# NCERT Solutions Class 10 Maths Chapter 9

## NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Trigonometry is a complex topic. The NCERT book for Class 10 Mathematics has explained Trigonometry in the simplest and most comprehensive way to students. There are practice questions at the end of the chapter to help students revise their concepts and prepare for the exams.

Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 9 to help students so that they can study from it independently without any assistance from the teachers or parents. It will ensure that even the minutest doubts are resolved. In fact the students would be interested in learning and mastering the topic with ease.  solutions are prepared by subject experts at Extramarks, who have explained answers in simple language along with the right examples and diagrams while ensuring that they are accurate and adhere to the guidelines laid by CBSE.

## NCERT Solutions for Class 10 Maths

CBSE Class 10 is a good balance of practical and theoretical subjects. And every student’s sole aim is to maximise their scores in every subject. Though securing full marks in theoretical subjects can be difficult, subjects like Mathematics give students a fair shot at improving their Class 10 Board score. Just like practising is key to good marks, having access to exceptional study material is equally important too. NCERT Solutions for Class 10 Mathematics Chapter 9 can help students to prepare for their board exams. Here are the chapters covered in Class 10 Mathematics:

•     Chapter 1 – Real Numbers
•     Chapter 2 – Polynomials
•     Chapter 3 – Pair of Linear Equations in Two Variables
•     Chapter 4 – Quadratic Equations
•     Chapter 5 – Arithmetic Progressions
•     Chapter 6 – Triangles
•     Chapter 7 – Coordinate Geometry
•     Chapter 8 – Introduction to Trigonometry
•     Chapter 9 – Some Applications of Trigonometry
•     Chapter 10 – Circles
•     Chapter 11 – Constructions
•     Chapter 12 – Areas Related to Circles
•     Chapter 13 – Surface Areas and Volumes
•     Chapter 14 – Statistics
•     Chapter 15 – Probability

In NCERT Class 10 Mathematics Chapter 9, a student will study how to use trigonometry to find out the height, distance, and related elements of different shapes or objects by only using formulas not actually measuring them. It is a vast chapter with a lot of tricky problems,  difficult concepts, and formulas, and it requires patience, perseverance, and practice to get it right.

### Some Applications of Trigonometry

The chapter is broadly divided into three sections:

• The first section or Section 9.1: Introduction to trigonometry where the student will learn all about instances from the past, why’s, what’s and how’s of trigonometry and its application in practical life.
• Second section or Section 9.2 and  Exercise 9.1: This part explains in-depth heights and distances along with the junctures where trigonometry basics can be applied to make measuring height and distance accurate. This section is also followed by an elaborate exercise at the end.
• Third section or Section 9.3: The third section comprises a summary of the chapter. This part is beneficial for last-minute revision.

NCERT Solutions for Class 10 Mathematics Chapter 9 includes simple explanations of these sections along with stepwise answers to exercises in the second section

### What is Trigonometry?

Trigonometry is a chapter that is dreaded by most Class 10 students. This is because of the numerous formulas and complicated concepts involved in it. But Trigonometry is one of those subjects that holds prime importance in history. It is a subject widely studied by different scholars around the globe. Trigonometry was founded to cater to the requirement of calculating the distance from Earth to the stars and other planets. Eventually, Trigonometry was also used to find out the height of Mount Everest and is now widely used for navigation purposes like map construction.

As also conveyed in NCERT Class 10 Mathematics Chapter 9, trigonometry was a surveyor’s best tool for centuries. In fact, using trigonometry, the “ Great Trigonometric Survey” of British India fabricated the two largest theodolites to ever exist. These theodolites are now displayed in a museum in Dehradun. Furthermore, trigonometry also led to the discovery of the largest mountain Mount Everest in India in 1852.

### Theodolites

Theodolite is a surveying instrument used to measure angles with a rotating telescope.

### Heights and Distances

The height and distance of an object are derived with the help of trigonometric ratios.

