Refrigeration and air conditioning professionals often encounter practical problems involving angles, distances, and forces that require precise calculation. Trigonometry and hyperbolic functions are essential mathematical tools for analyzing and solving these problems, such as determining pipe slopes, duct angles, and stress in curved components. This chapter focuses on the fundamental trigonometric rules, their derivations, and applications within refrigeration and air conditioning contexts in Kenya.

1.1 Trigonometric Rules Are Stated and Derived

Trigonometric rules provide relationships between the sides and angles of triangles, enabling calculation of unknown dimensions critical in refrigeration and air conditioning system installations. Understanding these rules helps professionals accurately design piping layouts, ductwork, and structural supports. The main trigonometric rules include the sine rule, cosine rule, and tangent rule, which apply to both right-angled and oblique triangles.

1.1.1 The Sine Rule

The sine rule relates the ratios of sides of a triangle to the sines of their opposite angles. It is applicable to any triangle, making it useful for complex duct layouts or pipe runs where angles and lengths vary.

The sine rule states:

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

where \(a, b, c\) are the lengths of the sides and \(A, B, C\) are the opposite angles.

Worked Examples

Example 1: A refrigeration pipe forms a triangle with sides \(a = 4\,m\), angle \(A = 30^\circ\), and angle \(B = 45^\circ\). Find side \(b\).

Given: \(a = 4\,m\), \(A = 30^\circ\), \(B = 45^\circ\)

Using the sine rule:

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

$$\frac{4}{\sin 30^\circ} = \frac{b}{\sin 45^\circ}$$

Calculate sines:

$$\sin 30^\circ = 0.5, \quad \sin 45^\circ = 0.7071$$

Substitute:

$$\frac{4}{0.5} = \frac{b}{0.7071}$$

$$8 = \frac{b}{0.7071}$$

Multiply both sides by 0.7071:

$$b = 8 \times 0.7071$$

$$b = 5.6568\,m$$

Answer: \(b = 5.66\,m\)

Example 2: In a ductwork layout, angles \(A = 60^\circ\), \(B = 50^\circ\), and side \(a = 3.5\,m\) are known. Calculate side \(b\).

Given: \(A = 60^\circ\), \(B = 50^\circ\), \(a = 3.5\,m\)

First find angle \(C\):

$$C = 180^\circ - (A + B) = 180^\circ - (60^\circ + 50^\circ) = 70^\circ$$

Apply sine rule:

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

$$\frac{3.5}{\sin 60^\circ} = \frac{b}{\sin 50^\circ}$$

Calculate sines:

$$\sin 60^\circ = 0.8660, \quad \sin 50^\circ = 0.7660$$

Substitute:

$$\frac{3.5}{0.8660} = \frac{b}{0.7660}$$

$$4.0407 = \frac{b}{0.7660}$$

Multiply both sides by 0.7660:

$$b = 4.0407 \times 0.7660$$

$$b = 3.095\,m$$

Answer: \(b = 3.10\,m\)

Example 3: A triangle has sides \(a = 5\,m\), \(b = 7\,m\), and angle \(A = 40^\circ\). Find angle \(B\).

Given: \(a = 5\,m\), \(b = 7\,m\), \(A = 40^\circ\)

Using sine rule:

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

Rearranged to find \(\sin B\):

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

Calculate \(\sin A\):

$$\sin 40^\circ = 0.6428$$

Substitute:

$$\sin B = \frac{7 \times 0.6428}{5}$$

$$\sin B = \frac{4.4996}{5} = 0.8999$$

Find angle \(B\):

$$B = \sin^{-1} 0.8999 = 64.16^\circ$$

Answer: \(B = 64.16^\circ\)

Example 4: In an air conditioning duct triangle, side \(a = 6\,m\), angle \(A = 35^\circ\), and angle \(C = 75^\circ\). Find side \(c\).

