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Chapter 4 of 14
Practice Quiz

Some Applications of Trigonometry

Karnataka Board · Class 10 · Mathematics

Practice quiz for Some Applications of Trigonometry — Karnataka Board Class 10 Mathematics. MCQs and questions with answers to test your preparation.

45 questions20 flashcards5 concepts

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Quick Quiz: Some Applications of Trigonometry

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1

A ladder 10 m long reaches a window 8 m high from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.

2

From a point 20 m away from the foot of a tower, the angle of elevation of the top is 30°. What is the height of the tower?

3

A kite is flying at a height of 75 m from the ground. The string makes an angle of 60° with the ground. Find the length of the string.

4

The shadow of a tower is √3 times its height. What is the angle of elevation of the sun?

45 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

A man 1.8 m tall casts a shadow of 1.8 m. What is the angle of elevation of the sun?

Show answer

45°

Step 1: Height = 1.8 m, shadow = 1.8 m, angle of elevation = ? Step 2: tan(angle) = height/shadow = 1.8/1.8 = 1. Step 3: We know tan 45° = 1. Step 4: Therefore, angle of elevation = 45°.

2multiple choice
1 marks

From the top of a 60 m high building, the angle of depression of a car on the road is 30°. How far is the car from the building?

Show answer

60√3 m

Step 1: Building height = 60 m, angle of depression = 30°, horizontal distance = ? Step 2: Angle of depression equals angle of elevation from car to top. Step 3: tan 30° = height/distance = 60/distance. Step 4: 1/√3 = 60/distance, so distance = 60√3 m.

3multiple choice
1 marks

A tree breaks and bends to touch the ground 12 m away from its base, making 30° with the ground. What was the original height of the tree?

Show answer

8 + 4√3 m

Step 1: Let broken part length = l, standing part = h. Distance from base = 12 m, angle = 30°. Step 2: In right triangle: cos 30° = 12/l, so √3/2 = 12/l, giving l = 24/√3 = 8√3 m. Step 3: sin 30° = h/l, so 1/2 = h/(8√3), giving h = 4√3 m. Step 4: Original height = h + l = 4√3 + 8√3 = 12√3 m. Wait, let me recalculate. Step 2: sin 30° = h/l = 1/2, and cos 30° = 12/l = √3/2. Step 3: From cos equation: l = 12/(√3/2) = 24/√3 = 8√3 m. Step 4: From sin equation: h = l/2 = 4√3 m. Actually, let me be more careful. The standing part is h, broken part makes hypotenuse l. We have base = 12, angle = 30°. S

4multiple choice
1 marks

An airplane flying at 3000 m height observes the angle of depression of an airport to be 30°. What is the horizontal distance from the plane to the airport?

Show answer

3000√3 m

Step 1: Height = 3000 m, angle of depression = 30°, horizontal distance = ? Step 2: tan 30° = height/horizontal distance = 3000/distance. Step 3: 1/√3 = 3000/distance. Step 4: distance = 3000√3 m.

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What are the important topics in Some Applications of Trigonometry for Karnataka Board Class 10 Mathematics?
Some Applications of Trigonometry covers several key topics that are frequently asked in Karnataka Board Class 10 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Some Applications of Trigonometry — Karnataka Board Class 10 Mathematics?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

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