By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems and make precise measurements in your trade.
Engineering surveyors rely heavily on trigonometry to measure distances, heights, and angles that are often inaccessible directly. This chapter focuses on applying trigonometric ratios to solve practical problems involving triangles and angles of elevation and depression encountered in land surveying tasks across Kenya. Mastery of these concepts enables surveyors to accurately determine positions, elevations, and gradients essential for mapping and construction projects.
Understanding the relationships between the sides of a triangle is fundamental in land surveying, where measuring inaccessible distances depends on trigonometric principles. Surveyors frequently use right-angled triangles formed by ground features and instrumental setups to calculate unknown sides from measured angles and known lengths.
In right-angled triangles, the hypotenuse is the longest side opposite the right angle. The Pythagorean theorem relates the hypotenuse \(c\) to the other sides \(a\) and \(b\) as:
$$c = \sqrt{a^{2} + b^{2}}$$
Example 1: A surveyor measures a horizontal distance of 40 m and a vertical height of 30 m to a point on a hill. Find the direct line-of-sight distance (hypotenuse).
Given: \(a = 40\,m\), \(b = 30\,m\)
$$c = \sqrt{40^{2} + 30^{2}}$$
$$c = \sqrt{1600 + 900}$$
$$c = \sqrt{2500}$$
$$c = 50\,m$$
Answer: 50 m
Example 2: A right-angled triangle has legs measuring 15 m and 36 m. Calculate the hypotenuse.
Given: \(a = 15\,m\), \(b = 36\,m\)
$$c = \sqrt{15^{2} + 36^{2}}$$
$$c = \sqrt{225 + 1296}$$
$$c = \sqrt{1521}$$
$$c = 39\,m$$
Answer: 39 m
Example 3: A surveyor finds a triangle with legs 7.5 m and 24 m. Determine the hypotenuse length.
Given: \(a = 7.5\,m\), \(b = 24\,m\)
$$c = \sqrt{7.5^{2} + 24^{2}}$$
$$c = \sqrt{56.25 + 576}$$
$$c = \sqrt{632.25}$$
$$c = 25.15\,m$$
Answer: 25.15 m
Example 4: A slope distance is unknown, but the vertical height is 18 m and the horizontal distance 80 m. Find the slope distance.
Given: \(a = 80\,m\), \(b = 18\,m\)
$$c = \sqrt{80^{2} + 18^{2}}$$
$$c = \sqrt{6400 + 324}$$
$$c = \sqrt{6724}$$
$$c = 82\,m$$
Answer: 82 m
Example 5: A surveyor records two perpendicular sides as 12.4 m and 9.6 m. Calculate the hypotenuse.
Given: \(a = 12.4\,m\), \(b = 9.6\,m\)
$$c = \sqrt{12.4^{2} + 9.6^{2}}$$
$$c = \sqrt{153.76 + 92.16}$$
$$c = \sqrt{245.92}$$
$$c = 15.68\,m$$
Answer: 15.68 m
When the hypotenuse and one leg are known, the other leg can be found by rearranging the Pythagorean theorem:
$$a = \sqrt{c^{2} - b^{2}} \quad \text{or} \quad b = \sqrt{c^{2} - a^{2}}$$
Example 1: A slope distance (hypotenuse) is 60 m, and the vertical height is 16 m. Find the horizontal distance.
Given: \(c = 60\,m\), \(b = 16\,m\)
$$a = \sqrt{60^{2} - 16^{2}}$$
$$a = \sqrt{3600 - 256}$$
$$a = \sqrt{3344}$$
$$a = 57.82\,m$$
Answer: 57.82 m
Example 2: The hypotenuse of a triangle is 25 m, and one leg is 7 m. Find the other leg.
Given: \(c = 25\,m\), \(a = 7\,m\)
$$b = \sqrt{25^{2} - 7^{2}}$$
$$b = \sqrt{625 - 49}$$
$$b = \sqrt{576}$$
$$b = 24\,m$$
Answer: 24 m
Example 3: A surveyor measures slope distance 50 m and horizontal distance 48 m. Find the vertical height.
