written as per special scheme of assessment for board

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Written as per Special Scheme of Assessment for Board Examination of Class X for the Session 2021-22 prescribed by CBSE

CBSE

Class X

Mathematics MCQsChapterwise & Subtopicwise for Term - I & II

P.O. No. 3192TEID: 2253

Subtopic wise and Question Type wise segregation of MCQs for efficient practice

Covers Textual Exercise Questions and Exemplar Questions

‘1367’ MCQs Including Questions from previous years board papers

Extensive coverage of all types of MCQs included in latest paper pattern; Stand-Alone,

Assertion-Reason and Case-study based Questions

Quick Review of each chapter to facilitate quick revision

Topic Test along with solutions at the end of every chapter for self-evaluation

Hints provided as pertinent

symbol represents Topics/Subtopics/Questions that are not part of the rationalized

syllabus

Salient Features

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Sample Content

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Target’s “CBSE Class X Mathematics MCQs” is a complete, thorough, critically analysed and extensively drafted book to cater to Special Scheme of Assessment for Class X Board Examination for the session 2021-22 announced by CBSE.

The book contains MCQs based on all the textual chapters (Both Term - I and Term - II). To bring sense of familiarity in the students, the flow of subtopics within the chapters is purposely kept aligned with the NCERT textbook. This book aims to provide comprehensive and thorough preparation material of MCQs to excel in exam. Complete coverage of topics in this book would prove to be a strong source of foundation of practising for Class X Board Examination. To keep students focused on preparation, topics/subtopics cited in the syllabus of Special Scheme of Assessment 2021-22 are mapped against the topics/subtopics of the textbook in the beginning of every chapter. Subtopics which are not part of Rationalized syllabus are marked as .

The Subtopic-wise segregation for each chapter of this book helps the students to practice questions smoothly and as per their own pace.

Each chapter begins with Synopsis to offer crisp revision to students in efficient form of pointers, tables, charts etc. followed by Quick review. Section of ‘Multiple Choice Questions’ consists of an adequate practice of Stand-Alone, Assertion- Reason and Case-study based questions. These questions have been specially created and compiled keeping following objectives in mind – to help students revise concepts as well as to prepare them to solve complex questions which require strenuous effort and understanding of multiple-concepts. The assortment of MCQs is a beautiful blend of straight forward, average and higher order thinking questions.

To aid students, hints are provided for questions wherever deemed necessary. ‘Things to Remember’ help the students gain knowledge required to understand different concepts.

Topic Test along with solutions at the end of the chapter allows students to gauge their preparedness of chapter.

We hope that the book builds up necessary knowledge and skillset in the students and boost their confidence required to succeed in the examination.

- PublisherEdition: First

PREFACE

Sample Content

No. Unit Name Marks I Number Systems 6 II Algebra 10 III Coordinate Geometry 6 IV Geometry 6 V Trigonometry 5 VI Mensuration 4 VII Statistics & Probability 3

Total 40 Internal Assessment 10 Total 50

UNIT-NUMBER SYSTEMS 1. REAL NUMBERS Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and

motivating through examples. Decimal representation of rational numbers in terms of terminating/non-terminating recurring decimals.

UNIT-ALGEBRA 2. POLYNOMIALS Zeroes of a polynomial. Relationship between zeroes and coefficients of quadratic polynomials only. 3. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically - by substitution and by elimination. Simple situational problems. Simple problems on equations reducible to linear equations.

UNIT-COORDINATE GEOMETRY 4. COORDINATE GEOMETRY LINES (In two-dimensions)

Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division)

UNIT-GEOMETRY 5. TRIANGLES Definitions, examples, counter examples of similar triangles. 1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct

points, the other two sides are divided in the same ratio. 2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side. 3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are

proportional and the triangles are similar. 4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are

equal and the two triangles are similar. 5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including

these angles are proportional, the two triangles are similar. 6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the

hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other.

