Trigonometry is a fundamental mathematical tool for land surveyors in Kenya. Accurate measurement of distances and angles on uneven terrain requires applying trigonometric principles to solve problems involving triangles formed by survey lines and landmarks. This chapter focuses on practical trigonometric applications relevant to field measurements, such as determining unknown sides and angles using direct observations and calculating angles of elevation and depression, which are common in topographic surveys and construction layout.

2.1 Sides of a triangle

Determining the lengths of sides in a triangle is a frequent task in land surveying when direct measurement is impractical due to obstacles or terrain. By applying trigonometric laws and direct angular observations, surveyors can calculate unknown distances from known baseline measurements and observed angles. This section covers the formulas and methods used to compute side lengths in triangles encountered on survey sites.

2.1.1 Using the Law of Cosines to Find a Side from Two Sides and Included Angle

The Law of Cosines relates the lengths of three sides of a triangle with the cosine of one of its angles. It is essential when two sides and the included angle are known, allowing calculation of the third side.

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

Worked Examples

Example 1: A surveyor measures two sides of a triangular plot as \(a = 50\,m\) and \(b = 70\,m\), with the included angle \(C = 60^\circ\). Find the length of side \(c\).

Given: \(a = 50\,m\), \(b = 70\,m\), \(C = 60^\circ\)

$$ c^{2} = 50^{2} + 70^{2} - 2 \times 50 \times 70 \times \cos 60^\circ $$

$$ c^{2} = 2500 + 4900 - 7000 \times 0.5 $$

$$ c^{2} = 7400 - 3500 = 3900 $$

$$ c = \sqrt{3900} = 62.45\,m $$

Answer: 62.45 m

Example 2: On a hillside survey, two sides of a triangle are \(a = 120\,m\) and \(b = 85\,m\) with an included angle \(C = 45^\circ\). Calculate side \(c\).

Given: \(a = 120\,m\), \(b = 85\,m\), \(C = 45^\circ\)

$$ c^{2} = 120^{2} + 85^{2} - 2 \times 120 \times 85 \times \cos 45^\circ $$

$$ c^{2} = 14400 + 7225 - 20400 \times 0.7071 $$

$$ c^{2} = 21625 - 14445.24 = 7180 $$

$$ c = \sqrt{7180} = 84.74\,m $$

Answer: 84.74 m

Example 3: A survey triangle has sides \(a = 200\,m\), \(b = 150\,m\) and the included angle \(C = 120^\circ\). Find side \(c\).

Given: \(a = 200\,m\), \(b = 150\,m\), \(C = 120^\circ\)

$$ c^{2} = 200^{2} + 150^{2} - 2 \times 200 \times 150 \times \cos 120^\circ $$

$$ c^{2} = 40000 + 22500 - 60000 \times (-0.5) $$

$$ c^{2} = 62500 + 30000 = 92500 $$

$$ c = \sqrt{92500} = 304.13\,m $$

Answer: 304.13 m

2.1.2 Using the Law of Sines to Find a Side from One Side and Two Angles

The Law of Sines relates the sides of a triangle to the sines of their opposite angles. It is useful when one side and two angles are known.

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

Worked Examples

Example 1: In a survey triangle, side \(a = 80\,m\) is opposite angle \(A = 40^\circ\), and angle \(B = 60^\circ\). Find side \(b\).

Given: \(a = 80\,m\), \(A = 40^\circ\), \(B = 60^\circ\)

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

$$ b = a \times \frac{\sin B}{\sin A} = 80 \times \frac{\sin 60^\circ}{\sin 40^\circ} $$

$$ b = 80 \times \frac{0.8660}{0.6428} = 80 \times 1.347 = 107.76\,m $$

Answer: 107.76 m

Example 2: A triangle has side \(a = 100\,m\) opposite angle \(A = 50^\circ\), and angle \(C = 70^\circ\). Find side \(c\).

Given: \(a = 100\,m\), \(A = 50^\circ\), \(C = 70^\circ\)

First find angle \(B\):

$$ B = 180^\circ - (A + C) = 180^\circ - 120^\circ = 60^\circ $$

Apply Law of Sines:

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

$$ c = a \times \frac{\sin C}{\sin A} = 100 \times \frac{\sin 70^\circ}{\sin 50^\circ} $$

$$ c = 100 \times \frac{0.9397}{0.7660} = 100 \times 1.227 = 122.7\,m $$

Answer: 122.7 m

Example 3: In a plot survey, angle \(A = 35^\circ\), angle \(B = 65^\circ\), and side \(b = 90\,m\). Find side \(a\).

