CBSE Class 10 Maths Some Applications of Trigonometry RD Sharma Solutions
Mathematics in Class 10 introduces you to trigonometry. In the simplest terms, it is the study of the relationships between the sides and angles of a triangle. You now know about trigonometric ratios of different angles and trigonometric identities. Trigonometry also has some useful applications in real life that are covered in Chapter 12. Further, RD Sharma Solutions for Class 10 Maths Chapter 12 – Some Applications of Trigonometry offers assistance in solving problems related to the real-life applications of trigonometry.
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Now that you understand the basics of trigonometry, it is interesting to learn how it is used in our daily lives. Chapter 12 is all about calculating heights and distances of objects.
RD Sharma Solutions for Class 10 Chapter 10 will help you understand these applications and solve problems based on them with ease.
Sub-topics covered under RD Sharma Solutions for Class 10 Maths Chapter 12
All the Exercise questions with solutions in Chapter 12 – Some Applications of Trigonometry are given below:
- Class 10 Chapter 12 Some Applications of Trigonometry Exercise 12.1
Solved Examples from RD Sharma Class 10 Solutions – Chapter 12
Question 1: A tower is 100√3 m. high. The angle of elevation of its top from a point 100 m. away from its foot is ……..
Answer:
Let’s draw the diagram first:
We know that,
tanA = BC/AC = 100√3/100 = √3
∴ tanA = √3
Further, from trigonometric ratios, we know that
tan60 = √3
Therefore,
A = 60°
Question 2: A bridge across a river makes an angle of 45° with the river bank as shown in the figure. If the length of the bridge across the river is 150 m, what is the width of the river?
Answer:
Let’s consider the right-angled triangle ABC. We have,
sin45° = BC/AC
We know that,
sin45° = 1/√2
Therefore, we have
1/√2 = BC/150
⇒ BC = 150/√2
⇒ BC = 75√2
Therefore, the width of the river is 75√2 meters.
Question 3: A person, standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank is 60°. When he retreats 20 m from the bank, he finds the angle to be 30°. Find the height of the tree and the breadth of the river.
Answer:
Let’s draw the diagram first:
Let AB be the breadth of the river and BC be the height of the tree which makes an angle of 60° at a point A on the opposite bank. Further, let D be the position of the person after retreating 20 m from the bank.
Also, let
AB = x meters
BC = h meters.
We know that,
tan(θ) = Opposite Side / Adjacent Side
Therefore, from the right-angled triangle ABC, we have
tan60° = BC/AB
⇒ √3 = h/x
⇒ h = x√3
From the right-angled triangle DBC, we have
tan30° = BC/DB
⇒ 1/√3 = h/(20 + x)
⇒ h = (x + 20) / √3
Comparing the two values of ‘h’, we get
x√3 = (x + 20)/√3
⇒ 3x = x + 20
⇒ 2x = 20
⇒ x = 10
Therefore,
h = x√3 = 10√3 = 17.32
Hence,
The height of the tree = 17.32 meters and the breadth of the river = 10 meters.
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