “SOH CAH TOA”, where:

1. SOH stands for $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$
2. CAH stands for $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$
3. TOA stands for $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$

Worked Examples

Example 1: In a right-angled triangle $XYZ$, the angle at $Y$ is $90^\circ$. If the length of side $XY = 3\text{ cm}$ and the length of side $YZ = 4\text{ cm}$, calculate the length of the hypotenuse $XZ$, and find the values of $\sin \theta$, $\cos \theta$, and $\tan \theta$ where $\theta$ is the angle at $Z$.

Solution:

First, apply Pythagoras’ theorem to find the length of the hypotenuse $XZ$:

$$XZ^2 = XY^2 + YZ^2$$

$$XZ^2 = 3^2 + 4^2$$

$$XZ^2 = 9 + 16 = 25$$

$$XZ = \sqrt{25} = 5\text{ cm}$$

With respect to the angle $\theta$ at $Z$:

1. $\text{Opposite} = XY = 3\text{ cm}$
2. $\text{Adjacent} = YZ = 4\text{ cm}$
3. $\text{Hypotenuse} = XZ = 5\text{ cm}$

Now, calculate the trigonometric ratios:

$$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{5} = 0.6$$

$$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5} = 0.8$$

$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} = 0.75$$

Example 2: A right-angled triangle has a hypotenuse of length $13\text{ cm}$. If one of the acute angles is $\alpha$ and the side adjacent to $\alpha$ is $12\text{ cm}$, find the length of the opposite side and calculate $\sin \alpha$ and $\tan \alpha$.

Solution:

Let the opposite side be $x$. Apply Pythagoras’ theorem:

$$13^2 = x^2 + 12^2$$

$$169 = x^2 + 144$$

$$x^2 = 169 – 144 = 25$$

$$x = \sqrt{25} = 5\text{ cm}$$

The length of the opposite side is $5\text{ cm}$. Now, calculate the ratios:

$$\sin \alpha = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13}$$

$$\tan \alpha = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{5}{12}$$

Derivation of Trigonometric Ratios of Special Angles

Special angles in trigonometry are angles that appear frequently in geometric problems. These angles are $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, and $90^\circ$. The exact values of their trigonometric ratios can be derived geometrically without using calculators.

Derivation for $45^\circ$

To derive the ratios for $45^\circ$, consider an isosceles right-angled triangle where the two equal sides are of length $1$ unit each. Since the triangle is isosceles and right-angled, the other two angles must be equal to $45^\circ$ each.

Let the adjacent side be $1$ unit and the opposite side be $1$ unit. By Pythagoras’ theorem, the hypotenuse $h$ is calculated as:

$$h^2 = 1^2 + 1^2 = 2$$

$$h = \sqrt{2}$$

Look at the isosceles right-angled triangle shown below.

Using the definitions of the basic trigonometric ratios for $\theta = 45^\circ$:

$$\sin 45^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

$$\cos 45^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

$$\tan 45^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1}{1} = 1$$

Derivation for $30^\circ$ and $60^\circ$

To derive the ratios for $30^\circ$ and $60^\circ$, consider an equilateral triangle $ABC$ with side lengths of $2$ units. Each interior angle of an equilateral triangle is $60^\circ$.

Draw a perpendicular line from vertex $A$ to the base $BC$ at point $D$. This line bisects the base $BC$, making $BD = 1$ unit. It also bisects the angle at $A$, creating two angles of $30^\circ$ each. The triangle $ABD$ is a right-angled triangle with angles $30^\circ$, $60^\circ$, and $90^\circ$.

The hypotenuse $AB = 2$ units, and the base $BD = 1$ unit. The height $AD$ is calculated using Pythagoras’ theorem:

$$AB^2 = AD^2 + BD^2$$

$$2^2 = AD^2 + 1^2$$

$$4 = AD^2 + 1$$

$$AD^2 = 3$$

$$AD = \sqrt{3}$$

Observe the split equilateral triangle below.

Using the right-angled triangle $ABD$ with the $60^\circ$ angle at $B$:

$$\sin 60^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\sqrt{3}}{2}$$

$$\cos 60^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{2}$$

$$\tan 60^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{3}}{1} = \sqrt{3}$$

