# Chapter-Test Heights and Distances ML Aggarwal Solutions ICSE Class-10

Chapter-Test Heights and Distances ML Aggarwal Solutions ICSE Class-10 Chapter-20. We Provide Step by Step Answer of Exercise-20 with MCQs and Chapter-Test of **Heights and Distances** Questions / Problems related for ICSE Class-10 APC Understanding Mathematics . Visit official Website **CISCE ** for detail information about ICSE Board Class-10.

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 10th |

Chapter-20 | Heights and Distances |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exercise – 20 , MCQS , and Chapter-Test |

Academic Session | 2021-2022 |

## Chapter-Test Heights and Distances ML Aggarwal Solutions ICSE Class-10

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**Chapter-Test **

** ****How to Solve Heights and Distances Problems/Questions / Exercise of ICSE Class-10 Mathematics**

Before viewing Answer of Chapter-20 **Heights and Distances** of ML Aggarwal Solutions. Read the Chapter Carefully with formula and then solve all example of Exe-20 given in your text book . For more practice on **Heights and Distances** related problems /Questions / Exercise try to solve **Heights and Distances** exercise of other famous publications also such as Goyal Brothers Prakshan (RS Aggarwal ICSE) / Concise Selina Publications ICSE Mathematics. Get the Formula of **Heights and Distances** for ICSE Class 10 Maths to understand the topic more clearly in effective way.

**Chapter-Test ML Aggarwal Solutions Heights and Distances for ICSE Maths Chapter 20**

(Page 481)

**Question 1**

**The angle of elevation of the top of a tower from a point A (on the ground) is 30°. On walking 50 m towards the tower, the angle of elevation is found to be 60°. Calculate**

**(i) the height of the tower (correct to one decimal place).**

**(ii) the distance of the tower from A.**

**Answer 1**

**
**Let TR be the tower and A is a point on the ground

and angle of elevation of the top of tower = 30°

AB = 50 m

and from B, the angle of elevation is 60°

Let TR = h and AR = x

BR = x – 50

In right triangle ATR

tan θ = TR/AR

tan 30^{0} = h/x

1/√3 = h/x

⇒ x = √3h **…(1)**

In right triangle BTR

tan θ = TR/BR

tan 60^{0} = h/(x – 50)

√3 = h/(x – 50)

⇒ h = √3 (x – 50) **…(2)**

Using both the equations

⇒ h = √3 (√3h – 50)

h = 3h – 50√3

⇒ 2h = 50√3

⇒ h = 25 √3

h = 25 × 1.732 = 43.3

Now substituting the values of h in equation (1)

x = √3 × 25√3

⇒ x = 25 × 3

⇒ x = 75

Height of the tower = 43.3 m

Distance of A from the foot of the tower = 75 m

**Question 2**

**An aeroplane 3000 m high passes vertically above another aeroplane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the two planes.**

**Answer 2**

Let A and B are two aeroplanes

and P is a point on the ground such that

angles of elevations from A and B are 60° and 45° respectively.

AC = 3000 m

Let AB = x

∴ BC = 3000 – x

Let PC = y

In right triangle APC

tan θ = AC/PC

tan 60^{0} = 3000/y

√3 = 3000/y

⇒ y = 3000/√3 **…(1)**

In right triangle BPC

tan θ = BC/PC

tan 45^{0} = (3000 – x)/y

1 = (3000 – x)/ y

⇒ y = 3000 – x

Using equation (1)

3000/√3 = 3000 – x

x = 3000 – 3000/√3

Multiply and divide by √3

x = 3000 – (3000 × √3)/(√3× √3)

So we get

x = 3000 – 1000 (1.732)

⇒ x = 3000 – 1732

⇒ x = 1268 m

**Question 3**

**A 7m long flagstaff is fixed on the top of a tower. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 45° and 36° respectively. Find the height of the tower correct to one place of demical.**

**Answer 3**

Let TR be the tower and PT is the flag on it such that PT = 7m

Let TR = h and AR = x

Angles of elevation from P and T are 45° and 36° respectively.

Now in right ∆PAR

tan θ = PR/AR

tan 45^{0} = (7 + h)/x

1 = (7 + h)/x

⇒ x = 7 + h **…(1)**

In right triangle TAR

tan θ = TR/AR

tan 36^{0} = h/

0.7265 = h/x

⇒ h = x (0.7265) **…(2)**

Using both the equations

h = (7 + h) (0.7265)

h = 7 × 0.7265 + 0.7265h

⇒ h – 0.7265h = 7 × 0.7265

0.2735h = 7 × 0.7265

By division

h = (7 × 0.7265)/0.2735

h = (7 × 7265)/2735

⇒ h = 18.59 = 18.6 m

**Question 4 Chapter-Test Heights and Distances ML Aggarwal **

**A boy 1.6 m tall is 20 m away from a tower and observes that the angle of elevation of the top of the tower is 60°. Find the height of the tower.**

**Answer 4**

Let AB be the boy and TR be the tower

∴ AB = 1.6 m

Let TR = h

from A, show AE || BR

∴ ER = AB = 1.6 m

TE = h – 1.6

AE = BR = 20 m

In right triangle TAE

tan θ = TE/AE

Substituting the values

tan 60^{0} = (h – 1.6)/20

So we get

√3 = (h – 1.6)/20

⇒ h – 1.6 = 20√3

⇒ h = 20√3 + 1.6

⇒ h = 20 (1.732) + 1.6

By further calculation

h = 34.640 + 1.6

h = 36.24

∴ Height of tower = 36.24 m

**Question 5**

**A boy 1.54 m tall can just see the sun over a wall 3.64 m high which is 2.1 m away from him. Find the angle of elevation of the sun.**

**Answer 5**

Let AB be the boy and CD be the wall which is at a distance of 2.1 m

Hence, the angle of elevation of the sun is 45^{0}.

**Question 6**

**An aeroplane is flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.**

**Answer 6**

A and D are the two positions of the aeroplane ;

AB is the height and P is the point

∴ AB = 1 km,

Let AD = x and PB = y

and angles of elevation from A and D at point P are 60° and 30° respectively.

x = √3 – 1/√3

⇒ x = (3 – 1)/√3

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