Trigonometry and hyperbolic functions are crucial mathematical tools used across diverse professional fields in Kenya, from construction and architecture to healthcare technology and agricultural planning. Understanding these functions enables accurate measurement of angles, distances, and forces in practical situations such as surveying land for a cooperative farm or designing structural elements in a hospital expansion project. This chapter develops competence in applying trigonometric ratios, functions, and the unit circle concept, laying the foundation for more advanced analysis involving hyperbolic functions later in the course.

2.1 Importance of trigonometry in engineering

Trigonometry provides the mathematical foundation for analyzing and solving problems involving angles and distances in various professional applications. In Kenyan institutions such as county government offices planning infrastructure layout or banks calculating interest rate models, trigonometric methods offer precise computational tools. This section focuses on practical uses and problem-solving techniques relevant to diploma-level students across all sectors.

2.1.1 Role of trigonometry in measurement and surveying

Trigonometry allows accurate calculation of inaccessible distances and heights using angle measurements and baseline lengths. Surveyors in land management offices use trigonometric principles to determine property boundaries without physically measuring every side.

Worked Examples

Example 1: A surveyor at a county government office measures a baseline of 50 m and an angle of elevation of 30° to the top of a water tower. Find the height of the tower.

Given: Baseline \(= 50\,m\), Angle \(\theta = 30^\circ\)

$$ \text{Height} = \text{Baseline} \times \tan \theta $$

$$ = 50 \times \tan 30^\circ $$

$$ = 50 \times 0.577 $$

$$ = 28.85\,m $$

Answer: \(28.85\,m\)

Example 2: A technician at a hospital measures a 40 m distance between two points and the angle between the line of sight and the ground is 45°. Calculate the height of an antenna mounted on the hospital roof.

Given: Distance \(= 40\,m\), Angle \(\theta = 45^\circ\)

$$ \text{Height} = 40 \times \tan 45^\circ $$

$$ = 40 \times 1 $$

$$ = 40\,m $$

Answer: \(40\,m\)

Example 3: A farmer wants to fence a triangular plot. One side is 80 m, and the angles adjacent to this side are 50° and 60°. Find the length of the side opposite to the 50° angle.

Given: Side \(a=80\,m\), Angles \(A=50^\circ\), \(B=60^\circ\)

Use Law of Sines:

$$ \frac{a}{\sin A} = \frac{b}{\sin B} $$

$$ b = \frac{a \times \sin B}{\sin A} $$

$$ = \frac{80 \times \sin 60^\circ}{\sin 50^\circ} $$

$$ = \frac{80 \times 0.8660}{0.7660} $$

$$ = \frac{69.28}{0.7660} $$

$$ = 90.5\,m $$

Answer: \(90.5\,m\)

2.1.2 Application in structural design and safety analysis

Civil and structural engineers apply trigonometric functions to calculate forces, slopes, and angles in beams, roofs, and supports. For example, architects designing a hotel in Nairobi use trigonometry to determine roof pitch and load distribution.

Worked Examples

Example 1: A roof truss makes a 35° angle with the horizontal. If the horizontal span is 12 m, find the length of the sloping rafter.

Given: Horizontal span \(= 12\,m\), Angle \(\theta=35^\circ\)

$$ \text{Rafter length} = \frac{\text{Horizontal span}}{\cos \theta} $$

$$ = \frac{12}{\cos 35^\circ} $$

$$ = \frac{12}{0.8192} $$

$$ = 14.65\,m $$

Answer: \(14.65\,m\)

Example 2: A support beam is inclined at 40° to the floor and carries a vertical load of 5000 N. Calculate the component of the load along the beam.

Given: Load \(= 5000\,N\), Angle \(\theta=40^\circ\)

$$ \text{Load component} = 5000 \times \cos 40^\circ $$

$$ = 5000 \times 0.7660 $$

$$ = 3830\,N $$

Answer: \(3830\,N\)

Example 3: A hotel engineer needs to calculate the slope of a ramp that rises 1.5 m over a horizontal distance of 12 m. Find the angle of elevation.

