NCERT Solutions for Class 10 Mathematics Chapter 9 Some Applications of Trigonometry 

This worksheet provides NCERT Solutions for Class 10 Mathematics Chapter Some Applications of Trigonometry. This chapter focuses on solving real-life problems using trigonometric ratios such as sine, cosine, and tangent. It helps students understand how to calculate heights, distances, and angles using mathematical concepts. This worksheet includes complete and accurate NCERT Solutions that follow the exact structure and methods used in the chapter, making it easier for students to learn and revise effectively. 

This chapter is concept-based and does not include any stories or poems. It focuses on real-life situations where trigonometry is applied to find heights and distances. The problems are numerical and activity-based, helping students apply formulas to practical scenarios like poles, trees, buildings, and towers. 

What this NCERT chapter covers? 

• Application of trigonometric ratios in real-life problems 
• Understanding heights and distances 
• Use of sine, cosine, and tangent ratios 
• Problem-solving using right-angled triangles 
• Concept-based numerical calculations 

How to use these NCERT solutions? 

Students should first try solving each question on their own and then use these NCERT Solutions to check their answers. These solutions follow the exact NCERT order and method, making it easy to compare steps. Parents and teachers can use this worksheet to guide students and ensure they understand each concept clearly. It is also helpful for revision and practice before exams. 

Student tips & learning tricks 

• Carefully identify the given values like height, distance, and angle before solving 
• Choose the correct trigonometric ratio based on the problem 
• Avoid calculation mistakes while simplifying values like √3 
• Draw diagrams wherever needed to understand the problem better 
• Follow step-by-step solutions as per NCERT format 

Why NCERT solutions are important? 

NCERT Solutions help students build a strong understanding of fundamental concepts. They follow the official NCERT pattern, which is important for exams. Practicing these solutions improves accuracy, confidence, and problem-solving skills. It also helps students prepare better for school assessments and board exams. 

Complete answer key – NCERT solutions 

Exercise 9.1 

1 
Given: Length of rope = 20 m, angle = 30° 
To find: Height of pole 

Solution: 
sin 30° = Height / Hypotenuse 
1/2 = h / 20 
h = 20 × 1/2 = 10 m 

Answer: Height = 10 m 

2 
Given: Distance between foot and point where top touches ground = 8 m 
Angle = 30° 
Let height of tree = h 
Let the tree break at point B and touch ground at C. 
Let BC = x and AC = 8 m 

In right ∆, 
tan 30° = AB / AC 
1/√3 = AB / 8 
AB = 8/√3 

Also, BC = remaining part of tree 

Better approach: 
cos 30° = AC / total broken part 
√3/2 = 8 / (h − AB) 

h − AB = 16/√3 
Total height: 
h = AB + (h − AB) 
h = 8/√3 + 16/√3 = 24/√3 = 8√3 m 

Answer: Height = 8√3 m 

3 
Case (i): 
Height = 1.5 m, angle = 30° 

sin 30° = Height / Length 
1/2 = 1.5 / L 
L = 3 m 

Case (ii): 
Height = 3 m, angle = 60° 

sin 60° = Height / Length 
√3/2 = 3 / L 
L = 2√3 m 

Answer: 
(i) 3 m 
(ii) 2√3 m 

4 
Given: Distance = 30 m, angle = 30° 

tan 30° = Height / Base 
1/√3 = h / 30 
h = 30/√3 = 10√3 m 

Answer: Height = 10√3 m 

5 
Given: Height = 60 m, angle = 60° 

sin 60° = Height / Length 
√3/2 = 60 / L 
L = 120/√3 = 40√3 m 

Answer: Length = 40√3 m 

6 
Let initial distance = x 

tan 30° = (30 − 1.5)/x 
1/√3 = 28.5/x → x = 28.5√3 

tan 60° = 28.5/(x − d) 
√3 = 28.5/(x − d) 

x − d = 28.5/√3 

d = 28.5√3 − 28.5/√3 
d = 28.5 ( (3 − 1) / √3 ) 
d = 28.5 × (2/√3) 
d = 19√3 

Answer: Distance walked = 19√3 m 

7 
Let tower height = h 

tan 60° = (20 + h)/x 
tan 45° = 20/x 

From second: x = 20 

Substitute: 
√3 = (20 + h)/20 

h = 20√3 − 20 

Answer: Height = 20(√3 + 1) m 

8 
Let pedestal height = h 

tan 45° = h/x → x = h 
tan 60° = (h + 1.6)/x 

√3 = (h + 1.6)/h 

h√3 = h + 1.6 
h(√3 − 1) = 1.6 

h = 1.6(√3 + 1) m 

Answer: Height = 1.6 (√3 + 1) m 

9 
Let building height = h 

tan 30° = h/x → h = x/√3 
tan 60° = 50/x → x = 50/√3 

So h = (50/√3)/√3 = 50/3 

Answer: Height = 50/3 m 

10 
Let distance from first pole = x 
Other distance = 80 − x 

tan 60° = h/x → h = √3x 
tan 30° = h/(80 − x) → h = (80 − x)/√3 

Equate: 
√3x = (80 − x)/√3 
3x = 80 − x 

x = 20 
h = 20√3 

Answer: 
Height = 20√3 m 
Distances = 20 m and 60 m 

11 
Let width = x 

tan 60° = h/x → h = √3x 
tan 30° = h/(x + 20) → h = (x + 20)/√3 

Equate: 
√3x = (x + 20)/√3 
3x = x + 20 

x = 10 
h = 10√3 

Answer: 
Height = 10√3 m 
Width = 10 m 

12 
Let tower height = h 

tan 45° = 7/x → x = 7 
tan 60° = (h − 7)/7 

√3 = (h − 7)/7 

h = 7 + 7√3 

Answer: Height = 7 + 7√3 m 

13 
tan 45° = 75/x → x = 75 
tan 30° = 75/y → y = 75√3 

Distance = y − x = 75(√3 − 1) 

Answer: 75(√3 − 1) m 

14 
Initial distance: 
tan 60° = (88.2 − 1.2)/x 
√3 = 87/x → x = 87/√3 

Final distance: 
tan 30° = 87/y 
1/√3 = 87/y → y = 87√3 

Distance travelled = y − x = 58.8√3 m 

Answer: 58.8√3 m 

15 
Let initial distance = x 

tan 30° = h/x → x = h√3 
tan 60° = h/(x − d) 

√3 = h/(x − d) 

x − d = h/√3 

d = h√3 − h/√3 = 2h/√3 

Speed = d/6 

Time to reach = x / speed 
= (h√3) / (2h/√3 ÷ 6) 
= 9 seconds 

Answer: Time = 9 seconds 

Master Class 10 Mathematics Chapter Some Applications of Trigonometry with accurate NCERT Solutions and improve your problem-solving skills for exams and real-life applications. 

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