### Basic Terminologies related to Trigonometry Ratios

The trigonometric ratio of a specific angle is the ratio of the sides of a right-angle triangle in terms of any of its acute angle triangles.

In terms of ∠C, the ratio of trigonometry is given as:

Sine – The sine of an angle is the ratio of the perpendicular side to that of the hypotenuse side of the  angle.

Cosine- The cosine of an angle is the ratio of the adjacent side to the hypotenuse side of the angle.

Tangent – The tan of an angle is the ratio of the perpendicular side adjacent to the angle.

Cosecant- Cosecant is the reciprocal of sine.

Therefore, Cosec C = Hypotenuse side/Opposite side

Secant- Secant is the reciprocal of cosine.

Therefore, Sec C= Hypotenuse side/Adjacent side

Cotangent- Cotangent is the reciprocal of tangent.

Therefore, Cot C = Adjacent side/Opposite side

### Trigonometric Ratio Table

 ∠C 0° 30° 45° 60° 90° Sin C 0 1/2 1/√2 √3/2 1 Cos C 1 √3/2 1/2 1/2 0 Tan C 0 1/√3 1 √3 Not defined Cosec C Not defined 2 √2 2/√3 1 Sec C 1 2/√3 √2 2 Not defined Cot C Not defined √3 1 1/√3 0

## Important features of the NCERT Solutions for Class 10 Maths Chapter 9- Some Applications of Trigonometry

The important features of the NCERT Solutions for Class 10 Mathematics Chapter 9 by Extramarks are given below:

• NCERT solutions for class 10 Mathematics chapter 9 are prepared by subject matter experts and experienced faculty thus students can be rest assured that they are referring to reliable and authentic learning material.
• The answers in NCERT solutions Class 10 Mathematics Chapter 9 are as per the CBSE guidelines.
• All answers are written in a step-by-step manner with detailed explanation. .. Followed by diagrams and clarifications on calculations if necessary.

.It helps to build their mathematical and problem-solving abilities and enhances their confidence in answering questions in the exams.

## Why are Some Applications of Trigonometry Important?

From the examination perspective, the chapter Applications of Trigonometry is important to solve the questions based on it. But from a practical point of view, the chapter is important to measure the height of towers/mountains, determine the distance of shore from sea or gauge the distance between two celestial bodies.

Solving such problems demands a thorough understanding of trigonometric formulas along with trigonometric ratios, relations, and angle values. This can be mastered by simply referring to NCERT Solutions for Class 10 Mathematics Chapter 9 by Extramarks. It provides complete information and guidance so that students don’t have to look for answers elsewhere.

#### Section 9.1 – Introduction

Section 9.1 – Introduction of class 10 Mathematics Chapter 9 starts by explaining the applications of trigonometry i.e., where and why it is used by scholars around the globe. It also conveys why trigonometry is a more practical and cost-effective way to find the height and distance of objects instead of physically measuring them.

#### Section 9.2: Height and Distance

Section 9.2: Height and Distance of class 10 Mathematics chapter 9 has extensive concepts like the line of sight, horizontal level, angle of elevation, and angle of depression. It also includes examples and diagrams for students so that they will be able to understand every concept and answer any question easily.. This in turn helps students to solve questions with fewer mistakes.

#### Exercise 9.1: “Heights and Distances”

As stated above, Exercise 9.1 of class 10 Mathematics chapter 9 includes 16 questions based on the concept of Height and Distance.

#### Section 9.3: Summary

Section 9.3: Summary of class 10 Mathematics chapter 9 is a synopsis of the entire chapter. A really helpful section to brush up on the basics and try to get the solution based on your understanding and knowledge. .

## Key Terminology Related to  Height and Distance

Line of Sight – A line drawn from the eye of an observer to the point on the object is called a line of sight.

The Angle of Elevation – The angle formed between the horizontal line and line of sight(when upward from the horizontal line) is called the angle of elevation.

