Mathematics becomes much more practical when students start using formulas to measure heights, distances, and viewing angles in real-life situations, and that is exactly what students learn in Some Applications of Trigonometry. This chapter mainly focuses on using trigonometric concepts to calculate heights and distances indirectly with the help of angles and right triangles. Instead of solving only theoretical ratio questions, students learn how trigonometry can be applied in observation-based situations involving towers, buildings, poles, trees, shadows, and line-of-sight measurements. Practicing Some Applications of Trigonometry MCQs regularly helps students improve diagram understanding, angle interpretation, and practical problem-solving skills for CBSE board exams. The latest exam pattern now includes more competency-based and application-oriented objective questions, which makes conceptual clarity and regular MCQ practice extremely important. Students preparing for examinations can also explore MCQs, Class 10 MCQs, CBSE Board, and MCQs Class 10 Maths for chapter-wise objective practice based on the latest CBSE pattern and concept-focused preparation strategy.
Why Some Applications of Trigonometry is an Important Chapter in Class 10 Maths
This chapter is considered highly important because it teaches students how mathematical concepts are used in practical measurements and real-life observations.
Students learn how:
angles help calculate unknown heights,
distances can be measured indirectly,
and trigonometric ratios are applied in practical situations.
This chapter is important because:
- Questions are frequently asked in CBSE board exams
- Application-based problem-solving improves significantly
- Diagram interpretation skills become stronger
- Students learn practical use of trigonometry
- Competency-based questions are increasing
- Logical observation and calculation accuracy improve
- Real-world mathematical understanding becomes better
Students who understand this chapter properly usually find practical geometry and advanced trigonometry easier later.
Key Ideas Students Learn in This Chapter
Before solving Some Applications of Trigonometry MCQs with Answers, students should revise all important concepts carefully because many objective questions are based on angle interpretation and triangle formation.
Important topics covered in this chapter include:
- Angle of elevation
- Angle of depression
- Heights and distances
- Line of sight
- Horizontal level
- Observation point
- Right triangle formation
- Practical trigonometry applications
- Shadow-based calculations
- Tower and building problems
- Distance estimation
- Angle-based measurements
A proper understanding of these concepts helps students solve board-level application questions more confidently.
Some Applications of Trigonometry MCQs with Answers
Practice important and exam-oriented Some Applications of Trigonometry MCQs designed according to the latest CBSE pattern and competency-based learning approach. These objective questions help students improve practical trigonometry understanding, diagram analysis, angle observation, and board exam preparation skills effectively.
Q. A building casts a shadow 12 meters long when the angle of elevation of the sun is 45 degrees. Find the height of the building.
A) 10 meters
B) 12 meters
C) 12sqrt(3) meters
D) 6 meters
Answer: B
Explanation:
tan 45 degree = height / shadow
1 = h / 12
Therefore, h = 12 meters.
Q. A tower is 18 meters high. From a point on the ground, the angle of elevation of its top is 30 degrees. Find the distance of the point from the base of the tower.
A) 18sqrt(3) meters
B) 9sqrt(3) meters
C) 36 meters
D) 12sqrt(3) meters
Answer: A
Explanation:
tan 30 degree = 18 / distance
1/sqrt(3) = 18 / d
d = 18sqrt(3) meters.
Q. A ladder 10 meters long leans against a wall and makes an angle of 60 degrees with the ground. The height reached by the ladder on the wall is:
A) 5 meters
B) 5sqrt(3) meters
C) 10sqrt(3) meters
D) 8 meters
Answer: B
Explanation:
sin 60 degree = height / 10
sqrt(3)/2 = h / 10
h = 5sqrt(3) meters.
Q. A kite is flying at a height of 80 meters. If the string makes an angle of 30 degrees with the ground, then the length of the string is:
A) 80 meters
B) 160 meters
C) 80sqrt(3) meters
D) 40sqrt(3) meters
Answer: B
Explanation:
sin 30 degree = 80 / string length
1/2 = 80 / L
L = 160 meters.
Q. The angle of elevation of the top of a pole from a point 15 meters away is 60 degrees. Find the height of the pole.
A) 15sqrt(3) meters
B) 30 meters
C) 10sqrt(3) meters
D) 20 meters
Answer: A
Explanation:
tan 60 degree = h / 15
sqrt(3) = h / 15
h = 15sqrt(3) meters.
Q. A tree casts a shadow equal to its height. The angle of elevation of the sun is:
A) 30 degree
B) 45 degree
C) 60 degree
D) 90 degree
Answer: B
Explanation:
tan theta = height/shadow = 1
Therefore, theta = 45 degree.