Given: \(A=35^\circ\), \(B=65^\circ\), \(b=90\,m\)

Calculate angle \(C\):

$$ C = 180^\circ - (35^\circ + 65^\circ) = 80^\circ $$

Apply Law of Sines:

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

$$ a = b \times \frac{\sin A}{\sin B} = 90 \times \frac{\sin 35^\circ}{\sin 65^\circ} $$

$$ a = 90 \times \frac{0.574}{0.906} = 90 \times 0.634 = 57.06\,m $$

Answer: 57.06 m

2.1.3 Finding Unknown Sides in Right-Angled Triangles Using Pythagoras’ Theorem

In surveying, right-angled triangles frequently occur, especially when measuring heights or horizontal distances. Pythagoras’ theorem is fundamental for calculating unknown sides when two sides are known.

$$ c^{2} = a^{2} + b^{2} $$

Worked Examples

Example 1: A surveyor measures a horizontal distance \(a = 60\,m\) and vertical height \(b = 25\,m\). Find the slope distance \(c\).

Given: \(a = 60\,m\), \(b = 25\,m\)

$$ c^{2} = 60^{2} + 25^{2} = 3600 + 625 = 4225 $$

$$ c = \sqrt{4225} = 65\,m $$

Answer: 65 m

Example 2: In a topographic survey, the slope distance \(c = 100\,m\) and vertical height \(b = 40\,m\) are measured. Find the horizontal distance \(a\).

Given: \(c = 100\,m\), \(b = 40\,m\)

$$ a^{2} = c^{2} - b^{2} = 100^{2} - 40^{2} = 10000 - 1600 = 8400 $$

$$ a = \sqrt{8400} = 91.65\,m $$

Answer: 91.65 m

Example 3: A slope distance \(c = 75\,m\) and horizontal distance \(a = 45\,m\) are observed. Calculate the vertical height \(b\).

Given: \(c = 75\,m\), \(a = 45\,m\)

$$ b^{2} = c^{2} - a^{2} = 75^{2} - 45^{2} = 5625 - 2025 = 3600 $$

$$ b = \sqrt{3600} = 60\,m $$

Answer: 60 m

2.1.4 Calculating Perimeter and Area Using Side Lengths in Survey Triangles

Once side lengths are known, perimeter and area calculations are important for land valuation and planning. Heron's formula allows area calculation from sides only.

Heron's formula:

$$ s = \frac{a + b + c}{2} $$

$$ \text{Area} = \sqrt{s (s - a)(s - b)(s - c)} $$

Worked Examples

Example 1: A triangle has sides \(a=40\,m\), \(b=50\,m\), and \(c=60\,m\). Calculate its perimeter and area.

Given: \(a=40\,m\), \(b=50\,m\), \(c=60\,m\)

$$ s = \frac{40 + 50 + 60}{2} = \frac{150}{2} = 75\,m $$

$$ \text{Area} = \sqrt{75 (75 - 40)(75 - 50)(75 - 60)} = \sqrt{75 \times 35 \times 25 \times 15} $$

$$ = \sqrt{984375} = 992.16\,m^{2} $$

Perimeter:

$$ P = 40 + 50 + 60 = 150\,m $$

Answer: Perimeter = 150 m, Area = 992.16 m²

Example 2: Calculate the area and perimeter of a triangle with sides \(a=80\,m\), \(b=100\,m\), \(c=120\,m\).

Given: \(a=80\,m\), \(b=100\,m\), \(c=120\,m\)

$$ s = \frac{80 + 100 + 120}{2} = 150\,m $$

$$ \text{Area} = \sqrt{150 (150-80)(150-100)(150-120)} = \sqrt{150 \times 70 \times 50 \times 30} $$

$$ = \sqrt{15,750,000} = 3968.63\,m^{2} $$

Perimeter:

$$ P = 80 + 100 + 120 = 300\,m $$

Answer: Perimeter = 300 m, Area = 3968.63 m²

Example 3: For a triangle with sides \(a=100\,m\), \(b=85\,m\), and \(c=75\,m\), find the perimeter and area.

Given: \(a=100\,m\), \(b=85\,m\), \(c=75\,m\)

$$ s = \frac{100 + 85 + 75}{2} = 130\,m $$

$$ \text{Area} = \sqrt{130 (130-100)(130-85)(130-75)} = \sqrt{130 \times 30 \times 45 \times 55} $$

$$ = \sqrt{9,682,500} = 3111.37\,m^{2} $$

Perimeter:

$$ P = 100 + 85 + 75 = 260\,m $$

Answer: Perimeter = 260 m, Area = 3111.37 m²

Chapter Summary

This chapter focused on applying trigonometry to solve practical engineering problems involving triangles. It began with methods to determine the sides of a triangle through direct observation and measurement techniques. The study then progressed to angles of elevation and depression, explaining how to calculate these angles using the fundamental trigonometric ratios of sine, cosine, and tangent. Each ratio was explored with examples on how they relate the sides and angles of right-angled triangles in real-world contexts. The chapter concluded by introducing the calculation of triangle areas using trigonometric rules, enabling the determination of areas when certain sides and angles are known. Together, these topics provide essential tools for analyzing and solving geometric problems encountered in engineering tasks.