The Angle of Depression – The angle drawn between the horizontal line and line of sight(if it lies downward) is called an angle of depression.

### Benefits of NCERT Solutions for Class 10 Maths

• NCERT Solutions have answers to all the practice questions given in Class 10 NCERT textbook.
• Solutions are helpful material from a revision and last-minute preparation perspective.
• It is a reliable study material, prepared by subject matter experts who thoroughly researched the topic and curated it in accordance with CBSE examination guidelines. When students study from it, they will get an edge over their peers. .

### Related Questions

1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. (see fig. 9.11)

Solution:

The length of the rope is 20 m and the angle made by the rope with the ground level is 30°.

Given: AC = 20 m and angle C = 30°

To Find: Height of the pole

Let AB be the vertical pole

In right ΔABC, using the sine formula

sin 30° = AB/AC

Using the value of sin 30 degrees is ½, we have

1/2 = AB/20

AB = 20/2

AB = 10

Therefore, the height of the pole is 10 m.

Q.1 A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see the following figure). Ans. $In the given figure above, we have sin 30°= AB AC = AB 20 or 1 2 = AB 20 or AB= 20 2 =10 Therefore, height of the pole is 10 m. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@A550@$

Q.2 A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Ans.

Let the tree AB bends at point C. Foot of the tree is at A and CB’ is the broken part of the tree i.e, BC = CB’. $\begin{array}{l}\text{Now, in the above figure, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}30°}=\frac{\text{AC}}{\text{AB’}}=\frac{\text{AC}}{8}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{AC}}{8}\\ \text{or AC}=\frac{8}{\sqrt{3}}\\ \text{Also,}\\ \text{cos\hspace{0.17em}30°}=\frac{\text{AB’}}{\text{CB’}}=\frac{\text{8}}{\text{CB’}}\\ \text{or}\frac{\sqrt{3}}{2}=\frac{\text{8}}{\text{CB’}}\\ \text{or CB’}=\frac{16}{\sqrt{3}}\\ \text{Therefore, height of the tree}=\text{AC}+\text{CB’}=\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}=8\sqrt{3}\text{\hspace{0.17em}\hspace{0.17em}m}\end{array}$

Q.3 A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Ans.

As per given information, we have the following figure. $\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin\hspace{0.17em}30°}=\frac{\text{AB}}{\text{AC}}=\frac{1.5}{\text{AC}}\\ \text{or}\frac{1}{2}=\frac{1.5}{\text{AC}}\\ \text{or AC}=1.5×2=3\text{m}\\ \text{Therefore, length of the slide for the children below}\\ \text{the age of 5 years is}3\text{m.}\end{array}$

Also, for elder children, we have the following slide. $\begin{array}{l}\text{In}\mathrm{\Delta }\text{PQR, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin\hspace{0.17em}}6\text{0°}=\frac{\text{PQ}}{\text{PR}}=\frac{3}{\text{PR}}\\ \text{or}\frac{\sqrt{3}}{2}=\frac{3}{\text{PR}}\\ \text{or PR}=\frac{3×2}{\sqrt{3}}=2\sqrt{3}\text{m}\\ \text{Therefore, length of the slide for the elder children is}2\sqrt{3}\text{m.}\end{array}$

Q.4 The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Ans.

Let PQ is a tower with foot at Q and R is a point on the ground such that distance between Q and R is 30 m. $\begin{array}{l}\text{In}\mathrm{\Delta }\text{PQR, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}30°}=\frac{\text{PQ}}{\text{QR}}=\frac{\text{PQ}}{\text{30}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{PQ}}{\text{30}}\\ \text{or PQ}=\frac{30}{\sqrt{3}}=10\sqrt{3}\text{m}\\ \text{Therefore, height of the tower is }10\sqrt{3}\text{m}.\end{array}$

Q.5 A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Ans.