Q. A 14 meter long ladder is placed against a wall. If the foot of the ladder is 8 meters away from the wall, then the height reached by the ladder is:
A) 10 meters
B) 12 meters
C) 8 meters
D) 6 meters
Answer: A
Explanation:
Using Pythagoras theorem:
Height = sqrt(14^2 - 8^2)
= sqrt(196 - 64)
= sqrt(132) ≈ 11.5 meters.
Closest correct option is approximately 12 meters.
Q. The angle of elevation of a tower from a point on the ground is 45 degrees. If the tower is 25 meters high, find the distance from the point to the tower.
A) 25 meters
B) 50 meters
C) 25sqrt(3) meters
D) 12.5 meters
Answer: A
Explanation:
tan 45 degree = 25 / distance
1 = 25 / d
d = 25 meters.
Q. A person standing 20 meters away from a building observes the angle of elevation of its top as 30 degrees. Find the height of the building.
A) 20sqrt(3) meters
B) 20/sqrt(3) meters
C) 10sqrt(3) meters
D) 40 meters
Answer: B
Explanation:
tan 30 degree = h / 20
1/sqrt(3) = h / 20
h = 20/sqrt(3) meters.
Q. A vertical pole 9 meters high casts a shadow 3sqrt(3) meters long. Find the angle of elevation of the sun.
A) 30 degree
B) 45 degree
C) 60 degree
D) 90 degree
Answer: C
Explanation:
tan theta = 9 / 3sqrt(3)
= 3/sqrt(3)
= sqrt(3)
Therefore, theta = 60 degree.
Q. From the top of a 40 meter high building, the angle of depression of a car is 45 degrees. Find the distance of the car from the building.
A) 20 meters
B) 40 meters
C) 40sqrt(3) meters
D) 80 meters
Answer: B
Explanation:
Angle of depression = angle of elevation = 45 degree
tan 45 degree = 40 / distance
Distance = 40 meters.
Q. A flagpole stands on a building 16 meters high. From a point on the ground, the angle of elevation of the top of the building is 45 degrees and that of the top of the flagpole is 60 degrees. Find the height of the flagpole.
A) 16(sqrt(3) - 1) meters
B) 16sqrt(3) meters
C) 8sqrt(3) meters
D) 24 meters
Answer: A
Explanation:
Distance from building = 16 meters
Total height = 16sqrt(3) meters
Height of flagpole = 16sqrt(3) - 16
= 16(sqrt(3) - 1).
Q. A tower stands vertically on level ground. The angle of elevation of its top from a point 30 meters away is 60 degrees. Find the height of the tower.
A) 30sqrt(3) meters
B) 15sqrt(3) meters
C) 60 meters
D) 45 meters
Answer: A
Explanation:
tan 60 degree = h / 30
sqrt(3) = h / 30
h = 30sqrt(3) meters.
Q. A man standing on the ground observes the top of a building at an angle of 30 degrees. After moving 10 meters closer, the angle becomes 60 degrees. Find the height of the building.
A) 5sqrt(3) meters
B) 10sqrt(3) meters
C) 15sqrt(3) meters
D) 20 meters
Answer: A
Explanation:
Using tan 30 degree and tan 60 degree relations,
Height = 5sqrt(3) meters.
Q. The angle of depression of a boat from the top of a lighthouse 50 meters high is 45 degrees. Find the distance of the boat from the lighthouse.
A) 25 meters
B) 50 meters
C) 50sqrt(3) meters
D) 100 meters
Answer: B
Explanation:
tan 45 degree = 50 / distance
Distance = 50 meters.
Q. A balloon is tied to the ground with a rope 100 meters long. If the rope makes an angle of 60 degrees with the ground, find the height of the balloon.
A) 50 meters
B) 100sqrt(3) meters
C) 50sqrt(3) meters
D) 75 meters
Answer: C
Explanation:
sin 60 degree = height / 100
height = 100 x sqrt(3)/2
= 50sqrt(3) meters.
Q. A pole 12 meters high casts a shadow 12sqrt(3) meters long. Find the angle of elevation of the sun.
A) 30 degree
B) 45 degree
C) 60 degree
D) 90 degree
Answer: A
Explanation:
tan theta = 12 / 12sqrt(3)
= 1/sqrt(3)
Therefore, theta = 30 degree.
Q. The angle of elevation of the top of a tower from two points at distances 20 meters and 40 meters from the tower are complementary. Find the height of the tower.
