By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems involving angles and measurements with precision and ease.
Refrigeration and air conditioning systems often require precise angular measurements and calculations to design ductwork, piping, and component placement. Accurate use of trigonometric rules ensures correct interpretation of forces, angles, and dimensions in system layouts and installation. This chapter focuses on applying trigonometric rules relevant to refrigeration and air conditioning engineering, enabling technicians and engineers to solve practical problems involving angles and lengths in their work.
In refrigeration and air conditioning, understanding the relationships between sides and angles of triangles is essential for tasks such as calculating pipe slopes, duct angles, and component alignments. The primary trigonometric rules include the sine rule, cosine rule, and tangent rule, which relate angles to side lengths in triangles. Mastery of these rules allows professionals to solve unknown dimensions and angles accurately on site or during design.
The sine rule relates the lengths of sides of a triangle to the sines of its opposite angles. It is especially useful in non-right-angled triangles common in piping and ductwork layouts where direct measurement of angles or sides is difficult.
The sine rule formula is:
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$
where \(a, b, c\) are the sides opposite angles \(A, B, C\) respectively.
Example 1: A technician needs to find the length of a pipe \(a\) opposite angle \(A = 40^\circ\) in a triangular duct section where side \(b = 5 \, m\) opposite angle \(B = 60^\circ\).
Given: \(A = 40^\circ\), \(B = 60^\circ\), \(b = 5 \, m\)
$$ \frac{a}{\sin 40^\circ} = \frac{5}{\sin 60^\circ} $$
$$ a = \frac{5 \times \sin 40^\circ}{\sin 60^\circ} $$
$$ a = \frac{5 \times 0.6428}{0.8660} $$
$$ a = \frac{3.214}{0.8660} = 3.71 \, m $$
Answer: 3.71 m
Example 2: In a refrigeration system layout, a triangular duct has angles \(A = 50^\circ\), \(B = 70^\circ\), and side \(a = 4 \, m\). Find side \(b\).
Given: \(A=50^\circ\), \(B=70^\circ\), \(a=4 \, m\)
$$ \frac{4}{\sin 50^\circ} = \frac{b}{\sin 70^\circ} $$
$$ b = \frac{4 \times \sin 70^\circ}{\sin 50^\circ} $$
$$ b = \frac{4 \times 0.9397}{0.7660} $$
$$ b = \frac{3.7588}{0.7660} = 4.91 \, m $$
Answer: 4.91 m
Example 3: A triangular support frame for an air conditioning unit has sides \(a = 6 \, m\), \(b = 8 \, m\), and angle \(A = 45^\circ\). Find angle \(B\).
Given: \(a=6 \, m\), \(b=8 \, m\), \(A=45^\circ\)
$$ \frac{6}{\sin 45^\circ} = \frac{8}{\sin B} $$
$$ \sin B = \frac{8 \times \sin 45^\circ}{6} $$
$$ \sin B = \frac{8 \times 0.7071}{6} = \frac{5.6568}{6} = 0.9428 $$
$$ B = \sin^{-1} 0.9428 = 70.6^\circ $$
Answer: 70.6°
The cosine rule is used to calculate unknown sides or angles in any triangle, especially when two sides and the included angle are known. This is common in determining pipe lengths and clearance distances where right angles do not apply.
The cosine rule formula for side \(c\) is:
$$ c^2 = a^2 + b^2 - 2ab \cos C $$
To find angle \(C\):
$$ \cos C = \frac{a^2 + b^2 - c^2}{2ab} $$
Example 1: Calculate the length of duct \(c\) when sides \(a = 3 \, m\), \(b = 4 \, m\), and included angle \(C = 60^\circ\).
Given: \(a=3 \, m\), \(b=4 \, m\), \(C=60^\circ\)
$$ c^2 = 3^2 + 4^2 - 2 \times 3 \times 4 \times \cos 60^\circ $$
$$ c^2 = 9 + 16 - 24 \times 0.5 = 25 - 12 = 13 $$
$$ c = \sqrt{13} = 3.61 \, m $$
Answer: 3.61 m
Example 2: Find angle \(C\) when sides \(a=5 \, m\), \(b=7 \, m\), and \(c=8 \, m\).
Given: \(a=5 \, m\), \(b=7 \, m\), \(c=8 \, m\)
$$ \cos C = \frac{5^2 + 7^2 - 8^2}{2 \times 5 \times 7} $$
$$ \cos C = \frac{25 + 49 - 64}{70} = \frac{10}{70} = 0.1429 $$
$$ C = \cos^{-1} 0.1429 = 81.8^\circ $$
Answer: 81.8°
Example 3: A refrigeration pipe forms a triangle with sides \(a=6 \, m\), \(b=9 \, m\), and angle \(C=120^\circ\). Find side \(c\).
Given: \(a=6 \, m\), \(b=9 \, m\), \(C=120^\circ\)
$$ c^2 = 6^2 + 9^2 - 2 \times 6 \times 9 \times \cos 120^\circ $$
$$ c^2 = 36 + 81 - 108 \times (-0.5) = 117 + 54 = 171 $$
$$ c = \sqrt{171} = 13.08 \, m $$
Answer: 13.08 m
The tangent rule helps determine unknown angles or sides in triangles, particularly useful in calculating slopes and angles of ductwork or piping runs.