Let the kite is flying at point A and AC is the string of the kite tied at point C on the ground. $\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sin\hspace{0.17em}}6\text{0°}=\frac{\text{AB}}{\text{AC}}=\frac{60}{\text{AC}}\\ \text{or}\frac{\sqrt{3}}{2}=\frac{60}{\text{AC}}\\ \text{or AC}=\frac{60×2}{\sqrt{3}}=40\sqrt{3}\text{m}\\ \text{Therefore, length of the string is}40\sqrt{3}\text{m.}\end{array}$

Q.6 A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Ans.

Let AB be a building of height 30 m. Let the boy is initially at D and then from D moves to point E. We have to find the distance DE $\begin{array}{l}\text{In}\mathrm{\Delta }\text{ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}6\text{0°}=\frac{\text{AC}}{\text{EC}}=\frac{28.5}{\text{EC}}\\ \text{or}\sqrt{3}=\frac{28.5}{\text{EC}}\\ \text{or EC}=\frac{28.5}{\sqrt{3}}\text{m}\\ \text{Now,}\\ \text{In}\mathrm{\Delta }\text{ACD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}3\text{0°}=\frac{\text{AC}}{\text{CD}}=\frac{28.5}{\text{EC}+\text{DE}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{28.5}{\text{EC}+\text{DE}}\\ \text{or EC}+\text{DE}=28.5\sqrt{3}\text{}\\ \text{or DE}=28.5\sqrt{3}\text{}-\text{EC}=28.5\sqrt{3}-\frac{28.5}{\sqrt{3}}\\ \text{or DE}=\frac{28.5×3-28.5}{\sqrt{3}}=\frac{57}{\sqrt{3}}=\frac{19×3}{\sqrt{3}}=19\sqrt{3}\text{m}\\ \text{Therefore, the required distance is}19\sqrt{3}\text{m.}\end{array}$

Q.7 From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Ans.

Let BC be a building on which a transmission tower AB is fixed. Let D be a point on the ground. $\begin{array}{l}\text{In}\mathrm{\Delta }\text{BCD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}45\text{°}=\frac{\text{BC}}{\text{DC}}=\frac{20}{\text{DC}}\\ \text{or}1=\frac{20}{\text{DC}}\\ \text{or DC}=20\text{m}\\ \text{Now, in}\mathrm{\Delta }\text{\hspace{0.17em}ACD, we have}\\ \text{tan\hspace{0.17em}}60\text{°}=\frac{\text{AC}}{\text{DC}}=\frac{\text{AC}}{20}\\ \text{or}\sqrt{3}=\frac{\text{AC}}{20}\\ \text{or AC}=20\sqrt{3}\text{}\\ \text{or AC}=\text{AB}+\text{BC}=20\sqrt{3}\text{}\\ \text{or AB}=20\sqrt{3}-\text{BC}=20\sqrt{3}-\text{20}=20\left(\sqrt{3}-1\right)\\ \text{Therefore, height of the tower is}20\left(\sqrt{3}-1\right)\text{m.}\end{array}$

Q.8 A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Ans.

Let AB be the statue on the pedestal BC. Let D be a point on the ground. $\begin{array}{l}\text{In}\mathrm{\Delta }\text{BCD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}45\text{°}=\frac{\text{BC}}{\text{DC}}\\ \text{or}1=\frac{\text{BC}}{\text{DC}}\\ \text{or DC}=\text{BC}\\ \text{Now, in}\mathrm{\Delta }\text{\hspace{0.17em}ACD, we have}\\ \text{tan\hspace{0.17em}}60\text{°}=\frac{\text{AC}}{\text{DC}}=\frac{\text{AB}+\text{BC}}{\text{BC}}\\ \text{or}\sqrt{3}=\frac{\text{1.6}+\text{BC}}{\text{BC}}\\ \text{or}\sqrt{3}\text{\hspace{0.17em}BC}=\text{1.6}+\text{BC}\\ \text{or BC}\left(\sqrt{3}-1\right)\text{\hspace{0.17em}}=\text{1.6}\\ \text{or BC}=\frac{\text{1.6}}{\left(\sqrt{3}-1\right)}=\frac{\text{1.6}}{\left(\sqrt{3}-1\right)}×\frac{\sqrt{3}+1}{\sqrt{3}+1}=0.8\left(\sqrt{3}+1\right)\text{m}\\ \text{Therefore, height of the pedestal is}0.8\left(\sqrt{3}+1\right)\text{m.}\end{array}$

Q.9 The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Ans.