Given: Rise \(= 1.5\,m\), Run \(= 12\,m\)

$$ \theta = \tan^{-1} (\frac{\text{Rise}}{\text{Run}}) $$

$$ = \tan^{-1} (\frac{1.5}{12}) $$

$$ = \tan^{-1} (0.125) $$

$$ = 7.13^\circ $$

Answer: \(7.13^\circ\)

2.1.3 Trigonometry in navigation and positioning

Trigonometric calculations help determine directions, bearings, and positions in mapping and GPS technology. County wildlife conservancies and farms use these methods for accurate location tracking and resource management.

Worked Examples

Example 1: A ranger at a conservancy observes an animal 200 m away at a bearing of 60° from north. Calculate the eastward and northward distances from the observer.

Given: Distance \(= 200\,m\), Bearing \(\theta=60^\circ\)

Eastward distance:

$$ 200 \times \sin 60^\circ = 200 \times 0.8660 = 173.2\,m $$

Northward distance:

$$ 200 \times \cos 60^\circ = 200 \times 0.5 = 100\,m $$

Answer: East \(173.2\,m\), North \(100\,m\)

Example 2: A surveyor at a county government office needs to find the resultant displacement of two successive movements: 100 m east and 150 m north.

Given: \(d_1=100\,m\) east, \(d_2=150\,m\) north

Resultant distance:

$$ \sqrt{100^2 + 150^2} = \sqrt{10000 + 22500} = \sqrt{32500} = 180.28\,m $$

Angle from east:

$$ \theta = \tan^{-1} (\frac{150}{100}) = \tan^{-1}(1.5) = 56.31^\circ $$

Answer: \(180.28\,m\) at \(56.31^\circ\) north of east

Example 3: A farmer walks 250 m at 30° east of north, then 300 m at 60° west of north. Find the farmer’s net displacement from the starting point.

Given:

\(d_1=250\,m, \theta_1=30^\circ\) east of north

\(d_2=300\,m, \theta_2=60^\circ\) west of north

Calculate north components:

$$ N_1 = 250 \times \cos 30^\circ = 250 \times 0.8660 = 216.5\,m $$

$$ N_2 = 300 \times \cos 60^\circ = 300 \times 0.5 = 150\,m $$

Calculate east components (east positive, west negative):

$$ E_1 = 250 \times \sin 30^\circ = 250 \times 0.5 = 125\,m $$

$$ E_2 = -300 \times \sin 60^\circ = -300 \times 0.8660 = -259.8\,m $$

Resultant north:

$$ N = 216.5 + 150 = 366.5\,m $$

Resultant east:

$$ E = 125 - 259.8 = -134.8\,m $$

Displacement magnitude:

$$ \sqrt{366.5^2 + (-134.8)^2} = \sqrt{134326 + 18176} = \sqrt{152502} = 390.5\,m $$

Direction (angle west of north):

$$ \theta = \tan^{-1} (\frac{134.8}{366.5}) = \tan^{-1}(0.368) = 20.2^\circ $$

Answer: \(390.5\,m\) at \(20.2^\circ\) west of north

Chapter Summary

This chapter explored the critical role of trigonometry in engineering by examining fundamental trigonometric ratios and functions, including sine, cosine, tangent, and their reciprocals, along with angle measurement using degrees and radians on the unit circle. It covered the graphs of trigonometric functions and introduced essential identities such as Pythagorean, reciprocal, quotient, co-function, even-odd, sum and difference, double-angle, and half-angle formulas. Techniques for solving trigonometric equations were presented, highlighting the use of basic functions, identities, and inverse trigonometric functions, with applications to periodic phenomena and mechanical oscillations. The chapter then applied trigonometry to both right-angle and non-right-angle triangles, using the Law of Sines and Law of Cosines to solve problems relevant to structural analysis and navigation. An introduction to hyperbolic functions followed, defining hyperbolic sine, cosine, tangent, and their reciprocals, alongside their graphs and properties. The study of hyperbolic identities mirrored that of trigonometric ones, including Pythagorean-like relationships and angle formulas. Finally, methods for solving hyperbolic equations were demonstrated using basic functions, identities, and inverse hyperbolic functions, with applications emphasizing their importance in engineering contexts.