TERM - I

COURSE STRUCTURE 2021 - 22

Sample Content

7. (Motivate) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

9. (Motivate) In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angle opposite to the first side is a right angle.

UNIT- TRIGONOMETRY 6. INTRODUCTION TO TRIGONOMETRY

Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined). Values of the trigonometric ratios of 30, 45 and 60. Relationships between the ratios.

TRIGONOMETRIC IDENTITIES Proof and applications of the identity sin2A + cos2A = 1. Only simple identities to be given. UNIT-MENSURATION 7. AREAS RELATED TO CIRCLES

Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60° and 90° only. Plane figures involving triangles, simple quadrilaterals and circle should be taken.)

UNIT- STATISTICS & PROBABILITY 8. PROBABILITY Classical definition of probability. Simple problems on finding the probability of an event.

No. Unit Name Marks I Algebra(Cont.) 10 II Geometry(Cont.) 9 III Trigonometry(Cont.) 7 IV Mensuration(Cont.) 6 V Statistics & Probability(Cont.) 8 Total 40 Internal Assessment 10 Total 50

UNIT-ALGEBRA 1. QUADRATIC EQUATIONS

Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by factorization, and by using quadratic formula. Relationship between discriminant and nature of roots. Situational problems based on quadratic equations related to day to day activities (problems on equations reducible to quadratic equations are excluded)

2. ARITHMETIC PROGRESSIONS

Motivation for studying Arithmetic Progression. Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems.

(Applications based on sum to n terms of an A.P. are excluded)

TERM - II

Sample Content

UNIT- GEOMETRY 3. CIRCLES Tangent to a circle at, point of contact 1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact. 2. (Prove) The lengths of tangents drawn from an external point to a circle are equal. 4. CONSTRUCTIONS 1. Division of a line segment in a given ratio (internally). 2. Tangents to a circle from a point outside it. UNIT-TRIGONOMETRY 5. SOME APPLICATIONS OF TRIGONOMETRY HEIGHTS AND DISTANCES-Angle of elevation, Angle of Depression.

Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45°, 60°.

UNIT-MENSURATION 6. SURFACE AREAS AND VOLUMES 1. Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres,

hemispheres and right circular cylinders/cones. 2. Problems involving converting one type of metallic solid into another and other mixed problems.

(Problems with combination of not more than two different solids be taken). UNIT-STATISTICS & PROBABILITY 7. STATISTICS

Mean, median and mode of grouped data (bimodal situation to be avoided). Mean by Direct Method and Assumed Mean Method only. Internal Assessment for Term I and Term II:

Internal Assessment Marks Total Marks

Periodic Tests 3 10 marks for the term

Multiple Assessments 2

Portfolio 2 Student Enrichment Activities-practical work 3

Sample Content

Term I

Unit Chapter No. Chapter Name Page No.

I 01 Real Numbers 1

I 02 Polynomials 20

I 03 Pair of Linear Equations in Two Variables 41

IV 06 Triangles 99

III 07 Coordinate Geometry 131

V 08 Introduction to Trigonometry 155

VI 12 Areas Related to Circles 219

VII 15 Probability 306

Term II

Unit Chapter No. Chapter Name Page No.

I 04 Quadratic Equations 65

I 05 Arithmetic Progressions 82

III 09 Some Applications of Trigonometry 176

II 10 Circles 193

II 11 Constructions 214

IV 13 Surface Areas and Volumes 239

V 14 Statistics 270

Note: 1. * mark represents Textual question. 2. mark represents NCERT Exemplar question.

3. symbol represents Topics/Subtopics/Questions that are not part of the

rationalized syllabus for the Special Evaluation Scheme 2021-22.

CONTENTS

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Sample Content

176 176

1

Subtopic No. Subtopic Name Syllabus 9.1 Heights and Distances Heights and Distances : Angle of elevation,

Angle of depression. Simple problems on heights and distances. Problems should not involve more than two right angles. Angles of elevation/depression should be only 30, 45, 60.