Given: \(a = 6\,m\), \(A = 35^\circ\), \(C = 75^\circ\)

Find angle \(B\):

$$B = 180^\circ - (A + C) = 180^\circ - (35^\circ + 75^\circ) = 70^\circ$$

Apply sine rule:

$$\frac{a}{\sin A} = \frac{c}{\sin C}$$

$$\frac{6}{\sin 35^\circ} = \frac{c}{\sin 75^\circ}$$

Calculate sines:

$$\sin 35^\circ = 0.574, \quad \sin 75^\circ = 0.9659$$

Substitute:

$$\frac{6}{0.574} = \frac{c}{0.9659}$$

$$10.45 = \frac{c}{0.9659}$$

Multiply both sides by 0.9659:

$$c = 10.45 \times 0.9659$$

$$c = 10.10\,m$$

Answer: \(c = 10.10\,m\)

Example 5: A refrigeration pipe triangle has sides \(b = 8\,m\), \(c = 10\,m\), and angle \(B = 50^\circ\). Find angle \(C\).

Given: \(b = 8\,m\), \(c = 10\,m\), \(B = 50^\circ\)

Using sine rule:

$$\frac{b}{\sin B} = \frac{c}{\sin C}$$

Rearranged to find \(\sin C\):

$$\sin C = \frac{c \sin B}{b}$$

Calculate \(\sin B\):

$$\sin 50^\circ = 0.7660$$

Substitute:

$$\sin C = \frac{10 \times 0.7660}{8} = \frac{7.66}{8} = 0.9575$$

Find angle \(C\):

$$C = \sin^{-1} 0.9575 = 73.5^\circ$$

Answer: \(C = 73.5^\circ\)

1.1.2 The Cosine Rule

The cosine rule calculates a side or an angle in any triangle when two sides and the included angle or all three sides are known. It is particularly useful in refrigeration pipework where direct measurement is difficult but lengths and angles are partially known.

The cosine rule is:

$$c^{2} = a^{2} + b^{2} - 2ab \cos C$$

where \(a, b, c\) are sides and \(C\) is the included angle opposite side \(c\).

Worked Examples

Example 1: Find side \(c\) of a triangle with sides \(a = 5\,m\), \(b = 7\,m\), and included angle \(C = 60^\circ\).

Given: \(a = 5\,m\), \(b = 7\,m\), \(C = 60^\circ\)

Formula:

$$c^{2} = a^{2} + b^{2} - 2ab \cos C$$

Calculate \(\cos 60^\circ = 0.5\)

Substitute:

$$c^{2} = 5^{2} + 7^{2} - 2 \times 5 \times 7 \times 0.5$$

$$c^{2} = 25 + 49 - 35$$

$$c^{2} = 39$$

$$c = \sqrt{39} = 6.2449\,m$$

Answer: \(c = 6.24\,m\)

Example 2: In an air conditioning duct triangle, sides \(a = 8\,m\), \(b = 6\,m\), and side \(c = 10\,m\). Find angle \(C\).

Given: \(a = 8\,m\), \(b = 6\,m\), \(c = 10\,m\)

Rearranged cosine rule to find \(\cos C\):

$$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2ab}$$

Calculate squares:

$$a^{2} = 64, \quad b^{2} = 36, \quad c^{2} = 100$$

Substitute:

$$\cos C = \frac{64 + 36 - 100}{2 \times 8 \times 6} = \frac{0}{96} = 0$$

Find angle \(C\):

$$C = \cos^{-1} 0 = 90^\circ$$

Answer: \(C = 90^\circ\)

Example 3: A pipe bend forms a triangle with sides \(a = 9\,m\), \(b = 12\,m\), and included angle \(C = 120^\circ\). Find side \(c\).

Given: \(a = 9\,m\), \(b = 12\,m\), \(C = 120^\circ\)

Calculate \(\cos 120^\circ = -0.5\)

Apply formula:

$$c^{2} = 9^{2} + 12^{2} - 2 \times 9 \times 12 \times (-0.5)$$

$$c^{2} = 81 + 144 + 108 = 333$$

$$c = \sqrt{333} = 18.2483\,m$$

Answer: \(c = 18.25\,m\)

Example 4: Find angle \(C\) in a triangle with sides \(a = 7\,m\), \(b = 9\,m\), and \(c = 11\,m\).

Given: \(a = 7\,m\), \(b = 9\,m\), \(c = 11\,m\)

Calculate \(\cos C\):

$$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2ab} = \frac{49 + 81 - 121}{2 \times 7 \times 9} = \frac{9}{126} = 0.07143$$

Calculate angle:

$$C = \cos^{-1} 0.07143 = 85.9^\circ$$

Answer: \(C = 85.9^\circ\)

Example 5: For a refrigeration pipe triangle with sides \(a = 10\,m\), \(b = 6\,m\), and angle \(C = 45^\circ\), find side \(c\).