Using the same triangle $ABD$ with the $30^\circ$ angle at $A$:

$$\sin 30^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{2}$$

$$\cos 30^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{3}}{2}$$

$$\tan 30^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

Derivation for $0^\circ$ and $90^\circ$

As the angle $\theta$ in a right-angled triangle approaches $0^\circ$, the length of the opposite side approaches $0$, and the length of the adjacent side becomes equal to the hypotenuse. Therefore:

$$\sin 0^\circ = \frac{0}{\text{Hypotenuse}} = 0$$

$$\cos 0^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = 1$$

$$\tan 0^\circ = \frac{0}{\text{Adjacent}} = 0$$

As the angle $\theta$ approaches $90^\circ$, the length of the adjacent side approaches $0$, and the length of the opposite side becomes equal to the hypotenuse. Therefore:

$$\sin 90^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = 1$$

$$\cos 90^\circ = \frac{0}{\text{Hypotenuse}} = 0$$

$$\tan 90^\circ = \frac{\text{Opposite}}{0} \quad (\text{Undefined})$$

Summary of Special Angles

The exact values of the trigonometric ratios for these special angles are summarized in the table below:

Worked Examples

Example 1: Evaluate the expression $\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ$ without using a calculator.

Solution:

Substitute the exact values of the special angles into the expression:

$$\sin 30^\circ = \frac{1}{2}$$

$$\cos 60^\circ = \frac{1}{2}$$

$$\cos 30^\circ = \frac{\sqrt{3}}{2}$$

$$\sin 60^\circ = \frac{\sqrt{3}}{2}$$

Now, multiply and add the terms:

$$\left(\frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right)$$

$$= \frac{1}{4} + \frac{3}{4}$$

$$= \frac{4}{4} = 1$$

Example 2: Find the value of $x$ in the equation $x \tan 45^\circ = \sin 60^\circ \cos 30^\circ$ without using calculating aids.

Solution:

Substitute the exact values of the trigonometric ratios:

$$\tan 45^\circ = 1$$

$$\sin 60^\circ = \frac{\sqrt{3}}{2}$$

$$\cos 30^\circ = \frac{\sqrt{3}}{2}$$

Substitute these values into the equation:

$$x(1) = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$$

$$x = \frac{3}{4}$$

Sine and Cosine Ratios on the Unit Circle

A unit circle is a circle with a radius of exactly one unit, centered at the origin $(0,0)$ of the Cartesian coordinate plane.

The unit circle is used to extend the definitions of trigonometric ratios to any angle, including angles greater than $90^\circ$ and negative angles.

Let a point $P(x, y)$ lie on the circumference of the unit circle. The radius line $OP$ has a length of $r = 1$ and makes an angle $\theta$ with the positive x-axis. By dropping a perpendicular line from $P$ to the x-axis, a right-angled triangle is formed with a base of length $x$ and a height of length $y$.

Applying the basic trigonometric ratios to this triangle:

$$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{x}{1} = x$$

$$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{y}{1} = y$$

Therefore, the coordinates of any point $P$ on the unit circle can be written as $(\cos \theta, \sin \theta)$.

Observe the unit circle diagram below to see how coordinates relate to sine and cosine.

The Cartesian plane is divided into four quadrants, which affect the signs (positive or negative) of the trigonometric ratios:

1. First Quadrant ($0^\circ < \theta < 90^\circ$): Both $x$ and $y$ are positive. Therefore, all ratios ($\sin$, $\cos$, $\tan$) are positive.
2. Second Quadrant ($90^\circ < \theta < 180^\circ$): $x$ is negative and $y$ is positive. Therefore, only $\sin \theta$ is positive.
3. Third Quadrant ($180^\circ < \theta < 270^\circ$): Both $x$ and $y$ are negative. Therefore, only $\tan \theta$ is positive (since $\frac{-y}{-x}$ is positive).
4. Fourth Quadrant ($270^\circ < \theta < 360^\circ$): $x$ is positive and $y$ is negative. Therefore, only $\cos \theta$ is positive.

This sign distribution is remembered using the acronym CAST, starting from the fourth quadrant and moving counter-clockwise, or ASTC (All, Sine, Tangent, Cosine) starting from the first quadrant.

Graphs of Sine and Cosine Functions

The values of $\sin \theta$ and $\cos \theta$ can be plotted against the angle $\theta$ for values from $0^\circ$ to $360^\circ$ to show their wave-like patterns.

Table of Values

The table below shows the values of $\sin \theta$ and $\cos \theta$ at intervals of $45^\circ$:

Observe the graphs of the sine and cosine functions plotted below.

Key features of these graphs include:

1. Periodicity: Both graphs repeat their patterns every $360^\circ$. This interval is called the period.
2. Amplitude: The maximum displacement from the central axis is $1$. The range of both functions is $-1 \le y \le 1$.
3. Phase Difference: The cosine graph is a sine graph shifted to the left by $90^\circ$.

Angles of Elevation and Depression

Trigonometric ratios are used in real-world measurements to find heights and distances that cannot be measured directly.

An angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object that is above the observer’s eye level.