9.1 Heights and Distances

i. Line of sight: Line drawn from the eye of the observer to the point on the object being viewed.ii. Angle of elevation: The angle made by the line of sight with

the horizontal when the object being viewed is above thehorizontal level.

iii. Angle of depression: The angle made by the line of sightwith the horizontal when the object being viewed is belowthe horizontal level.

Height of the object from the horizontal and its distance from the observer can be calculated using trigonometric ratios.

Note: i. If an observer is at a height h above the ground viewing an

object O on the ground, then the angle of depression is CAO,and not OAB.

ii. CAO = AOB =

Synopsis

CHAPTER Some Applications of Trigonometry9 Content and Concepts

B Horizontal level

Angle of depression

O (Object)

(Observer)

A

O (Object)

Horizontal level(Observer)Angle of elevation

A C

B

h

Ground

(Observer)

O (Object)

Things to Remember

If the height of an object and the length of its shadow are equal, then the angle of elevation of thesource of light is 45.

If an observer moves towards an object of fixed height, then the angle of elevation of the top of the objectgoes on increasing.

If an observer moves away from an object of fixed height, then the angle of elevation of the top of theobject goes on decreasing.

Sample Content

177

Chapter 9: Some Applications of Trigonometry

9.1 Heights and Distances 1. A pole 6 m high casts a shadow 2 3 m long on

the ground, then the Sun’s elevation is (A) 60 (B) 45 (C) 30 (D) 90 2. If the height of a vertical pole is 3 times the

length of its shadow on the ground, then the angle of elevation of the Sun at that time is

[CBSE 2014] (A) 30 (B) 60 (C) 45 (D) 75 3. The angle of elevation of the top of a tower at a

point on the ground 50 m away from the foot of the tower is 45. Then the height of the tower (in metres) is [CBSE 2014]

(A) 50 3 (B) 50

(C) 502

(D) 503

4. A tower stands vertically on the ground. From a point on the ground which is 25 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 45. Then the height (in metres) of the tower is [CBSE 2011]

(A) 25 2 (B) 25 3 (C) 25 (D) 12.5 5. A kite is flying at a height 30 m from the

ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is [CBSE 2012]

(A) 45 (B) 30 (C) 60 (D) 90 6. A ladder makes an angle of 60 with the ground

when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder (in metres) is [CBSE 2014]

(A) 43

(B) 4 3

(C) 2 2 (D) 4

Quick Review

MULTIPLE CHOICE QUESTIONS

Stand Alone MCQs

Angle of elevation and angle of depression

A D

C

B

E

is angle of elevation of C

from A and is angle of depression of E from A.

Used to find i. Height of an object ii. Distance of an object from

the point of observation.

Angle of elevation

Angle made by the line of sight with the horizontal when the object is above the horizontal level.

Used to find i. Height of object ii. Distance of object from


O (Object)

Horizontal level

Angle of depression

Angle made by the line of sight with the horizontal when the object is below the horizontal level.

Used to find i. Height of an object ii. Distance of an object from


O (Object)

Horizontal level

Applications of Trigonometry

Sample Content

178

178

Class X: Mathematics MCQs

7. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60 with the wall, then the height of the wall is

[CBSE 2013]

(A) 15 3 m (B) 15 32

m

(C) 152

m (D) 15 m 8. A bridge, in the shape of a straight path, makes

an angle of 60o with the width of the river. If the length of the bridge is 100 m, then the width of the river is [CBSE 2012]

(A) 50 m (B) 173.2 m (C) 43.3 m (D) 100 m 9. If two towers of height h1 and h2 subtend angles

of 60 and 30 respectively at the mid-point of the line joining their feet, then h1 : h2 is

[CBSE 2012] (A) 3 : 1 (B) 3 :1 (C) 1: 3 (D) 1 : 3 10. If a 1.6 m tall girl stands at a distance of 3.2 m

from a lamp post and casts a shadow of 4.8 m on the ground, then the height of the lamp post is

(A) 2 m3

(B) 4 m3

(C) 8 m3

(D) 16 m3

11. 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. If the distance between the foot of the tree to the point where the top touches the ground is 8 m, then the height of the tree is

(A) 2 3 m (B) 4 3 m (C) 8 3 m (D) 16 3 m 12. 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 45. If the tower is 30 m high, then the height of the building is

(A) 4 3 m (B) 8 3 m

(C) 10 3 m (D) 20 3 m 13. The shadow of a tower is 30 m long when the

Sun’s elevation is 30, what is the length of the shadow when the Sun’s elevation is 60?