Given: \(a = 10\,m\), \(b = 6\,m\), \(C = 45^\circ\)

Calculate \(\cos 45^\circ = 0.7071\)

Apply formula:

$$c^{2} = 10^{2} + 6^{2} - 2 \times 10 \times 6 \times 0.7071$$

$$c^{2} = 100 + 36 - 84.852 = 51.148$$

$$c = \sqrt{51.148} = 7.152\,m$$

Answer: \(c = 7.15\,m\)

1.1.3 The Tangent Rule and Angle Sum Rule

The tangent rule relates the difference and sum of two sides of a triangle to the tangent of half the difference and half the sum of the opposite angles. It is useful in determining angles when two sides and one angle are known, especially in pipe slope adjustments or duct angle settings.

The tangent rule states:

$$\frac{a, b}{a + b} = \frac{\tan \frac{1}{2}(A - B)}{\tan \frac{1}{2}(A + B)}$$

Additionally, the angle sum rule for triangles is:

$$A + B + C = 180^\circ$$

Worked Examples

Example 1: Given sides \(a = 9\,m\), \(b = 6\,m\), and angle \(A = 70^\circ\), find angle \(B\).

Given: \(a = 9\,m\), \(b = 6\,m\), \(A = 70^\circ\)

Apply tangent rule:

$$\frac{a, b}{a + b} = \frac{\tan \frac{1}{2}(A - B)}{\tan \frac{1}{2}(A + B)}$$

Calculate left side:

$$\frac{9 - 6}{9 + 6} = \frac{3}{15} = 0.2$$

Set:

$$0.2 = \frac{\tan \frac{1}{2}(70^\circ - B)}{\tan \frac{1}{2}(70^\circ + B)}$$

Let \(x = \frac{1}{2}(70^\circ - B)\) and \(y = \frac{1}{2}(70^\circ + B)\).

Rewrite:

$$0.2 = \frac{\tan x}{\tan y} \Rightarrow \tan x = 0.2 \tan y$$

Since \(x + y = 70^\circ\), try values iteratively or use trial and error:

Try \(B = 40^\circ\):

• \(x = \frac{1}{2}(70 - 40) = 15^\circ\), \(\tan 15^\circ = 0.2679\)
• \(y = \frac{1}{2}(70 + 40) = 55^\circ\), \(\tan 55^\circ = 1.4281\)
• Ratio: \(0.2679 / 1.4281 = 0.1876\) close to 0.2

Try \(B = 38^\circ\):

• \(x = 16^\circ\), \(\tan 16^\circ = 0.2867\)
• \(y = 54^\circ\), \(\tan 54^\circ = 1.3764\)
• Ratio: \(0.2867 / 1.3764 = 0.2083\) slightly above 0.2

Interpolate between 38 and 40:

Estimate \(B \approx 39^\circ\).

Answer: \(B \approx 39^\circ\)

Example 2: In an HVAC duct triangle, sides \(a = 7\,m\), \(b = 5\,m\), and angle \(A = 60^\circ\). Find angle \(B\).

Given: \(a = 7\,m\), \(b = 5\,m\), \(A = 60^\circ\)

Calculate left side:

$$\frac{7 - 5}{7 + 5} = \frac{2}{12} = 0.1667$$

Set:

$$0.1667 = \frac{\tan \frac{1}{2}(60^\circ - B)}{\tan \frac{1}{2}(60^\circ + B)}$$

Try \(B = 40^\circ\):

• \(x = 10^\circ\), \(\tan 10^\circ = 0.1763\)
• \(y = 50^\circ\), \(\tan 50^\circ = 1.1918\)
• Ratio: \(0.1763 / 1.1918 = 0.1479\) less than 0.1667

Try \(B = 37^\circ\):

• \(x = 11.5^\circ\), \(\tan 11.5^\circ = 0.2032\)
• \(y = 48.5^\circ\), \(\tan 48.5^\circ = 1.1270\)
• Ratio: \(0.2032 / 1.1270 = 0.1803\) greater than 0.1667

Interpolate between 37 and 40:

Estimate \(B \approx 38.5^\circ\).

