Class 10 Maths Chapter 9: Applications of Trigonometry

Some Applications of Trigonometry is Chapter 9 of NCERT Class 10 Maths. It shows how the trigonometric ratios you learnt in the previous chapter can be used to find heights of towers, buildings and trees, and distances between objects, without actually measuring them. Watch the four video lessons below, then use these notes and solved examples to practise heights and distances questions for the board exam.

What you will learn

• The meaning of line of sight, angle of elevation and angle of depression
• How to turn a word problem into a neat right-angled triangle diagram
• Which ratio to pick: sin, cos or tan, depending on the sides given
• Solving two-position problems where the observer moves towards or away from an object
• Problems with two objects, such as a building and a tower seen from each other

Video Lessons

Applications of Trigonometry - Part 1

Applications of Trigonometry - Part 2

Applications of Trigonometry - Part 3

Applications of Trigonometry - Part 4

Key Concepts and Formulas

Line of Sight and Angle of Elevation

The line of sight is the straight line from the eye of the observer to the object being viewed. When the object is above the horizontal level of the eye, you have to raise your head to see it.

• Angle of elevation: the angle between the line of sight and the horizontal, when the object is above the eye level.
• The horizontal line is drawn through the eye of the observer, not along the ground, unless the observer's height is ignored.
• If the observer's height is given (for example 1.5 m), add it to the height found from the triangle.

Angle of Depression

When the object is below the horizontal level of the eye, you have to lower your head to see it. This happens when you look down from the top of a building, cliff or lighthouse.

• Angle of depression: the angle between the line of sight and the horizontal, when the object is below the eye level.
• The angle of depression from A to B equals the angle of elevation from B to A, because they are alternate angles between parallel horizontal lines.
• Always shift the angle of depression to the ground point using alternate angles before solving the triangle.

Choosing the Right Ratio

In almost every question the height is the perpendicular and the horizontal distance is the base of a right triangle. Choose the ratio that connects the side you know with the side you want.

tan θ = height ÷ horizontal distance
sin θ = height ÷ slant length
cos θ = horizontal distance ÷ slant length

• tan θ is used most often; sin θ is useful for ladders, kite strings and ropes.

Standard Values You Must Remember

Questions in the CBSE board exam mostly use 30°, 45° and 60°, so learn these values by heart.

tan 30° = 1/√3    tan 45° = 1      tan 60° = √3
sin 30° = ½       sin 45° = 1/√2   sin 60° = √3/2
cos 30° = √3/2    cos 45° = 1/√2   cos 60° = ½
√3 ≈ 1.732        √2 ≈ 1.414

• Use the decimal values of √3 and √2 only when the question asks for a decimal answer.

Step-by-Step Method

A neat diagram is half the solution in this chapter. Follow the same routine for every question.

• Draw the figure, mark the right angle, the given angle and the known length.
• Name the unknown side as h or x.
• Write one trigonometric equation for each right triangle in the figure.
• In two-triangle problems, solve the equations together by eliminating the common unknown.

Solved Examples

Example 1: From a point on the ground 25 m away from the foot of a tower, the angle of elevation of its top is 60°. Find the height of the tower.

tan 60° = h / 25
h = 25 × √3
h = 25 × 1.732
∴ Height = 25√3 m ≈ 43.3 m

Example 2: A 12 m long ladder leans against a wall and makes an angle of 60° with the ground. How high up the wall does it reach?

sin 60° = h / 12
h = 12 × √3/2
∴ h = 6√3 m ≈ 10.39 m

Example 3: From the top of a 50 m high lighthouse, the angle of depression of a boat is 30°. How far is the boat from the base of the lighthouse?

Angle of elevation from boat = 30° (alternate angles)
tan 30° = 50 / d
1/√3 = 50 / d
d = 50√3
∴ Distance ≈ 86.6 m

Example 4: A vertical pole of height 8√3 m casts a shadow 8 m long. Find the Sun's altitude.

tan θ = 8√3 / 8
tan θ = √3
∴ θ = 60°

Example 5: The angle of elevation of the top of a tower changes from 30° to 60° when an observer walks 40 m towards it. Find the height of the tower.

Let nearer distance = x, height = h
tan 60° = h / x       →  h = x√3
tan 30° = h / (x + 40) →  h = (x + 40)/√3
x√3 = (x + 40)/√3
3x = x + 40
x = 20 m
∴ h = 20√3 m ≈ 34.64 m

Example 6: From the top of a 12 m high building, the angle of elevation of the top of a tower is 45° and the angle of depression of its foot is 30°. Find the height of the tower.

tan 30° = 12 / d
d = 12√3 m
Height above building = d × tan 45°
Height above building = 12√3 m
Tower height = 12 + 12√3
∴ Tower height = 12(1 + √3) m ≈ 32.78 m

Common Mistakes to Avoid

• Marking the angle of depression at the observer's foot instead of moving it to the ground point using alternate angles.
• Forgetting to add the observer's own height when the eye is above the ground.
• Using sin when the slant side is not given, or tan when only the slant side and height are involved. Check which two sides you know first.
• Rounding √3 too early. Keep answers in surd form until the last step, then convert if asked.

Quick Revision

tan θ = height ÷ base
tan 30° = 1/√3, tan 45° = 1, tan 60° = √3

• Elevation: look up; depression: look down; both measured from the horizontal
• Angle of depression from A to B = angle of elevation from B to A
• In two-position problems, write two equations and eliminate the common unknown

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