(A) 10 m (B) 10 3 m (C) 20 m (D) 20 3 m

14. If the shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the sun is 60, then the angle of elevation of the sun at the time of the longer shadow is

(A) 30 (B) 45 (C) 60 (D) 90 15. The angle of elevation of the top of a tower

from a point A on the ground is 30. On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to 60. The distance of the tower from point A is

(A) 10 m (B) 20 m (C) 30 m (D) 40 m 16. The shadow of a tower standing on a level plane

is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. The height of the tower is

(A) 10 3 m (B) 15 3 m (C) 20 3 m (D) 25 3 m 17. The angle of elevation of the top of a tower at a

distance of 120 m from a point A on the ground is 45. If the angle of elevation of the top of a flagpole fixed at the top of the tower, at A is 60, then the height of the flagpole is

(Take 3 = 1.73) (A) 70.3 m (B) 84.5 m (C) 87.6 m (D) 104.9 m 18. An aeroplane, when 3000 m high, passes

vertically above another plane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 60 and 45 respectively. The vertical distance between the two aeroplanes is

(Take 3 = 1.732) (A) 1150 m (B) 1268 m (C) 1701 m (D) 2134 m 19. The angle of elevation of an aeroplane from a

point A on the ground is 60. After a flight of 15 seconds, the angle of elevation changes to 30. If the aeroplane is flying at a constant height of 1500 3 m, then the speed of the aeroplane is

(A) 670 km/hr (B) 720 km/hr (C) 750 km/hr (D) 780 km/hr 20. The angle of elevation of the top of a tower from

a point P on the ground is . After walking a distance d towards the foot of the tower, angle of elevation is found to be . Then

[CBSE 2012] (A) < (B) > (C) = (D) None of these

Sample Content

179


21. A ship is sighted at sea from the top of a lighthouse of 75 m height. If the angle of depression is found to be 30, then the distance of ship from the lighthouse (in metres) is

[CBSE 2012]

(A) 25 (B) 75 3 (C) 25 3 (D) 75 2 22. The angle of depression of a car parked on the

road from the top of a 150 m high tower is 30. The distance of the car from the tower (in metres) is [CBSE 2014]

(A) 50 3 (B) 150 3 (C) 150 2 (D) 75 23. In the given figure, if APB = 15 and

QAP = 60, the angle of depression of the object P from the point B is [CBSE 2012]

(A) 15 (B) 30 (C) 45 (D) 60 24. An aeroplane is flying at a height of 300 m

above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45 and 60 respectively. The width of the river is [Take 3 = 1.732]

(A) 173.2 m (B) 273.2 m (C) 473.2 m (D) 573.2 m 25. The horizontal distance between two poles is

15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30. If the height of the second pole is 24 m, then the height of the first pole is

(Take 3 = 1.732) (A) 12.34 m (B) 15.34 m (C) 15.66 m (D) 18.66 m 26. The angles of elevation and depression of the

top and the bottom of a tower from the top of a building, 60 m high, are 30 and 60 respectively. The difference between the heights of the building and the tower is

(A) 10 m (B) 15 m (C) 20 m (D) 25 m

27. From the top of a 60 m high building, the angles of depression of the top and the bottom of a tower are 45 and 60 respectively. The height of the tower is (Take 3 = 1.73)

(A) 15.4 m (B) 25.4 m (C) 34.6 m (D) 54.6 m 28. At a point A, 20 metres above the level of water

in a lake, the angle of elevation of a cloud is 30. The angle of depression of the reflection of the cloud in the lake, at A is 60. The distance of the cloud from A is

