# Class 10 RD SHARMA Solutions Maths Chapter 12 - Heights and Distances

You can expect the right type of step-by-step explanations from TopperLearning’s class 10 **R.D Sharma Solutions **for Chapter 12 - Heights And Distances. The solutions come with all the concept-wise questions covering important areas from the chapter. Because everything is readily available on the website and a dedicated TopperLearning App, you can start learning and revising answers on time.

Professionally trained subject matter experts have devised the question and answers discussed within, abiding by the latest **CBSE class 10 Maths **guidelines. Your grades in Maths will improve with practice. Also, all the helpful resources (video lessons, ask a doubt, revision notes, sample papers and previous years' question papers) made available here stand as the ultimate help, making learning super easy. Eventually, you will score better grades, standing out from the rest of the class.

TopperLearning’s R.D. Sharma Solutions is a one-stop guide for students, offering easy access to thoroughly illustrated examples and step-by-step answers in Heights And Distances. Whether you are stuck with any particular question while measuring the heights, or it is the distance part you find hard, easy-to-understand answers will help you make it through. Further, TopperLearning makes class 10 easy with a series of subject-wise important questions, fill-in-the-blanks, multiple-choice questions, subjective questions and important revision notes.

These solutions aim to thoroughly guide you through the essential concepts in Mathematics, with steady answers to the important concepts in this chapter. Because you will appear in class 10 board exams within a few months, preparation and continuous doubt-solving become mandatory, and things get easy with the best resources. Other than R.D Sharma solutions for class 10, at TopperLearning, you can also avail of R.D Sharma solutions for class 9, R.D Sharma solutions for class 11 and R.D Sharma solutions for class 12.

## Some Applications of Trigonometry Exercise Ex. 12.1

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Let BC be the building, AB be the transmission tower, and D be the point on ground from where elevation angles are to be measured.

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Let AB be the statue, BC be the pedestal and D be the point on ground from where elevation angles are to be measured.

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Let AB be the lighthouse and the two ships be at point C and D respectively.

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Let AC = h be the height of the chimney.

Height of the tower = DE = BC = 40 m

In ∆ABE,

∴AB = BE√3….(i)

In ∆CBE,

tan 30° =

Substituting BE in (i),

AB = 40√3 × √3

= 120 m

Height of the chimney = AB + BC = 120 + 40 = 160 m

Yes, the height of the chimney meets the pollution control norms.

### Solution 42

Let the ships be at B and C.

In D ABD,

∴ BD = 200 m

In D ADC,

Distance between the two ships = BC = BD + DC

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Here m∠CAB = m∠FEB = 30°.

Let BC = h m, AC = x m

In D ADE,

In D BAC,

Height of the second pole is 15.34 m

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Let AQ be the tower and R, S respectively be the points which are 4m, 9m away from base of tower.

As the height can not be negative, the height of the tower is 6 m.

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### Solution 56 (i)

Let AB be the cliff, so AB=150m.

C and D are positions of the boat.

DC is the distance covered in 2 min.

∠ACB = 60^{o} and ∠ADB = 45^{o}

∠ABC = 90^{o}

In ΔABC,

tan(∠ACB)=

In ΔABD,

tan(∠ADB)=

So, DC=BD - BC

=

Now,

### Solution 56 (ii)

Let AB be the lighthouse and C be the position of man initially.

Suppose, a man changes his position from C to D.

As per the question, we obtain the following figure

Let speed of the boat be x metres per minute.

Therefore, CD = 2x

Using trigonometry, we have

Also,

Hence, speed of the boat is 57.8 m.

### Solution 57

AB is the tower.

DC is the distance between cars.

AB=120m

In ΔABC,

tan(∠ACB) =

In ΔABD,

tan(∠ADB) =

So, DC=BD+BC

### Solution 58

Let CD be the tower.

So CD =15m

AB is the distance between the points.

∠CAD = 60^{o} and ∠CBD = 45^{o}

∠ADC = 90^{o}

In ΔADC,

tan(∠CAD)=

In ΔCBD,

tan(∠CBD)=

So AB=BD - AD

### Solution 59

Now, in triangle APB,

sin 60^{o} = AB/ BP

√3/2 = h/ BP

This gives

h = 14.64 km

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## Some Applications of Trigonometry Exercise 12.41

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Wire BD

ED || AC

So, EA = DC and ED = AC

EA = 14

AB = EA + EB

20 = 14 + EB

EB = 6

So, the correct option is (a).

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## Some Applications of Trigonometry Exercise 12.42

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If height of one person is x then height of another one is 2x. Also If angle of elevation of one is θ then for another it is 90 - θ.

AB = a

C is mid point.

So, the correct option is (d).

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If height of one pole is x then height of the other one is 2x. Also If the angle of elevation of one is θ then for the other it is

90 - θ.

AB = a

C is mid point.

So, the correct option is (b).

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EC || AB

Hence

EA = CB = 10

AD = AE + ED

ED = AD - AE

= 16 - 10 = 6

So, the correct option is (c).

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## Some Applications of Trigonometry Exercise 12.43

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From the figure, it is cleared that we have to find the length of BC.

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