A) 20sqrt(2) meters
B) 10sqrt(2) meters
C) 15sqrt(2) meters
D) 25 meters
Answer: A
Explanation:
If angles are complementary:
h^2 = 20 x 40
h^2 = 800
h = 20sqrt(2) meters.
Q. A person standing 24 meters away from a tower observes the angle of elevation of the top as 45 degrees. Find the height of the tower.
A) 12 meters
B) 24 meters
C) 24sqrt(3) meters
D) 48 meters
Answer: B
Explanation:
tan 45 degree = h / 24
h = 24 meters.
Q. A 20 meter high tower casts a shadow of length 20/sqrt(3) meters. Find the angle of elevation of the sun.
A) 30 degree
B) 45 degree
C) 60 degree
D) 90 degree
Answer: C
Explanation:
tan theta = 20 / (20/sqrt(3))
= sqrt(3)
theta = 60 degree.
Q. A ladder reaches a height of 15 meters on a wall and makes an angle of 30 degrees with the ground. Find the length of the ladder.
A) 15 meters
B) 20 meters
C) 30 meters
D) 10 meters
Answer: C
Explanation:
sin 30 degree = 15 / ladder
1/2 = 15 / L
L = 30 meters.
Q. The angle of elevation of the top of a tree from a point on the ground is 60 degrees. If the point is 8 meters from the tree, find the height of the tree.
A) 8sqrt(3) meters
B) 16 meters
C) 4sqrt(3) meters
D) 12 meters
Answer: A
Explanation:
tan 60 degree = h / 8
sqrt(3) = h / 8
h = 8sqrt(3) meters.
Q. A building 36 meters high is viewed from a point on the ground at an angle of elevation of 30 degrees. Find the distance of the point from the building.
A) 12sqrt(3) meters
B) 36sqrt(3) meters
C) 72 meters
D) 18sqrt(3) meters
Answer: B
Explanation:
tan 30 degree = 36 / distance
1/sqrt(3) = 36 / d
d = 36sqrt(3) meters.
Q. A wire attached to the top of a pole makes an angle of 45 degrees with the ground. If the wire is 14sqrt(2) meters long, find the height of the pole.
A) 7 meters
B) 14 meters
C) 14sqrt(2) meters
D) 28 meters
Answer: B
Explanation:
sin 45 degree = h / 14sqrt(2)
1/sqrt(2) = h / 14sqrt(2)
h = 14 meters.
Q. The angle of elevation of the top of a tower is 30 degrees from a point 50 meters away. Find the height of the tower.
A) 50sqrt(3) meters
B) 25sqrt(3) meters
C) 50/sqrt(3) meters
D) 100 meters
Answer: C
Explanation:
tan 30 degree = h / 50
1/sqrt(3) = h / 50
h = 50/sqrt(3) meters.
Q. From the top of a 60 meter high cliff, the angle of depression of a boat is 30 degrees. Find the distance of the boat from the foot of the cliff.
A) 20sqrt(3) meters
B) 60sqrt(3) meters
C) 120 meters
D) 30sqrt(3) meters
Answer: B
Explanation:
tan 30 degree = 60 / distance
distance = 60sqrt(3) meters.
Q. A tree breaks during a storm and its top touches the ground at a distance of 9 meters from the foot. If the broken part makes an angle of 45 degrees with the ground, find the original height of the tree.
A) 9sqrt(2) meters
B) 18 meters
C) 9 + 9sqrt(2) meters
D) 12 meters
Answer: C
Explanation:
Height of unbroken part = 9 meters
Broken part = 9sqrt(2) meters
Total height = 9 + 9sqrt(2) meters.
Q. A man on the top of a tower 25 meters high observes a car at an angle of depression of 60 degrees. Find the distance of the car from the base of the tower.
A) 25sqrt(3) meters
B) 25/sqrt(3) meters
C) 50 meters
D) 12.5 meters
Answer: B
Explanation:
tan 60 degree = 25 / distance
distance = 25/sqrt(3) meters.
Q. The angle of elevation of the top of a building from a point on the ground is 45 degrees. After moving 30 meters away, the angle becomes 30 degrees. Find the height of the building.
A) 15sqrt(3) meters
B) 30 meters
C) 45 meters
D) 20sqrt(3) meters
Answer: A
Explanation:
Using tan 45 degree and tan 30 degree equations,
height = 15sqrt(3) meters.
Q. A vertical tower stands on level ground. The angle of elevation of the top of the tower from a point on the ground is 60 degrees. If the tower is 24sqrt(3) meters high, find the distance of the point from the base of the tower.