Let AB be the height of the building and CD be the height of the tower. $\begin{array}{l}\text{In}\mathrm{\Delta }\text{BCD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}60\text{°}=\frac{\text{CD}}{\text{BD}}\\ \text{or}\sqrt{3}=\frac{\text{50}}{\text{BD}}\\ \text{or BD}=\frac{\text{50}}{\sqrt{3}}\\ \text{Now, in}\mathrm{\Delta }\text{\hspace{0.17em}ABC, we have}\\ \text{tan\hspace{0.17em}}30\text{°}=\frac{\text{AB}}{\text{BD}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{AB}}{\frac{\text{50}}{\sqrt{3}}}\\ \text{or AB}=\frac{1}{\sqrt{3}}×\frac{\text{50}}{\sqrt{3}}=\frac{\text{50}}{3}=16\frac{2}{3}\\ \text{Therefore, height of the building is}16\frac{2}{3}\text{m.}\end{array}$

Q.10 Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Ans.

Let AB and CD be the two poles of equal height. Let O be a point on the road from where elevation angles are measured. $\begin{array}{l}\text{In}\mathrm{\Delta }\text{\hspace{0.17em}ABO, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}}60\text{°}=\frac{\text{AB}}{\text{BO}}\\ \text{or}\sqrt{3}=\frac{\text{AB}}{\text{BO}}\\ \text{or BO}=\frac{\text{AB}}{\sqrt{3}}\\ \text{Now, in}\mathrm{\Delta }\text{\hspace{0.17em}CDO, we have}\\ \text{tan\hspace{0.17em}}30\text{°}=\frac{\text{CD}}{\text{OD}}\\ \text{or}\frac{1}{\sqrt{3}}=\frac{\text{CD}}{\text{OD}}=\frac{\text{CD}}{\text{BD}-\text{BO}}=\frac{\text{CD}}{\text{80}-\text{BO}}\\ \text{or CD}\sqrt{3}=\text{80}-\text{BO}=\text{80}-\frac{\text{AB}}{\sqrt{3}}\\ \text{or AB}\sqrt{3}=\text{80}-\frac{\text{AB}}{\sqrt{3}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{AB}=\text{CD as heights of the poles are equal]}\\ \text{or AB}\sqrt{3}+\frac{\text{AB}}{\sqrt{3}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{80}\\ \text{or}\frac{4\text{AB}}{\sqrt{3}}=80\\ \text{or AB}=20\sqrt{3}\\ \text{Therefore, height of each pole is}20\sqrt{3}\text{m.}\\ \text{BO}=\frac{\text{AB}}{\sqrt{3}}=\frac{20\sqrt{3}}{\sqrt{3}}=20\text{m}\\ \text{OD}=\text{BD}-\text{BO}=80-20=60\text{m}\\ \text{Therefore, the required distances are 20 m and 60 m.}\end{array}$

Q.11 A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the following figure). Find the height of the tower and the width of the canal. Ans.