Answer: \(B \approx 38.5^\circ\)

Example 3: For a pipe triangle with sides \(a = 10\,m\), \(b = 8\,m\), and angle \(A = 75^\circ\), find angle \(B\).

Given: \(a = 10\,m\), \(b = 8\,m\), \(A = 75^\circ\)

Calculate:

$$\frac{10 - 8}{10 + 8} = \frac{2}{18} = 0.1111$$

Set:

$$0.1111 = \frac{\tan \frac{1}{2}(75^\circ - B)}{\tan \frac{1}{2}(75^\circ + B)}$$

Try \(B = 60^\circ\):

• \(x = 7.5^\circ\), \(\tan 7.5^\circ = 0.1317\)
• \(y = 67.5^\circ\), \(\tan 67.5^\circ = 2.4142\)
• Ratio: \(0.1317 / 2.4142 = 0.0546\) less than 0.1111

Try \(B = 50^\circ\):

• \(x = 12.5^\circ\), \(\tan 12.5^\circ = 0.2217\)
• \(y = 62.5^\circ\), \(\tan 62.5^\circ = 1.9219\)
• Ratio: \(0.2217 / 1.9219 = 0.1153\) close to 0.1111

Estimate \(B \approx 49^\circ\).

Answer: \(B \approx 49^\circ\)

Example 4: Given sides \(a = 12\,m\), \(b = 9\,m\), and angle \(A = 80^\circ\), find angle \(B\).

Given: \(a = 12\,m\), \(b = 9\,m\), \(A = 80^\circ\)

Calculate:

$$\frac{12 - 9}{12 + 9} = \frac{3}{21} = 0.1429$$

Try \(B = 50^\circ\):

• \(x = 15^\circ\), \(\tan 15^\circ = 0.2679\)
• \(y = 65^\circ\), \(\tan 65^\circ = 2.1445\)
• Ratio: \(0.2679 / 2.1445 = 0.125\) less than 0.1429

Try \(B = 45^\circ\):

• \(x = 17.5^\circ\), \(\tan 17.5^\circ = 0.3153\)
• \(y = 62.5^\circ\), \(\tan 62.5^\circ = 1.9219\)
• Ratio: \(0.3153 / 1.9219 = 0.1641\) greater than 0.1429

Estimate \(B \approx 47^\circ\).

Answer: \(B \approx 47^\circ\)

Example 5: Find angle \(B\) when \(a = 15\,m\), \(b = 12\,m\), and \(A = 85^\circ\).

Given: \(a = 15\,m\), \(b = 12\,m\), \(A = 85^\circ\)

Calculate:

$$\frac{15 - 12}{15 + 12} = \frac{3}{27} = 0.1111$$

Try \(B = 50^\circ\):

• \(x = 17.5^\circ\), \(\tan 17.5^\circ = 0.3153\)
• \(y = 67.5^\circ\), \(\tan 67.5^\circ = 2.4142\)
• Ratio: \(0.3153 / 2.4142 = 0.1306\) greater than 0.1111

Try \(B = 55^\circ\):

• \(x = 15^\circ\), \(\tan 15^\circ = 0.2679\)
• \(y = 70^\circ\), \(\tan 70^\circ = 2.7475\)
• Ratio: \(0.2679 / 2.7475 = 0.0975\) less than 0.1111

Estimate \(B \approx 53^\circ\).

Answer: \(B \approx 53^\circ\)

Chapter Summary

This chapter introduced the fundamental trigonometric rules, including the sine, cosine, and tangent rules, and provided detailed derivations to explain their mathematical basis. Students learned how to apply these rules to solve problems involving triangles in engineering contexts. The chapter also covered the plotting and interpretation of trigonometric graphs, such as sine, cosine, and tangent curves, highlighting their periodic nature and key characteristics like amplitude and period. Practical calculations using the trigonometric rules were demonstrated through step-by-step worked examples, emphasizing accuracy in solving real-world engineering problems. The integration of these concepts enables students to analyze angles, distances, and forces effectively in various technical scenarios. Mastery of these skills is essential for advanced studies in engineering mathematics and practical applications in the field.