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

## Chapter 12 - Some Applications of Trigonometry Exercise Ex. 12.1

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.

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

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

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.

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

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

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.

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,

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.

AB is the tower.

DC is the distance between cars.

AB=120m

In ΔABC,

tan(∠ACB) =

In ΔABD,

tan(∠ADB) =

So, DC=BD+BC

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

Now, in triangle APB,

sin 60^{o} = AB/ BP

√3/2 = h/ BP

This gives

h = 14.64 km

## Chapter 12 - Some Applications of Trigonometry Exercise 12.41

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).

## Chapter 12 - Some Applications of Trigonometry Exercise 12.42

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).

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).

EC || AB

Hence

EA = CB = 10

AD = AE + ED

ED = AD - AE

= 16 - 10 = 6

So, the correct option is (c).

## Chapter 12 - Some Applications of Trigonometry Exercise 12.43

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

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

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