In this chapter, we apply trigonometric ratios to solve real-life problems involving heights and distances. This is one of the most practical uses of trigonometry in everyday engineering, architecture, navigation, and astronomy.
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Key Concepts and Definitions
Angle of Elevation: The angle formed between the horizontal line and the line of sight when looking upward at an object. If you look up at the top of a tower, the angle your line of sight makes with the horizontal is the angle of elevation.
Angle of Depression: The angle formed between the horizontal line and the line of sight when looking downward at an object. If you look down from a cliff at a boat, the angle formed is the angle of depression.
Important Rule: The angle of elevation from point A to point B equals the angle of depression from point B to point A (alternate interior angles with a transversal cutting parallel horizontal lines).
Line of Sight: The straight line drawn from the observer's eye to the object being viewed.
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Key Formulas
- For a right triangle with angle theta at the base:
- tan theta = Height / Base (used most often)
- sin theta = Height / Hypotenuse
- cos theta = Base / Hypotenuse
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Worked Examples
A tower stands vertically on the ground. From a point 20 m away from the base, the angle of elevation of the top is 60°. Find the height of the tower.
- tan 60° = height / 20
- sqrt3 = h / 20
- h = 20 sqrt3 m (approximately 34.64 m)
A 1.5 m tall boy stands 30 m away from a building 31.5 m tall. Find the angle of elevation from the boy's eyes to the top of the building.
- Effective height = 31.5 - 1.5 = 30 m, distance = 30 m
- tan theta = 30/30 = 1
- theta = 45°
From the top of a 75 m high lighthouse, the angle of depression of a boat is 30°. Find the distance of the boat from the base of the lighthouse.
- Angle of depression = 30° => angle of elevation from boat = 30°
- tan 30° = 75 / d
- 1/sqrt3 = 75/d => d = 75 sqrt3 m
Two poles of heights 6 m and 11 m stand on a plane ground. The distance between their feet is 12 m. Find the distance between their tops.
- The vertical difference in heights = 11 - 6 = 5 m
- Horizontal distance = 12 m
- Distance between tops = √(52 + 122) = √(25 + 144) = √(169) = 13 m
A person standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 40 m away, the angle becomes 30°. Find the width of the river.
- Let width = d, height of tree = h
- From first position: tan 60° = h/d => h = d sqrt3
- From second position: tan 30° = h/(d + 40) => h = (d + 40)/sqrt3
- Setting equal: d sqrt3 = (d + 40)/sqrt3 => 3d = d + 40 => 2d = 40 => d = 20 m
From the top of a building 30 m high, the angles of depression of the top and bottom of another building are 30° and 60°. Find the height of the second building.
- Let height of second building = h, horizontal distance = x
- tan 60° = 30/x => x = 30/sqrt3 = 10 sqrt3
- tan 30° = (30 - h)/x => 1/sqrt3 = (30 - h)/(10 sqrt3)
- 10 sqrt3 / sqrt3 = 30 - h => 10 = 30 - h => h = 20 m
A ladder 10 m long makes an angle of 60° with the ground. Find the height it reaches on a wall.
- sin 60° = height / 10
- sqrt3/2 = h/10 => h = 5 sqrt3 m
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Common mistakes
> Common mistakes: Students often mix up angle of elevation and angle of depression. Always draw a clear diagram before setting up equations. Remember that when two positions are involved, there will be two equations — solve them simultaneously. Do not forget to account for the observer's eye height if mentioned.
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Summary
Applications of trigonometry help find unknown heights and distances using angles. The key tool is tan theta = opposite / adjacent. Drawing a neat diagram with all given information labelled is the most critical step in solving these problems. Most problems use 30°, 45°, or 60° standard angles.