We have the following figure. $\begin{array}{l}\text{In ΔABD, we have}\\ \text{tan\hspace{0.17em}30° =}\frac{\text{AB}}{\text{DB}}\text{=}\frac{\text{AB}}{\text{BC+CD}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{AB}}{\text{BC+20}}\\ \text{or BC+20 = AB}\sqrt{\text{3}}\\ \text{or BC = AB}\sqrt{\text{3}}-\text{20\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(1)}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60° =}\frac{\text{AB}}{\text{BC}}\\ \text{or}\sqrt{\text{3}}\text{=}\frac{\text{AB}}{\text{BC}}\\ \text{or AB = BC}\sqrt{\text{3}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(2)}\\ \text{From equations (1) and (2), we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC = AB}\sqrt{\text{3}}-\text{20}\\ \text{or BC = BC}\sqrt{\text{3}}\text{\hspace{0.17em}×}\sqrt{\text{3}}-\text{20=3BC-20}\\ \text{or 3BC}-\text{BC = 20}\\ \text{or 2BC = 20}\\ \text{or BC = 10 m}\\ \text{On putting this value of BC in equation (2), we get}\\ \text{AB = 10}\sqrt{\text{3}}\text{m}\\ \text{Therefore, height of the tower is 10}\sqrt{\text{3}}\text{m and width of}\\ \text{the canal is 10 m.}\end{array}$

Q.12 From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Ans.

Let AB is a cable tower and CD is a building. $\begin{array}{l}\text{We have, CD = BE = 7 m, DB = CE}\\ \text{In Δ\hspace{0.17em}CDB, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}45° = }\frac{\text{CD}}{\text{DB}}\\ \text{or 1 = }\frac{\text{7}}{\text{DB}}\\ \text{or DB = 7 m}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60° = }\frac{\text{AE}}{\text{CE}}\\ \text{or}\sqrt{\text{3}}\text{ = }\frac{\text{AE}}{\text{CE}}\\ \text{or AE = CE}\sqrt{\text{3}}\text{ = DB}\sqrt{\text{3}}\text{ = 7}\sqrt{\text{3}}\text{m\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}[CE\hspace{0.17em}= DB]}\\ \text{Therefore,}\\ \text{height of the tower = AB = AE+BE = 7}\sqrt{\text{3}}\text{+7 = 7(1+}\sqrt{\text{3}}\text{) m}\end{array}$

Q.13 As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Ans. $\begin{array}{l}\text{Let AB be a lighthouse and ships be at points C and D.}\\ \text{It is given that AB = 75 m. We have to find the distance CD.}\\ \text{In Δ\hspace{0.17em}ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}45° =}\frac{\text{AB}}{\text{BC}}\\ \text{or 1 =}\frac{\text{AB}}{\text{BC}}\\ \text{or BC = AB = 75 m}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ABD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}30° =}\frac{\text{AB}}{\text{DB}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{75}}{\text{DB}}\\ \text{or DB = 75}\sqrt{\text{3}}\\ \text{or DB = BC+CD = 75}\sqrt{\text{3}}\\ \text{or CD = 75}\sqrt{\text{3}}-\text{BC = 75}\sqrt{\text{3}}-\text{75 = 75\hspace{0.17em}(}\sqrt{\text{3}}-\text{1)}\\ \text{Therefore,}\\ \text{distance between the two ships = 75\hspace{0.17em}(}\sqrt{\text{3}}-\text{1) m}\end{array}$

Q.14 A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see the following figure). Find the distance travelled by the balloon during the interval. Ans. $\begin{array}{l}\text{Let the balloon be initially at A and then it moves to B.}\\ \text{Let CD be the girl.}\\ \text{In Δ\hspace{0.17em}ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60° =}\frac{\text{AE}}{\text{CE}}\\ \text{or}\sqrt{\text{3}}\text{=}\frac{\text{AE}}{\text{CE}}\\ \text{or CE =}\frac{\text{AE}}{\sqrt{\text{3}}}\text{=}\frac{\text{AF}-\text{EF}}{\sqrt{\text{3}}}\text{=}\frac{\text{88.2}-\text{1.2}}{\sqrt{\text{3}}}\text{=}\frac{\text{87}}{\sqrt{\text{3}}}\text{= 29}\sqrt{\text{3}}\\ \text{In Δ\hspace{0.17em}BCG, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}30° =}\frac{\text{BG}}{\text{CG}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{88.2}-\text{1.2}}{\text{CG}}\\ \text{or CG = 87}\sqrt{\text{3}}\text{m}\\ \text{Therefore,}\\ \text{distance travelled by the balloon = EG}\\ \text{ = CG-CE}\\ \text{ = 87}\sqrt{\text{3}}-\text{29}\sqrt{\text{3}}\\ \text{ = 58}\sqrt{\text{3}}\text{m}\end{array}$