(A) 10 m (B) 20 m (C) 30 m (D) 40 m For question numbers 17 to 21, two

statements are given – one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below: (A) Both A and R are true and R is correct

explanation of the Assertion. (B) Both A and R are true but R is not the

correct explanation of the Assertion. (C) A is true but R is false. (D) A is false but R is true. 29. Assertion (A) : If the angle of elevation of the

top of a tower from a point on the ground, which is 50 m away from the foot of the tower, is 60°, then the height of the tower is 50 3 m.

Reason (R) : In a right angled triangle,

tan = side opposite to hypotenuse

. 30. Assertion (A) : A ladder 20 m long just reaches

the top of a vertical wall. If the ladder makes an angle of 60 with the wall, then the height of the wall is 9 m.

Reason (R) : In a right angled triangle,

cos = side adjacent to hypotenuse

.

31. A Satellite flying at height h is watching the top

of the two tallest mountains in Uttarakhand and Karnataka, them being Nanda Devi (height 7,816m) and Mullayanagiri (height 1,930 m). The angles of depression from the satellite, to the top of Nanda Devi and Mullayanagiri are 30 and 60 respectively. If the distance between the peaks of two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains,

Assertion & Reason

Case-Study Based MCQs

A B

C30

30

75m

A B X

P Q

60

15

Sample Content

180

180


i. The distance of the satellite from the top of

Nanda Devi is

(Take 3 = 1.732) (A) 1118.36 km (B) 1137.52 km (C) 1937 km (D) 1025.36 km ii. The distance of the satellite from the top of

Mullayanagiri is (A) 1139.4 km (B) 577.52 km (C) 1937 km (D) 1025.36 km iii. The distance of the satellite from the ground is

(Take 3 = 1.732) (A) 1139.4 km (B) 566.996 km (C) 1937 km (D) 1025.36 km iv. What is the angle of elevation if a man is

standing at a distance of 7816 m from Nanda Devi?

(A) 30 (B) 45 (C) 60 (D) 0 v. If a milestone very far away from Mullayanagiri

mountain, makes an angle of 45 with the top of the mountain, find the distance of this milestone from the mountain.

(A) 1.560 km (B) 1.830 km (C) 1.930 km (D) 2.1590 km

32. The angle of elevation of the top of a chimney from the foot of a tower is 60 and the angle of depression of the foot of the chimney from the top of the tower is 30.

i. Find the measure of SQR. (A) 30 (B) 45 (C) 60 (D) 90 ii. The distance between the foot of chimney and foot

of tower is (A) 10 3 m (B) 20 3 m

(C) 40 3 m (D) 60 3 m iii. Find the height of the chimney. (A) 60 m (B) 80 m (C) 100 m (D) 120 m iv. The ratio of the height of the tower and the

length of its shadow on the ground is 1 : 1. The angle of elevation of the Sun is

(A) 30 (B) 45 (C) 60 (D) 90 v. The angle formed by the line of sight with the

horizontal when the object viewed is above the horizontal level is

(A) corresponding angle (B) angle of elevation (C) angle of depression (D) complete angle

9.1 Heights and Distances 1. (A) 2. (B) 3. (B) 4. (C) 5. (B) 6. (D) 7. (C) 8. (A) 9. (A) 10. (C) 11. (C) 12. (C) 13. (A) 14. (A) 15. (C) 16. (D) 17. (C) 18. (B) 19. (B) 20. (A) 21. (B) 22. (B) 23. (C) 24. (C) 25. (B) 26. (C) 27. (B) 28. (D) 29. (C) 30. (D) 31. i. (A) ii. (C) iii. (B) iv. (B) v. (C) 32. i. (A) ii. (C) iii. (D) iv. (B) v. (B)

Answers to MCQs

RQ 60

P

T S 30

40 m (Chimney)

(Tower)

