Trigonometric functions are essential in various professional fields across Kenya, including construction, surveying, agriculture, and healthcare infrastructure planning. Understanding angles and their measurement enables professionals to calculate distances, heights, and slopes accurately, which is crucial for site layout, structural design, and resource management. This chapter focuses on the fundamental concept of angles, their types, and how to apply trigonometric principles to solve practical problems encountered in diverse workplaces such as county hospitals, universities, banks, and farms.

2.1 Angles

Angles are the measure of rotation between two intersecting lines or planes. They are fundamental in trigonometry, used to describe directions, slopes, and shapes in professional applications. Accurately identifying and calculating different types of angles allows for precise measurements and informed decision-making in various projects.

2.1.1 Acute Angles

Acute angles are angles less than 90 degrees. They frequently appear in architectural designs, roof pitches, and agricultural land slope measurements, where precise angle calculation is necessary for structural integrity or water drainage.

The measure of an acute angle is denoted by \( \theta \) such that

$$ 0^\circ < \theta < 90^\circ $$

Worked Examples

Example 1:
A county hospital plans to install a security camera on a wall. The camera needs to be tilted upwards at an acute angle of \(30^\circ\) to cover the entrance properly. What is the height reached by the camera beam 5 meters away horizontally?

Given:
Horizontal distance \( = 5\,m \),
Angle of elevation \( \theta = 30^\circ \)

Using the formula for height in right-angled triangle,$$ \text{Height} = \text{Horizontal distance} \times \tan \theta $$

$$ \text{Height} = 5 \times \tan 30^\circ $$

$$ \tan 30^\circ = 0.577 $$

$$ \text{Height} = 5 \times 0.577 = 2.885\,m $$

Answer: 2.885 meters

Example 2:
A farmer is measuring the slope of a field which rises 3 meters over a horizontal distance of 10 meters. Find the acute angle of the slope.

Given:
Vertical rise \( = 3\,m \),
Horizontal distance \( = 10\,m \)

$$ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{10} $$

$$ \theta = \tan^{-1} (\frac{3}{10}) $$

$$ \theta = \tan^{-1} (0.3) = 16.7^\circ $$

Answer: 16.7 degrees

Example 3:
A bank building has a triangular signboard with one angle measuring \( 45^\circ \), which is acute. The adjacent side to this angle is 4 meters. Calculate the length of the opposite side.

Given:
Angle \( \theta = 45^\circ \),
Adjacent side \( = 4\,m \)

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

$$ \text{Opposite} = \tan 45^\circ \times 4 $$

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

$$ \text{Opposite} = 1 \times 4 = 4\,m $$

Answer: 4 meters

Example 4:
A hotel manager wants to install a ramp with an incline angle of \( 25^\circ \). If the vertical height to overcome is 1.5 meters, find the length of the ramp.

Given:
Height \( = 1.5\,m \),
Angle \( \theta = 25^\circ \)

Using sine function,$$ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} $$

$$ \text{Hypotenuse} = \frac{\text{Opposite}}{\sin \theta} = \frac{1.5}{\sin 25^\circ} $$

$$ \sin 25^\circ = 0.4226 $$

$$ \text{Hypotenuse} = \frac{1.5}{0.4226} = 3.55\,m $$

Answer: 3.55 meters

Example 5:
A SACCO office needs to measure the angle of elevation to the top of a nearby tree from a point 20 meters away. The tree height is 12 meters. Find the acute angle of elevation.

Given:
Height \( = 12\,m \),
Horizontal distance \( = 20\,m \)

$$ \tan \theta = \frac{12}{20} = 0.6 $$

$$ \theta = \tan^{-1} (0.6) = 30.96^\circ $$

Answer: 30.96 degrees

2.1.2 Obtuse Angles

Obtuse angles are angles greater than 90 degrees but less than 180 degrees. They often appear in design layouts, such as road intersections, plot boundaries, or architectural features requiring wide angled turns or shapes.

An obtuse angle \( \theta \) satisfies

$$ 90^\circ < \theta < 180^\circ $$

Obtuse angles can be handled by finding their reference acute angle \( \alpha \), where

$$ \alpha = 180^\circ - \theta $$

and then applying trigonometric functions to \( \alpha \).

Worked Examples

Example 1:
A university campus has a curved road forming an obtuse angle of \(120^\circ\) at a junction. Find the reference acute angle.

