SOME APPLICATIONS OF TRIGONOMETRY-MCQs

This MCQ set has been meticulously designed to strengthen conceptual understanding and examination readiness for NCERT Class X Mathematics Chapter 9, Some Applications of Trigonometry. By covering definitions, principles, diagrams, angle relationships, and real-life height-and-distance scenarios, these questions help learners internalize the logic behind trigonometric applications rather than memorizing formulas mechanically. Each question is paired with a clear explanation to reinforce reasoning skills and eliminate common misconceptions found in board exams. Whether used for classroom practice, competitive preparation, assignments, or online learning platforms, this MCQ collection provides a comprehensive, structured, and assessment-focused toolset for mastering angles of elevation, depression, and right-triangle modeling in practical measurement contexts. Students gain improved accuracy, speed, and confidence—essential for scoring high in CBSE board examinations and beyond.

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Exercise • Jan 2026

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Exercise

SOME APPLICATIONS OF TRIGONOMETRY

by Academia Aeternum

1. The angle formed between the horizontal line of sight and the upward line of sight is called:
2. If \(\tan 45^\circ = 1\), then the height of a tower is equal to the distance of the observer when the angle of elevation is:
3. The angle of depression from a cliff to a boat is formed between:
4. If the angle of elevation to the top of a pole is \(30^\circ\) and its height is \(10\text{ m}\), the horizontal distance is:
5. In height-and-distance problems, the triangle formed is always:
6. A kite is \(50\text{ m}\) high. Angle of elevation is \(60^\circ\). Distance from observer is:
7. \(\tan 30^\circ =\)
8. A tower casts a \(20\text{ m}\) shadow at \(45^\circ\). Height is:
9. Which device is used for measuring angles in surveying?
10. If angle of depression is \(30^\circ\), angle of elevation from the lower point is:
11. Man sees top of tower at \(60^\circ\). Moves \(10\text{ m}\) closer ? angle becomes \(90^\circ\). Height =
12. For height \(h\), distance \(d\), elevation angle \(\theta\):
13. Balloon height \(100\text{ m}\). Angle \(30^\circ\). Distance =
14. \(\tan 60^\circ =\)
15. Cliff height \(80\text{ m}\). Depression \(45^\circ\). Distance =
16. Building height \(20\text{ m}\). Angle \(60^\circ\). Distance =
17. Maximum height for same distance occurs at:
18. Line of sight triangle in trigonometry is always:
19. Elevation = depression because of:
20. If height = shadow, angle =
21. Height \(h = d\tan\theta\) uses ratio:
22. If \(\tan\theta = 1\), then \(\theta\) is:
23. Which is NOT part of Class 10 applications?
24. Tower \(30\text{ m}\). Angle \(30^\circ\). Distance =
25. Height \(40\text{ m}\). Depression \(45^\circ\). Distance =
26. \(\sin 30^\circ =\)
27. Moving farther from object ? angle:
28. Right angle arises from:
29. Pole seen at \(45^\circ\). Distance \(12\text{ m}\). Height =
30. Height \(10\sqrt{3}\). Angle \(60^\circ\). Distance =
31. Most used ratio in height/distance:
32. If angle increases, observer is:
33. Balloon observed at two angles involves:
34. Tower \(50\text{ m}\). Shadow \(50\sqrt{3}\). Angle =
35. Airplane elevation increases means airplane is:
36. If object is above observer, angle is:
37. Distance is maximum when angle:
38. Height \(= d\sqrt{3}\) indicates angle:
39. Shadow = 0 means angle =
40. If angle = \(0^\circ\), height appears:
41. Two elevation angles ? form:
42. To compute height, we need:
43. Line of sight is drawn from:
44. \(\cos 60^\circ =\)
45. Angle of depression is measured from:
46. If tower height doubles, angle:
47. Looking down from balcony is:
48. \(\sin 60^\circ =\)
49. Maximum angle of elevation possible:
50. A tree has height \(h\). Angle = \(45^\circ\). Distance =

Frequently Asked Questions

To apply trigonometric ratios (sin, cos, tan) to real-life problems involving heights and distances using angles of elevation and depression.

The straight, imaginary line joining the observer’s eye to the object being viewed.

The angle formed between the horizontal line of sight and the upward line of sight when an observer looks at an object above eye level.

The angle formed between the horizontal line of sight and the downward line of sight when an observer views an object below eye level.

Because the horizontal distance and vertical height naturally form perpendicular lines, creating right triangles useful for applying trigonometric ratios.

Primarily tangent (tan ?), but sine (sin ?) and cosine (cos ?) are also used depending on known sides.

tan ? = Opposite side / Adjacent side.

When the vertical height corresponds to the opposite side and the given length is the hypotenuse.

When the horizontal distance corresponds to the adjacent side and the given length is the hypotenuse.

Only standard angles (30°, 45°, 60°) are used, whose trigonometric ratios are known.

sin 30°=½, sin 45°=v2/2, sin 60°=v3/2; cos 30°=v3/2, cos 45°=v2/2, cos 60°=½; tan 30°=1/v3, tan 45°=1, tan 60°=v3.

Draw a clear, labeled diagram converting the scenario into a right triangle.

It helps identify the unknown side, the angle given, and the correct trigonometric ratio to use.

The imaginary line parallel to the ground passing through the observer’s eye.

The person, point, or object from which sight or measurement is taken.

Towers, poles, trees, buildings, mountains, ships, airplanes, balloons, and bridges.

A distance that cannot be measured directly, requiring trigonometric methods.

Yes, by using angles of elevation or depression from a known point.

Two-point observation method, resulting in two different right triangles.

Use tan ? = height/distance ? height = distance × tan ?.

distance = height / tan ?.

Viewing the top of a tower or kite from ground level.

Seeing a car from a lighthouse balcony.

Observing the same object from two different positions, yielding two different elevation angles.

As the observer moves closer, the angle of elevation increases.

Right-triangle geometry and trigonometric ratios.

Finding height, distance, width, altitude, or length using given angles of elevation or depression.

Sometimes, when two sides are known or trigonometric ratios are insufficient.

tan 45° = 1, simplifying height = distance.

tan 30° = 1/v3, often appearing in height problems requiring rationalization.

The object is vertically above the observer—distance is zero (mostly theoretical).

Opposite is the vertical side from angle; adjacent is the horizontal side.

NCERT restricts problems to standard angles with known ratios.

Theodolite, used by surveyors to measure angles.

To estimate altitude and distance from the ground.

For determining distance of ships, lighthouses, and ports.

The syllabus focuses only on elementary applications—heights and distances.

Single angle of elevation with either height or distance known.

When two right triangles share a common vertical line or horizontal line.

A real-life scenario with one or two angles and one unknown distance, requiring diagram + calculation.

Convert angle of depression to angle of elevation at the lower point (alternate interior angles).

Because they form alternate interior angles with horizontal lines.

Use tan ? = shadow length / height or vice versa.

Yes, especially when forming equations from the triangle.

To provide practical, realistic measurements.

Yes, depending on the triangle setup and chosen trigonometric ratio.

Height represents perpendicular distance from ground to the object’s top.

Yes, by modelling below-ground or underwater observations.

It strengthens understanding of angles and their variation.

Neat diagram, correct ratio selection, accurate steps, and clear final answer with units.

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