By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world problems involving angles and curves, making you a more capable and versatile technician in your trade.
Water engineering professionals in Kenya frequently apply trigonometry and hyperbolic functions to solve practical problems such as calculating slopes for canals, determining pipe inclinations, and analyzing forces in hydraulic structures. Mastery of these mathematical tools ensures precise design and efficient operation of water systems. This chapter develops skills to perform trigonometric and hyperbolic calculations relevant to water engineering tasks, progressively building from fundamental ratios to complex applications.
Trigonometric ratios are essential in water engineering for measuring angles and distances that are not easily accessible. They help determine slopes, heights, and depths in terrain surveying, pipe gradient calculations, and structural analysis. This section covers the primary trigonometric ratios and their use in solving right-angled triangle problems.
Trigonometric ratios relate the angles of a right triangle to the ratios of its sides. The primary ratios are sine, cosine, and tangent, defined as follows for an angle \(\theta\):
$$ \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} $$
$$ \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} $$
$$ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} $$
Example 1: A water pipeline is laid on a slope forming a 30° angle with the horizontal. Find the ratio of the vertical rise to the length of the pipe.
Given: \(\theta = 30^\circ\)
$$ \sin 30^\circ = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\text{Vertical rise}}{\text{Pipe length}} $$
$$ \sin 30^\circ = 0.5 $$
Answer: The vertical rise is 0.5 times the pipe length.
Example 2: A surveyor measures the angle of elevation of a water tower as 45°. If the horizontal distance to the tower base is 40 m, find the height of the tower.
Given: \(\theta = 45^\circ\), Adjacent side = 40 m
$$ \tan 45^\circ = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\text{Height}}{40} $$
$$ \tan 45^\circ = 1 $$
$$ 1 = \frac{\text{Height}}{40} $$
$$ \text{Height} = 40 \text{ m} $$
Answer: The height of the tower is 40 m.
Example 3: A drainage channel is constructed with a hypotenuse length of 15 m and an angle of 60° with the horizontal. Calculate the vertical drop of the channel.
Given: Hypotenuse = 15 m, \(\theta = 60^\circ\)
$$ \sin 60^\circ = \frac{\text{Opposite}}{15} $$
$$ \sin 60^\circ = 0.866 $$
$$ 0.866 = \frac{\text{Opposite}}{15} $$
$$ \text{Opposite} = 0.866 \times 15 = 12.99 \text{ m} $$
Answer: The vertical drop is 12.99 m.
Example 4: A pipe runs at an angle of 25° to the horizontal with a horizontal length of 50 m. Find the actual length of the pipe.
Given: Adjacent side = 50 m, \(\theta = 25^\circ\)
$$ \cos 25^\circ = \frac{50}{\text{Hypotenuse}} $$
$$ \cos 25^\circ = 0.9063 $$
$$ 0.9063 = \frac{50}{\text{Hypotenuse}} $$
$$ \text{Hypotenuse} = \frac{50}{0.9063} = 55.18 \text{ m} $$
Answer: The length of the pipe is 55.18 m.
Example 5: Calculate the angle of inclination of a canal if its vertical drop is 8 m over a horizontal distance of 40 m.
Given: Opposite side = 8 m, Adjacent side = 40 m
$$ \tan \theta = \frac{8}{40} = 0.2 $$
$$ \theta = \tan^{-1} 0.2 = 11.31^\circ $$
Answer: The angle of inclination is 11.31°.
Water engineers often need to find unknown sides or angles in right triangles formed by terrain slopes or pipe layouts. This subtopic explains how to use trigonometric ratios to solve for missing measurements.
Example 1: A water tower is 25 m tall. The angle of elevation from a point on the ground is 53°. Find the horizontal distance from the point to the tower base.
Given: Opposite = 25 m, \(\theta = 53^\circ\)
$$ \tan 53^\circ = \frac{25}{\text{Adjacent}} $$
$$ \tan 53^\circ = 1.3270 $$
$$ 1.3270 = \frac{25}{\text{Adjacent}} $$
$$ \text{Adjacent} = \frac{25}{1.3270} = 18.84 \text{ m} $$
Answer: The horizontal distance is 18.84 m.
Example 2: An irrigation canal slopes down at 15°. If the canal length is 120 m, find the vertical drop and horizontal distance covered.
Given: Hypotenuse = 120 m, \(\theta = 15^\circ\)
Vertical drop:
$$ \sin 15^\circ = \frac{\text{Opposite}}{120} $$
$$ \sin 15^\circ = 0.2588 $$
$$ \text{Opposite} = 0.2588 \times 120 = 31.06 \text{ m} $$
Horizontal distance:
$$ \cos 15^\circ = \frac{\text{Adjacent}}{120} $$
$$ \cos 15^\circ = 0.9659 $$
$$ \text{Adjacent} = 0.9659 \times 120 = 115.91 \text{ m} $$
Answer: Vertical drop is 31.06 m and horizontal distance is 115.91 m.
Example 3: A pipe is laid at an angle of 40° and has a vertical height difference of 14 m between its ends. Find the length of the pipe.
Given: Opposite = 14 m, \(\theta = 40^\circ\)
$$ \sin 40^\circ = \frac{14}{\text{Hypotenuse}} $$
$$ \sin 40^\circ = 0.6428 $$
$$ 0.6428 = \frac{14}{\text{Hypotenuse}} $$
$$ \text{Hypotenuse} = \frac{14}{0.6428} = 21.78 \text{ m} $$
Answer: Length of the pipe is 21.78 m.
