NCERT Solutions for class 10 Mathematics Chapter 9
NCERT solutions for Class 10 Mathematics Chapter 9 Some Applications of Trigonometry – complete answers & explanations
Class 10 Mathematics Chapter 9 Some Applications of Trigonometry helps students understand how trigonometry is used in real-life situations such as finding heights and distances. This chapter builds a strong foundation for solving practical problems involving angles of elevation and depression. It is an important topic for exams as it requires both conceptual clarity and correct application of formulas. This blog provides clear and reliable NCERT solutions that help students understand the correct approach to answering each question. Students can improve accuracy and confidence by practicing regularly. Download the worksheet and practice alongside solutions for better clarity. Parents can also support their child’s learning journey effectively through guided practice. Book a free trial now to get expert guidance.
What this NCERT chapter covers?
1. Understanding practical applications of trigonometry in real-life problems
2. Use of trigonometric ratios like sin, cos, and tan in solving questions
3. Concepts of height and distance in different scenarios
4. Solving problems involving angles of elevation and depression
5. Application of right triangle properties in real situations
6. Step-by-step approach to solving trigonometry-based word problems
7. Use of diagrams to visualize and interpret questions correctly
8. Learning to identify appropriate trigonometric ratios for each problem
9. Improving logical thinking and mathematical reasoning
10. Strengthening problem-solving skills for board exams
How to use these NCERT solutions?
1. First, read each question carefully and try solving it on your own
2. Use rough work to apply trigonometric formulas step by step
3. After attempting, compare your answer with the given solutions
4. Focus on understanding each step rather than just checking the final answer
5. Revise the formulas used in each question for better retention
6. Practice similar problems to improve speed and accuracy
7. Parents and teachers can guide students in understanding mistakes
8. Follow the same order of questions as given in the worksheet for consistency
Important tips & tricks for students
1. Always draw a neat diagram to understand the question clearly
2. Identify the correct trigonometric ratio before solving
3. Avoid calculation mistakes while simplifying values
4. Learn standard trigonometric values thoroughly
5. Double-check units in the final answer
6. Practice solving problems step by step to avoid confusion
7. Do not skip intermediate steps during exams
8. Read the question carefully to avoid misinterpretation
9. Practice regularly to build confidence
NCERT solutions – complete answer key
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
Using Pythagoras:
(h − AB)² + 8² = (h − AB)² (Not needed directly)
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
15.Distance travelled = y − x = 58.8√3 m
Answer: 58.8√3 m
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
Why NCERT solutions help students?
NCERT solutions help students prepare effectively for exams by providing the correct approach to solving each question. They improve concept clarity, ensure accuracy in answers, and build confidence through consistent practice. With structured solutions, students can easily understand how to apply formulas and solve problems step by step.
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