The tangent rule for angles \(A\) and \(B\) is:
$$ \frac{a, b}{a + b} = \frac{\tan \frac{1}{2}(A - B)}{\tan \frac{1}{2}(A + B)} $$
Alternatively, it can be used to find angles when sides are known, but it is less commonly used than sine and cosine rules in refrigeration applications.
Example 1: Given sides \(a=7 \, m\), \(b=5 \, m\), and angle \(A=70^\circ\), find angle \(B\) using the tangent rule.
Given: \(a=7 \, m\), \(b=5 \, m\), \(A=70^\circ\)
Apply the formula:
$$ \frac{7-5}{7+5} = \frac{\tan \frac{1}{2}(70^\circ - B)}{\tan \frac{1}{2}(70^\circ + B)} $$
$$ \frac{2}{12} = \frac{\tan \frac{1}{2}(70^\circ - B)}{\tan \frac{1}{2}(70^\circ + B)} $$
$$ \frac{1}{6} = \frac{\tan \frac{1}{2}(70^\circ - B)}{\tan \frac{1}{2}(70^\circ + B)} $$
Solving this requires iterative or graphical methods, but for practical purposes, approximate \(B\) by trial:
Try \(B=50^\circ\):
$$ \frac{\tan \frac{1}{2}(70-50)}{\tan \frac{1}{2}(70+50)} = \frac{\tan 10^\circ}{\tan 60^\circ} = \frac{0.1763}{1.732} = 0.1017 $$
Close to \(1/6 = 0.1667\), so try \(B=40^\circ\):
$$ \frac{\tan 15^\circ}{\tan 55^\circ} = \frac{0.2679}{1.428} = 0.1876 $$
Slightly over, so \(B \approx 45^\circ\) is close.
Answer: Approximate \(B = 45^\circ\)
Example 2: For sides \(a=10 \, m\), \(b=6 \, m\), and angle \(A=80^\circ\), estimate angle \(B\).
Given: \(a=10 \, m\), \(b=6 \, m\), \(A=80^\circ\)
Calculate:
$$ \frac{10-6}{10+6} = \frac{4}{16} = 0.25 $$
Try \(B=50^\circ\):
$$ \frac{\tan 15^\circ}{\tan 65^\circ} = \frac{0.2679}{2.1445} = 0.125 $$
Try \(B=40^\circ\):
$$ \frac{\tan 20^\circ}{\tan 60^\circ} = \frac{0.3640}{1.732} = 0.210 $$
Try \(B=35^\circ\):
$$ \frac{\tan 22.5^\circ}{\tan 57.5^\circ} = \frac{0.4142}{1.551} = 0.267 $$
Close to 0.25. So \(B \approx 36^\circ\).
Answer: Approximate \(B = 36^\circ\)
Example 3: In a triangular duct layout with sides \(a=8 \, m\), \(b=6 \, m\), and angle \(A=60^\circ\), find angle \(B\).
Given: \(a=8 \, m\), \(b=6 \, m\), \(A=60^\circ\)
Calculate:
$$ \frac{8-6}{8+6} = \frac{2}{14} = 0.1429 $$
Try \(B=40^\circ\):
$$ \frac{\tan 10^\circ}{\tan 50^\circ} = \frac{0.1763}{1.1918} = 0.148 $$
Close to 0.1429, so \(B \approx 39^\circ\).
Answer: Approximate \(B = 39^\circ\)
In a triangular refrigeration duct, angle \(A=45^\circ\), side \(a=6 \, m\), and side \(b=8 \, m\). Find angle \(B\) using the sine rule. (5 marks)
Calculate side \(c\) of a triangular pipe run where sides \(a=7 \, m\), \(b=9 \, m\), and included angle \(C=110^\circ\) using the cosine rule. (5 marks)
Using the tangent rule, estimate angle \(B\) in a triangle with sides \(a=9 \, m\), \(b=7 \, m\), and angle \(A=75^\circ\). (5 marks)
A triangular support frame has sides \(a=5 \, m\), \(b=12 \, m\), and angle \(C=100^\circ\). Find side \(c\) using the cosine rule. (5 marks)
Find side \(b\) in a duct triangle where angle \(A=55^\circ\), angle \(B=65^\circ\), and side \(a=10 \, m\) using the sine rule. (5 marks)
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Create a free accountThis chapter began by stating and deriving key trigonometric rules, providing the foundation for understanding relationships between angles and sides in triangles. It then covered the drawing and interpretation of trigonometric graphs, illustrating how functions such as sine, cosine, and tangent vary with angle measures. Through graphical analysis, students learned to identify important features like amplitude, period, and phase shift. The chapter progressed to practical calculations using trigonometric rules, including solving problems involving angles and distances relevant to civil engineering contexts. These calculations applied the laws of sines and cosines, as well as basic trigonometric identities, to determine unknown measurements. Overall, the chapter integrated theoretical concepts with applied problem-solving to enhance mastery of trigonometry in engineering situations.
Calculate the value of \(\sin 30^\circ\) and \(\cos 60^\circ\). (2 marks)
A refrigeration pipe is installed at an angle of \(45^\circ\) to the horizontal. If the pipe length is 5 m, calculate the vertical height from the base to the pipe's upper end. (3 marks)
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