Example 4: Calculate the angle of inclination of a water slide if the slide length is 30 m and the vertical drop is 12 m.
Given: Hypotenuse = 30 m, Opposite = 12 m
$$ \sin \theta = \frac{12}{30} = 0.4 $$
$$ \theta = \sin^{-1} 0.4 = 23.58^\circ $$
Answer: The angle of inclination is 23.58°.
Example 5: A surveyor measures a slope with a horizontal distance of 80 m and an angle of elevation of 18°. Find the vertical height and slope length.
Given: Adjacent = 80 m, \(\theta = 18^\circ\)
Vertical height:
$$ \tan 18^\circ = \frac{\text{Opposite}}{80} $$
$$ \tan 18^\circ = 0.3249 $$
$$ \text{Opposite} = 0.3249 \times 80 = 25.99 \text{ m} $$
Slope length:
$$ \sec 18^\circ = \frac{\text{Hypotenuse}}{80} $$
$$ \cos 18^\circ = 0.9511 $$
$$ \sec 18^\circ = \frac{1}{0.9511} = 1.0515 $$
$$ \text{Hypotenuse} = 1.0515 \times 80 = 84.12 \text{ m} $$
Answer: Vertical height is 25.99 m, slope length is 84.12 m.
Trigonometry helps calculate slopes, flow angles, and component forces in water engineering. This subtopic focuses on practical formula applications for these tasks.
Example 1: Calculate the slope percentage of a canal that drops 5 m over a horizontal distance of 200 m.
Given: Vertical drop = 5 m, Horizontal distance = 200 m
Slope percentage formula:
$$ \text{Slope \%} = \frac{\text{Vertical drop}}{\text{Horizontal distance}} \times 100 $$
$$ = \frac{5}{200} \times 100 = 2.5\% $$
Answer: The slope percentage is 2.5%.
Example 2: A water jet is projected at an angle of 60° with the horizontal. Calculate the horizontal and vertical components of a velocity of 20 m/s.
Given: Velocity \(v = 20\) m/s, \(\theta = 60^\circ\)
Horizontal component:
$$ v_x = v \cos \theta = 20 \times \cos 60^\circ = 20 \times 0.5 = 10 \text{ m/s} $$
Vertical component:
$$ v_y = v \sin \theta = 20 \times \sin 60^\circ = 20 \times 0.866 = 17.32 \text{ m/s} $$
Answer: Horizontal velocity = 10 m/s, vertical velocity = 17.32 m/s.
Example 3: Determine the angle of flow if the horizontal velocity is 12 m/s and vertical velocity is 5 m/s.
Given: \(v_x = 12\) m/s, \(v_y = 5\) m/s
$$ \tan \theta = \frac{v_y}{v_x} = \frac{5}{12} = 0.4167 $$
$$ \theta = \tan^{-1} 0.4167 = 22.62^\circ $$
Answer: The flow angle is 22.62°.
Example 4: A retaining wall leans backward forming an angle of 75° with the horizontal. Find the horizontal and vertical components of a 5000 N force acting perpendicular to the wall.
Given: Force \(F = 5000\) N, \(\theta = 75^\circ\)
Horizontal component:
$$ F_x = F \cos 75^\circ = 5000 \times 0.2588 = 1294 \text{ N} $$
Vertical component:
$$ F_y = F \sin 75^\circ = 5000 \times 0.9659 = 4829.5 \text{ N} $$
Answer: Horizontal force = 1294 N, vertical force = 4829.5 N.
Example 5: Calculate the angle of inclination of a sluice gate if the vertical force component is 2000 N and the total force is 2500 N.
Given: \(F_y = 2000\) N, \(F = 2500\) N
$$ \sin \theta = \frac{F_y}{F} = \frac{2000}{2500} = 0.8 $$
$$ \theta = \sin^{-1} 0.8 = 53.13^\circ $$
Answer: The angle of inclination is 53.13°.
A water pipe is inclined at 35° to the horizontal. If the pipe length is 40 m, find the vertical height difference between its ends. (4 marks)
A surveyor measures a slope with a horizontal distance of 100 m and an angle of elevation of 20°. Calculate the vertical height and slope length. (5 marks)
Calculate the slope percentage of a canal that drops 7 m over a horizontal distance of 280 m. (3 marks)
A water jet is projected at 45° with a velocity of 25 m/s. Find the horizontal and vertical components of the velocity. (5 marks)
A retaining wall experiences a force of 6000 N acting at 60° to the horizontal. Calculate the horizontal and vertical components of this force. (5 marks)
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Create a free accountThis chapter covered the fundamental concepts of trigonometric ratios, introducing the basic functions used to relate angles and sides in right-angled triangles. It progressed to detailed calculations involving these ratios, enabling the solving of practical problems in various civil engineering contexts. The study then shifted focus to hyperbolic functions, starting with the definition and properties of the hyperbolic sine function, sinh \(x\). It continued by exploring the hyperbolic cosine function, cosh \(x\), and their applications through exercises. Further, the chapter examined additional hyperbolic functions including cosecant (cosec \(x\)), hyperbolic cotangent (coth \(x\)), hyperbolic tangent (tanh \(x\)), and hyperbolic secant (sech \(x\)). Finally, the chapter demonstrated how to perform calculations involving these hyperbolic functions, ensuring a comprehensive understanding of their use in solving engineering problems.
Calculate the value of \(\sin 30^\circ\) using a calculator. (1 mark)
A water pipeline is laid at an angle of \(45^\circ\) to the horizontal. If the pipeline length is 100 m, find the vertical height difference between the two ends. (2 marks)
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