Given: \(c = 50\,m\), \(a = 48\,m\)
$$b = \sqrt{50^{2} - 48^{2}}$$
$$b = \sqrt{2500 - 2304}$$
$$b = \sqrt{196}$$
$$b = 14\,m$$
Answer: 14 m
Example 4: A surveyor measures the hypotenuse as 100 m and one leg as 60 m. Calculate the missing leg.
Given: \(c = 100\,m\), \(b = 60\,m\)
$$a = \sqrt{100^{2} - 60^{2}}$$
$$a = \sqrt{10000 - 3600}$$
$$a = \sqrt{6400}$$
$$a = 80\,m$$
Answer: 80 m
Example 5: The hypotenuse is 13 m, and one leg is 12 m. Find the other leg.
Given: \(c = 13\,m\), \(a = 12\,m\)
$$b = \sqrt{13^{2} - 12^{2}}$$
$$b = \sqrt{169 - 144}$$
$$b = \sqrt{25}$$
$$b = 5\,m$$
Answer: 5 m
Surveyors use the known ratios of triangle sides to calculate distances indirectly when direct measurement is impossible. This method is particularly useful in river crossing measurements or inaccessible terrain.
Example 1: A surveyor measures two sides of a triangle as 25 m and 40 m. Find the ratio of the shorter side to the longer side.
Given: \(a = 25\,m\), \(b = 40\,m\)
Ratio \(= \frac{25}{40} = 0.625\)
Answer: 0.625
Example 2: A right triangle has sides 9 m and 12 m. Find the hypotenuse and express the side ratios.
Given: \(a = 9\,m\), \(b = 12\,m\)
$$c = \sqrt{9^{2} + 12^{2}} = \sqrt{81 + 144} = \sqrt{225} = 15\,m$$
Ratios:
\(\frac{a}{c} = \frac{9}{15} = 0.6\)
\(\frac{b}{c} = \frac{12}{15} = 0.8\)
Answer: Hypotenuse = 15 m, ratios 0.6 and 0.8
Example 3: Calculate the ratio of sides for a triangle with legs 5 m and 12 m.
Given: \(a = 5\,m\), \(b = 12\,m\)
$$c = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13\,m$$
Ratios:
\(\frac{a}{c} = \frac{5}{13} \approx 0.385\)
\(\frac{b}{c} = \frac{12}{13} \approx 0.923\)
Answer: Hypotenuse = 13 m, ratios approximately 0.385 and 0.923
Example 4: A triangle has sides 8 m and 15 m. Find the hypotenuse and side ratios.
Given: \(a = 8\,m\), \(b = 15\,m\)
$$c = \sqrt{8^{2} + 15^{2}} = \sqrt{64 + 225} = \sqrt{289} = 17\,m$$
Ratios:
\(\frac{a}{c} = \frac{8}{17} \approx 0.471\)
\(\frac{b}{c} = \frac{15}{17} \approx 0.882\)
Answer: Hypotenuse = 17 m, ratios approximately 0.471 and 0.882
Example 5: Find the hypotenuse and ratios for a triangle with legs 3 m and 4 m.
Given: \(a = 3\,m\), \(b = 4\,m\)
$$c = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5\,m$$
Ratios:
\(\frac{a}{c} = \frac{3}{5} = 0.6\)
\(\frac{b}{c} = \frac{4}{5} = 0.8\)
Answer: Hypotenuse = 5 m, ratios 0.6 and 0.8
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Create a free accountThis chapter focused on applying trigonometry to solve problems involving triangles. It began with determining the lengths of sides in triangles using given angles and trigonometric principles. The concept of angles of elevation and depression was introduced to relate horizontal and vertical distances through trigonometric ratios. The chapter then explored the three primary trigonometric ratios: sine, cosine, and tangent, explaining how each ratio connects the sides of a right-angled triangle to its angles. Finally, the chapter covered methods for calculating the area of triangles using trigonometric rules, enabling the determination of area when certain sides and angles are known. Throughout, practical steps and calculations were emphasized to build proficiency in solving real-world engineering problems.
A surveyor measures a horizontal distance of 50 m between two points. If the angle of elevation from the lower point to the top of a hill is \(30^\circ\), calculate the height of the hill. (2 marks)
In a land parcel, a triangle has two sides measuring 40 m and 60 m with an included angle of \(45^\circ\). Calculate the length of the third side. (3 marks)
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