Given:
\( \theta = 120^\circ \)

$$ \alpha = 180^\circ - 120^\circ = 60^\circ $$

Answer: 60 degrees

Example 2:
At a retail business site, two walls meet forming an obtuse angle of \(135^\circ\). Find the sine of this angle using the reference angle.

Given:
\( \theta = 135^\circ \),
Reference angle \( \alpha = 180^\circ - 135^\circ = 45^\circ \)

Since \( \sin \theta = \sin \alpha \),$$ \sin 135^\circ = \sin 45^\circ = 0.707 $$

Answer: 0.707

Example 3:
A cooperative plans a fence that includes an obtuse angle of \(150^\circ\). If the length of one side adjacent to the angle is 12 meters, calculate the length of the projection of the other side forming the angle on the first side using cosine.

Given:
\( \theta = 150^\circ \),
Side length \( = 12\,m \)

Reference angle:$$ \alpha = 180^\circ - 150^\circ = 30^\circ $$

Cosine of obtuse angle:$$ \cos 150^\circ = -\cos 30^\circ = -0.866 $$

Projection length:$$ 12 \times \cos 150^\circ = 12 \times (-0.866) = -10.392\,m $$

The negative sign indicates direction opposite to the side.

Answer: 10.392 meters (opposite direction)

Example 4:
A county referral hospital plans a walkway forming an obtuse angle of \(110^\circ\). Find the length of the opposite side if the adjacent side is 7 meters and the hypotenuse is 10 meters.

Given:
\( \theta = 110^\circ \),
Adjacent side \( = 7\,m \),
Hypotenuse \( = 10\,m \)

Reference angle:$$ \alpha = 180^\circ - 110^\circ = 70^\circ $$

Using sine for opposite side:$$ \sin 110^\circ = \sin 70^\circ = 0.9397 $$

Opposite side:$$ 10 \times \sin 110^\circ = 10 \times 0.9397 = 9.397\,m $$

Answer: 9.397 meters

Example 5:
At a TVET college, a triangular sign has an obtuse angle of \(140^\circ\). Calculate the other two angles if one of them is \(20^\circ\).

Given:
Sum of angles in triangle = \(180^\circ\)
One angle \( = 140^\circ \),
Second angle \( = 20^\circ \)

Third angle:$$ 180^\circ - 140^\circ - 20^\circ = 20^\circ $$

Answer: 20 degrees

2.1.3 Reflex Angles

Reflex angles are angles greater than 180 degrees but less than 360 degrees. These angles are commonly encountered in land surveying, large-scale site planning, and navigation where turns exceed a straight line.

A reflex angle \( \theta \) satisfies

$$ 180^\circ < \theta < 360^\circ $$

The reference angle \( \alpha \) for reflex angles is

$$ \alpha = 360^\circ - \theta $$

Trigonometric functions for reflex angles can be found using the reference angle and sign rules.

Worked Examples

Example 1:
A county government office is positioned such that the angle between two roads is a reflex angle of \(210^\circ\). Find the reference acute angle.

Given:
\( \theta = 210^\circ \)

$$ \alpha = 360^\circ - 210^\circ = 150^\circ $$

Since \(150^\circ\) is obtuse, find the acute reference angle for \(150^\circ\):

$$ 180^\circ - 150^\circ = 30^\circ $$

Answer: 30 degrees

Example 2:
A hotel swimming pool has a curved boundary forming a reflex angle of \(270^\circ\). Find the cosine of this angle.

Given:
\( \theta = 270^\circ \)

Reference angle:$$ \alpha = 360^\circ - 270^\circ = 90^\circ $$

Cosine of reflex angle:$$ \cos 270^\circ = \cos 90^\circ = 0 $$

Answer: 0

Example 3:
A cooperative’s land boundary turns through a reflex angle of \(300^\circ\). Calculate \( \sin 300^\circ \).

Given:
\( \theta = 300^\circ \)

Reference angle:$$ \alpha = 360^\circ - 300^\circ = 60^\circ $$

Since \( \sin 300^\circ = -\sin 60^\circ \),$$ \sin 300^\circ = -0.866 $$

Answer: -0.866

Example 4:
A SACCO office building has a corner measuring a reflex angle of \(225^\circ\). Find the sine and cosine of this angle.