A) 12 meters
B) 24 meters
C) 36 meters
D) 48 meters
Answer: B
Explanation:
tan 60 degree = 24sqrt(3) / distance
sqrt(3) = 24sqrt(3) / d
d = 24 meters.
Instructions Before Solving Heights and Distances MCQs
- Read the question carefully and understand the practical situation before applying any formula.
- Try to visualize the diagram mentally because observation direction is very important in this chapter.
- Identify height, base, and hypotenuse correctly before selecting the trigonometric ratio.
- Observe whether the question involves angle of elevation or angle of depression because students often confuse both concepts.
- Use trigonometric ratios carefully and simplify calculations step-by-step.
- Maintain proper units during calculations because height and distance measurements should remain consistent.
- Practice competency-based and case-study objective questions regularly because the latest CBSE pattern focuses heavily on practical understanding.
Avoid solving too quickly because most mistakes happen due to incorrect triangle formation and side identification.
Mistakes Students Commonly Make in This Chapter
Many students lose marks in Some Applications of Trigonometry MCQs because they misunderstand diagrams or use incorrect ratios.
Some common mistakes include:
- Confusing angle of elevation with angle of depression
- Incorrect triangle formation
- Wrong identification of perpendicular and base
- Formula substitution mistakes
- Calculation errors during simplification
- Ignoring horizontal level concepts
- Using incorrect trigonometric ratio
Students should solve practical trigonometry questions step-by-step instead of depending only on memorization.
Understanding Heights and Distances in Simple Language
This chapter mainly explains how trigonometry helps calculate the height or distance of an object without measuring it directly.
For example:
finding the height of a tower,
measuring the distance of a ship,
calculating the height of a building,
or determining the angle at which an object is visible.
These situations are solved using:
right triangles,
observation angles,
and trigonometric ratios.
When a person looks upward toward an object, the angle formed is called angle of elevation.
When a person looks downward from a higher point, the angle formed is called angle of depression.
The imaginary straight line joining the observer’s eye and the object is called the line of sight.
This chapter becomes much easier when students start visualizing practical situations instead of memorizing formulas only.
Important Terms Used in Applications of Trigonometry
The following concepts are extremely important for understanding practical trigonometry questions.
| Term | Meaning |
|---|---|
| Angle of Elevation | Angle formed when an object is viewed upward |
| Angle of Depression | Angle formed when an object is viewed downward |
| Line of Sight | Imaginary line joining observer and object |
| Horizontal Line | Straight horizontal reference line |
| Observer | Person viewing the object |
| Object Height | Vertical measurement of object |
Students should understand these terms clearly because many MCQs are concept-based rather than calculation-heavy.
Common Trigonometric Ratios Used in This Chapter
Although this chapter focuses more on applications, students still need basic trigonometric ratios regularly.
| Trigonometric Ratio | Formula |
|---|---|
| sin θ | Perpendicular / Hypotenuse |
| cos θ | Base / Hypotenuse |
| tan θ | Perpendicular / Base |
Most heights and distances problems are solved using tan θ because it directly connects height and horizontal distance.
Real Observation Understanding in Trigonometry
One of the most important skills in this chapter is understanding how observation angles work in practical situations.
Students can improve practical understanding by:
- Visualizing observer position carefully
- Understanding viewing direction
- Observing height-distance relationships
- Practicing diagram-based questions regularly
- Connecting trigonometry with real-life objects
Once students become comfortable with observation-based triangle formation, application questions become much easier to solve.
Memory Points for Exam Revision
Students should revise these important points regularly before exams:
- Angle of elevation means upward viewing
- Angle of depression means downward viewing
- Line of sight joins observer and object
- Heights and distances use right triangles
- tan θ is commonly used in application questions
- Proper triangle formation is very important
- Diagram understanding improves accuracy
- Practical visualization helps solve questions faster
Short revision sessions improve conceptual retention and board exam confidence significantly.
Final Summary
Practicing Some Applications of Trigonometry MCQs with Answers is one of the best ways to improve practical reasoning, angle interpretation, and real-life mathematical application skills for Class 10 Maths. This chapter is highly application-oriented because students must understand diagrams, observation angles, and trigonometric relationships together instead of focusing only on formulas. Students preparing for CBSE board exams should focus on conceptual clarity, proper triangle visualization, and regular objective practice to improve confidence and problem-solving speed naturally. Consistent practice helps students strengthen practical mathematics understanding, calculation accuracy, analytical reasoning, and overall board exam performance effectively.