Satellite

GA

D B Nanda Devi

C S Q

RI

H P

Mullayanagiri

Sample Content

181


9.1 Heights and Distances 1. Let the angle of elevation be . In ABC, B = 90 and A =

tan = BCAB

= 62 3

= 3 tan = tan 60 = 60o

2. Let the angle of elevation be . In ABC, B = 90 and A =

tan = BCAB

= 3xx

= 3 tan = tan 60 = 60o

3. In ABC, B = 90 and C = 45

tan 45 = ABBC

1 = AB50

AB = 50 m 4. In ABC, B = 90 and C = 45

tan 45 = ABBC

1 = AB25

AB = 25 m 5. Let the angle of elevation be . In ABC, B = 90 and A =

sin = BCAC

= 3060

= 12

sin = sin 30 = 30o

6. In ABC, B = 90 and C = 60

BCcos60AC

1 22 AC

AC = 4 m 7. In ABC, B = 90 and A = 60

cos 60 = ABAC

1 AB2 15

AB = 152

m

Hints to MCQs

C

A

B 60

Ladder Wall

2 m

C B

60

A 15 m

(Ladder)

Wall

A

C

B

Pole

Shadow

A

C

B

30 m

String

(Kite)

C

A

B 50 m 45

Tower

Tower

C

A

B 25 m 45

A

C

B

6m (Pole)

2 3

(Shadow)

Sample Content

182

182


8. Let BC be the width of the river and AC be the bridge.

In ABC, B = 90 and C = 60

cos 60 = BCAC

1 BC2 100

BC = 50 m 9. tan 60o = 1h

x

3 = 1hx

h1 = 3 x

tan 30o = 2hx

213

hx

h2 = 3

x

h1 : h2 = 3 : 1 10. Let AB represent the lamp post, PQ represent

the girl and QC be the shadow of girl. In PQC, PQC = 90 and C =

PQtanQC

1.6tan4.8

tan = 13

…(i)

In ABC, B = 90 and C =

ABtanBC

tanBQ QC

h

13 3.2 4.8

h

…[From (i)]

13 8

h

h = 8 m3

11. Let SQ represent the height of the tree. Let the tree break at point P. SP is the broken part of the tree which takes

position PR such that PRQ = 30o. SP = PR ...(i) In PQR, Q = 90 and R = 30

tan 30 = PQQR

1 PQ83

8PQ3

cos 30 = QRPR

3 82 PR

16PR3

Height of the tree = PQ + SP = PQ + PR …[From (i)]

= 8 163 3

= 243

= 24 33 3

= 8 3 m 12.

R Q 30

P

8 m

S

A

B C Q

P

4.8 m3.2 m

1.6 m

(lamp post) h

C 30

A

B

h (Building)

45

30 m (Tower)

D

C

A

B 25 m 60

h1

30 60h2

x x

Sample Content

183


In ABC, ABC = 90 and ACB = 45

tan 45 = ABBC

1 = 30BC

BC = 30 m In BCD, BCD = 90 and DBC = 30

tan 30 = DCBC

1303

h

303

h = 30 33 3

= 10 3 m

13. In ABD, B = 90 and D = 30

ABtan30BD

1 AB303

30AB3

= 30 33 3

AB 10 3 m In ABC, B = 90 and ACB = 60

ABtan60BC

10 33BC

BC = 10 m 14. Let AB represent the tower of height h m and

BC = x be the shadow when the angle of elevation of the sun is 60.

In ABC, B = 90 and ACB = 60

ABtan60BC

3 hx

h = 3 x … (i) In ABD, B = 90 and D =

tan = 3hx

tan = 33

xx

…[From (i)]

tan = 13

tan = tan 30 = 30 15. In ACD, C = 90 and A = 30

DCtan30AC

13

hx

3

xh …(i)

In BCD, C = 90 and DBC = 60

DCtan60BC

3 = DCAC AB

320

h

x

3 20 h x

3 203

x x …[From (i)]

x = 3(x – 20) x = 3x 60 2x = 60 x = 30 m 16.