Q.15 A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Ans. $\begin{array}{l}\text{Let AB be a man and BC be a tower. Let the car is initially}\\ \text{at D and after 6 seconds it moves to point E.}\\ \text{In Δ\hspace{0.17em}ACE, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}60° =}\frac{\text{AC}}{\text{CE}}\\ \text{or}\sqrt{\text{3}}\text{=}\frac{\text{AC}}{\text{CE}}\\ \text{or AC =}\sqrt{\text{3}}\text{\hspace{0.17em}CE}...\text{(1)}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ACD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}30° =}\frac{\text{AC}}{\text{CD}}\\ \text{or}\frac{\text{1}}{\sqrt{\text{3}}}\text{=}\frac{\text{AC}}{\text{CD}}\\ \text{or CD = AC}\sqrt{\text{3}}\\ \text{or CD =}\sqrt{\text{3}}\text{×}\sqrt{\text{3}}\text{\hspace{0.17em}CE = 3CE}\left[\text{from (1), AC =}\sqrt{\text{3}}\text{\hspace{0.17em}CE}\right]\\ \text{or DE+CE = 3CE}\\ \text{or DE = 2CE}\\ \text{or CE =}\frac{\text{DE}}{\text{2}}\\ \text{The car takes 6 seconds to cover the distance DE.}\\ \text{Therefore, it will take 3 seconds to cover the half of the}\\ \text{distance DE i.e., the car will take 3 seconds to reach the}\\ \text{foot of the tower from the point E.}\end{array}$

Q.16 The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Ans.

Let AB is a tower. Let C and D respectively are two points at a distance of 4 m and 9 m from the base of the tower. $\begin{array}{l}\text{Let angles of elevation of the top of the tower from D be θ}\\ \text{and from C be 90°}-\text{θ.}\\ \text{In Δ\hspace{0.17em}ABC, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}(90°}-\text{θ) = }\frac{\text{AB}}{\text{BC}}\\ \text{or cotθ = }\frac{\text{AB}}{\text{4}}\\ \text{or AB = 4cotθ}...\text{(1)}\\ \text{Now,}\\ \text{In Δ\hspace{0.17em}ABD, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}tan\hspace{0.17em}θ = }\frac{\text{AB}}{\text{BD}}\\ \text{or \hspace{0.17em}tan\hspace{0.17em}θ = }\frac{\text{AB}}{\text{9}}\\ \text{or \hspace{0.17em}\hspace{0.17em}AB = 9 tanθ}...\text{(2)}\\ \text{From (1) and (2), we have}\\ \text{9 tanθ = 4cotθ}\\ \text{or \hspace{0.17em}\hspace{0.17em}}\frac{\text{tanθ}}{\text{cotθ}}\text{ = }\frac{\text{4}}{\text{9}}\\ {\text{or \hspace{0.17em}\hspace{0.17em}tan}}^{\text{2}}\text{θ = }\frac{\text{4}}{\text{9}}\\ \text{or \hspace{0.17em}\hspace{0.17em}tanθ = }\frac{\text{2}}{\text{3}}\\ \text{On putting this value of tanθ in (2), we get}\\ \text{AB = 9 tanθ = 9×}\frac{\text{2}}{\text{3}}\text{ = 6 m}\\ \text{Therefore, the height of the tower is 6 m.}\end{array}$