Given:
\( \theta = 225^\circ \)

Reference angle:$$ \alpha = 360^\circ - 225^\circ = 135^\circ $$

Since \( \sin 225^\circ = -\sin 135^\circ \) and \( \cos 225^\circ = -\cos 135^\circ \):

Calculate sine:$$ \sin 135^\circ = \sin 45^\circ = 0.707 $$$$ \sin 225^\circ = -0.707 $$

Calculate cosine:$$ \cos 135^\circ = -\cos 45^\circ = -0.707 $$$$ \cos 225^\circ = -(-0.707) = 0.707 $$

Answer: \(\sin 225^\circ = -0.707\), \(\cos 225^\circ = 0.707\)

Example 5:
A TVET college campus has a circular walkway with an angle of \(315^\circ\). Find the tangent of this angle.

Given:
\( \theta = 315^\circ \)

Reference angle:$$ \alpha = 360^\circ - 315^\circ = 45^\circ $$

Since \( \tan 315^\circ = -\tan 45^\circ = -1 \),

Answer: -1

2.1.4 Right Angle

A right angle is exactly \(90^\circ\). It is a fundamental angle in trigonometry and geometry, representing perpendicularity. Right angles form the basis of many construction and design calculations, such as ensuring walls are square or setting out machinery correctly.

By definition,

$$ \theta = 90^\circ $$

Trigonometric values at \(90^\circ\) are:

• \( \sin 90^\circ = 1 \)
• \( \cos 90^\circ = 0 \)
• \( \tan 90^\circ \) is undefined.

Worked Examples

Example 1:
A university wants to build a staircase with steps rising vertically at a right angle. Calculate the height of the step if the horizontal run is 1 meter and the angle is \(90^\circ\).

Given:
Run \( = 1\,m \),
Angle \( = 90^\circ \)

Using tangent:$$ \tan 90^\circ = \frac{\text{rise}}{\text{run}} $$

Since \( \tan 90^\circ \) is undefined (infinite), the rise is vertical and cannot be measured by tangent.

Answer: The rise is vertical; height is undefined by tangent but equal to the length of the step

Example 2:
A hospital room wall is perpendicular to the floor. If the floor length is 6 meters, what is the cosine of the angle between the floor and wall?

Given:
\( \theta = 90^\circ \)

$$ \cos 90^\circ = 0 $$

Answer: 0

Example 3:
A bank installs a solar panel tilted at \(90^\circ\) to the ground. Calculate the sine of the angle.

Given:
\( \theta = 90^\circ \)

$$ \sin 90^\circ = 1 $$

Answer: 1

Example 4:
A cooperative sets a ladder against a wall forming a right angle with the ground. If the ladder length is 5 meters, and the base is 3 meters from the wall, calculate the height reached on the wall.

Given:
Ladder \( = 5\,m \),
Base distance \( = 3\,m \)

Using Pythagoras theorem,$$ \text{Height} = \sqrt{5^2 - 3^2} $$

$$ = \sqrt{25 - 9} = \sqrt{16} = 4\,m $$

Answer: 4 meters

Example 5:
A TVET college classroom has a sloping roof making a right angle with the wall. If the roof length is 8 meters and the vertical wall height is 6 meters, find the angle between the roof and the horizontal floor.

Given:
Roof length \( = 8\,m \),
Wall height \( = 6\,m \)

Using cosine for angle \( \theta \):

$$ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{6}{8} = 0.75 $$

$$ \theta = \cos^{-1} 0.75 = 41.41^\circ $$

Answer: 41.41 degrees

Chapter Summary

This chapter began by defining different types of angles including acute, obtuse, reflex, and right angles, laying the foundation for understanding geometric shapes. It then explored various triangles such as isosceles, equilateral, right angled, and scalene, highlighting their unique properties. The focus shifted to trigonometric ratios, where sine, cosine, tangent, cosecant, secant, and cotangent were introduced as essential tools for relating angles to side lengths in triangles. Trigonometric identities were examined next, including proofs and the important Pythagorean identities that simplify calculations. Methods to solve trigonometric equations were presented to enable finding unknown angle measures or side lengths. The chapter concluded with hyperbolic functions, explaining sinh, cosh, cosech, tanh, and sech, which extend trigonometry concepts into hyperbolic geometry. Overall, the chapter provided a comprehensive understanding of trigonometric principles and their applications in solving practical mathematical problems.