An angle of depression is the angle between the horizontal line of sight and the line of sight down to an object that is below the observer’s eye level.

Observe the diagram below to see the relationship between the horizontal line, the angle of elevation, and the angle of depression.

Because the horizontal lines of sight are parallel, the angle of elevation from an object $B$ to an observer $A$ is equal to the angle of depression from the observer $A$ to the object $B$. They form alternate angles, which are mathematically equal.

Worked Examples

Example 1: A student stands $20\text{ metres}$ away from the base of a vertical school building in Enugu. The angle of elevation from the ground level where the student stands to the top of the building is $30^\circ$. Calculate the height of the building, leaving the answer in surd form.

Solution:

Let the height of the building be $h$. The situation forms a right-angled triangle where:

1. The side opposite the $30^\circ$ angle is the height of the building, $h$.
2. The side adjacent to the $30^\circ$ angle is the distance from the building, $20\text{ m}$.

Use the tangent ratio:

$$\tan 30^\circ = \frac{\text{Opposite}}{\text{Adjacent}}$$

$$\tan 30^\circ = \frac{h}{20}$$

$$h = 20 \times \tan 30^\circ$$

Substitute the exact value of $\tan 30^\circ = \frac{\sqrt{3}}{3}$:

$$h = 20 \times \frac{\sqrt{3}}{3}$$

$$h = \frac{20\sqrt{3}}{3}\text{ metres}$$

Example 2: From the top of a vertical cliff in Idanre, which is $120\text{ metres}$ high, the angle of depression of a stationary car on the ground is $60^\circ$. Find the distance of the car from the foot of the cliff without using a calculator.

Solution:

Let the distance of the car from the foot of the cliff be $d$.

The angle of depression from the top of the cliff is $60^\circ$. This is equal to the angle of elevation from the car to the top of the cliff (alternate angles). This forms a right-angled triangle where:

1. The side opposite the $60^\circ$ angle is the height of the cliff, $120\text{ m}$.
2. The side adjacent to the $60^\circ$ angle is the distance $d$.

Use the tangent ratio:

$$\tan 60^\circ = \frac{\text{Opposite}}{\text{Adjacent}}$$

$$\tan 60^\circ = \frac{120}{d}$$

$$d = \frac{120}{\tan 60^\circ}$$

Substitute the exact value of $\tan 60^\circ = \sqrt{3}$:

$$d = \frac{120}{\sqrt{3}}$$

Rationalize the denominator:

$$d = \frac{120 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$$

$$d = \frac{120\sqrt{3}}{3}$$

$$d = 40\sqrt{3}\text{ metres}$$

Exercises

Multiple Choice Questions

The questions below are for practice. Choose the correct option for each question:

1. In a right-angled triangle, the ratio of the adjacent side to the hypotenuse is called:
   1. Sine
   2. Cosine
   3. Tangent
   4. Secant
2. What is the exact value of $\sin 60^\circ$?
   1. $\frac{1}{2}$
   2. $\frac{\sqrt{2}}{2}$
   3. $\frac{\sqrt{3}}{2}$
   4. $1$
3. If $\cos \theta = \frac{1}{2}$ for an acute angle $\theta$, what is the value of $\theta$?
   1. $30^\circ$
   2. $45^\circ$
   3. $60^\circ$
   4. $90^\circ$
4. On a unit circle, the y-coordinate of a point on the circumference represents:
   1. $\cos \theta$
   2. $\sin \theta$
   3. $\tan \theta$
   4. The radius
5. An observer at the top of a tower looks down at a ball on the ground. The angle between the observer’s horizontal line of sight and the line of sight to the ball is the:
   1. Angle of elevation
   2. Angle of depression
   3. Alternate angle
   4. Vertical angle

Theory Questions

Solve the following problems step by step:

1. Without using mathematical tables or calculators, show that $\cos^2 45^\circ + \sin^2 45^\circ = 1$.
2. A ladder of length $10\text{ metres}$ leans against a vertical wall in Lagos. If the ladder makes an angle of $60^\circ$ with the horizontal ground, calculate:
   1. the distance from the foot of the ladder to the wall;
   2. the height up the wall reached by the ladder.
3. Draw a table of values for $y = \sin \theta$ for values of $\theta$ from $0^\circ$ to $180^\circ$ at intervals of $30^\circ$. Use the values to sketch the graph of the function.
4. A surveyor stands at a distance of $50\text{ metres}$ from the base of a vertical telecommunications mast. The angle of elevation of the top of the mast from the surveyor’s eye level is $45^\circ$. If the surveyor’s eye level is $1.5\text{ metres}$ above the ground, calculate the total height of the mast.