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Some Applications of Trigonometry – Class 10 Mathematics notes, important questions and MCQs
NCERT Class 10 Mathematics, chapter 9. Notes, key terms, important questions with answers and MCQs.
Chapter overview
Trigonometry finds heights and distances that cannot be measured directly, using angles of elevation and depression.
Angles
- Angle of elevation: angle above the horizontal when looking up at an object.
- Angle of depression: angle below the horizontal when looking down at an object.
- Angle of elevation from A to B equals the angle of depression from B to A (alternate angles).
Method
- Draw a right triangle with the height as the perpendicular and the distance as the base.
- Use tan θ = height/distance most often.
- Use sin or cos when the line of sight (hypotenuse) is involved.
Key terms
- Line of sight
- the line from the observer's eye to the object
- Angle of elevation
- angle between the horizontal and the line of sight to a higher object
- Angle of depression
- angle between the horizontal and the line of sight to a lower object
Formulas
- tan θ = height / horizontal distance
- Values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3
Important questions with answers
A ladder makes 60° with the ground and its foot is 2 m from the wall. Find its length.2 marks
cos 60° = 2/L, so L = 2/(1/2) = 4 m.
The angle of elevation of the top of a tower from a point 30 m from its foot is 30°. Find the height.3 marks
tan 30° = h/30, so h = 30/√3 = 10√3 ≈ 17.32 m.
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.5 marks
Let the distance between building and tower be x. Angle of depression 45° to the foot: tan 45° = 7/x, so x = 7 m. Let the tower rise h metres above the building top: tan 60° = h/7, so h = 7√3. Tower height = 7 + 7√3 = 7(1 + √3) ≈ 19.12 m.
Quick revision
- Elevation looks up; depression looks down.
- Most problems need tan.
- Draw the figure first — marks are given for it.
Exam tips
- Add the observer's height when the angle is measured from the eye level, not the ground.
- Use √3 ≈ 1.732 only at the final step.
Practice MCQs
The angle of elevation of the sun when a pole's shadow equals its height is
- 30°
- 45°
- 60°
- 90°
Show answer
Answer: (b) 45°. tan θ = 1 ⇒ θ = 45°.
Looking down at a boat from a cliff, the angle formed is called
- elevation
- reflex
- depression
- right
Show answer
Answer: (c) depression. It is below the horizontal.
A tower 30 m high is seen at 45° from a point. The point is how far from the foot?
- 10√3 m
- 15 m
- 30√3 m
- 30 m
Show answer
Answer: (d) 30 m. tan 45° = 30/d ⇒ d = 30 m.
A kite string 100 m long makes 60° with the ground. Height of the kite is
- 50 m
- 50√3 m
- 100√3 m
- 100 m
Show answer
Answer: (b) 50√3 m. sin 60° = h/100 ⇒ h = 50√3 m.
A tower subtends 30° at a point 30 m from its foot. Its height is
- 10√3 m
- 30√3 m
- 15 m
- 10 m
Show answer
Answer: (a) 10√3 m. h = 30 × tan 30° = 30/√3 = 10√3 m.
Angle of elevation of a tower top from 20√3 m away is 30°. Height is
- 10 m
- 60 m
- 20√3 m
- 20 m
Show answer
Answer: (d) 20 m. h = 20√3 × 1/√3 = 20 m.
Shadow of a tower is 40 m longer when the sun's altitude changes from 60° to 30°. The height is
- 40 m
- 40√3 m
- 20 m
- 20√3 m
Show answer
Answer: (d) 20√3 m. h(√3 − 1/√3) = 40 ⇒ h × 2/√3 = 40 ⇒ h = 20√3 m.
From the top of a 75 m lighthouse, angles of depression of two ships on the same side are 30° and 45°. Distance between the ships is
- 75(√3 − 1) m
- 75√3 m
- 75 m
- 75(√3 + 1) m
Show answer
Answer: (a) 75(√3 − 1) m. Distances 75√3 and 75, difference 75(√3 − 1) m.
All chapters of Class 10 Mathematics
Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.