D

A

B C 30 60

30 m

Tower

Shadow

D

A

BC 60

x3x

h (Tower)

A B C

D

30 60

h (tower)

20 m (x – 20) m x

A B C

D

60o 30o

h (Tower)

50 m x

Sample Content

184

184


In BCD, C = 90o and DBC = 60o

tan 60o = DCBC

3 = DCAC AB

3 = 50

hx

h = 3 (x 50) …(i) In ACD, C = 90o and DAC = 30o

tan 30o = DCAC

13

= hx

h = 3

x …(ii)

3 (x 50) = 3

x ...[From (i)]

x = 3 (x 50) x = 3x – 150 2x = 150 x = 75

h = 3

x ... [From (ii)]

= 753

= 75 33 3

= 25 3 17. In ABC, B = 90 and A = 45

tan 45 = BCAB

1 = BC120

BC = 120 m In ABD, B = 90 and A = 60

tan 60 = BDAB

BC CD3120

1203120

h

120 3 = 120 + h h = (120 1.73) 120 h = 207.6 120 = 87.6 m 18. Let the vertical distance between the two

aeroplanes be h m. In ABC, B = 90 and CAB = 45

BCtan 45AB

1 = BD CDAB

30001AB

h

AB = 3000 – h …(i) In ABD, B = 90 and DAB = 60

BDtan60AB

30003AB

3000AB3

= 3000 33 3

AB 1000 3 3000 – h = 1000 3 …[From (i)] 3000 h = 1000 1.732 3000 – h = 1732 h = 1268 m 19. Let E represent the aeroplane, D be the position

of the aeroplane after 15 seconds. EAB = 60 and DAC = 30 In ABE, ABE = 90 and EAB = 60

tan 60 = BEAB

1500 33AB

...[ BE DC 1500 3 ]

A B

C

D

h

4560

3000 m

A B C

D

E

30o

60

1500 3 m

C

A B

D

4560

120 m

h (flagpole)

Tower

Sample Content

185


AB = 1500 m In ACD, C = 90 and DAC = 30

tan 30 = DCAC

13

= 1500 3AC

AC = 1500 3 = 4500 m BC = AC AB = 4500 1500 = 3000 m Distance travelled in 15 sec = 3000 m

Speed of the aeroplane = distance travelledtime required

= 300015

m/s

= 3000 360015 1000

km/hr

= 720 km/hr 20. If an observer moves towards an object of fixed

height, then the angle of elevation of the top of the object goes on increasing.

21. In ABC, B = 90o and A = 30o

tan 30o = BCAB

1 75AB3

AB = 75 3 m 22. In ABC, B = 90o and A = 30o

tan 30o = BCAB

1 150AB3

AB = 150 3 m 23. In AQP, AQP + APQ + QAP = 180

o

Sum of the anglesof a triangle is 180 .

90 + APQ + 60 = 180 APQ = 30 BPQ = APQ + BPA = 30 + 15 = 45 seg AX || seg QP PBX = BPQ = 45 ...[Alternate angles]

24. Let AB represent the height of aeroplane A

above the ground and C and D be the two points on the banks of the river.

In ABC, ABC = 90 and C = 45

tan 45 = ABBC

1 = 300y

y = 300 In ABD, ABD = 90 and D = 60

tan 60 = ABBD

3003 x

x = 3003

= 300 33 3

x = 100 3 Width of the river = x + y = 100 3 + 300 = 100 (1.732) + 300 = 173.2 + 300 = 473.2 m 25. Let AB and EC represent the two poles. Let the height of first pole be h m. seg AD || seg EF and AE intersects them FEA = EAD = 30 …[Alternate angles] ABCD is a rectangle. AD = BC = 15 m and AB = DC = h m In EAD, EDA = 90 and EAD = 30

EDtan30AD

13

= EC CD15

1 24153

h

300 m

45 60

60 45

A

B C D x y

A

B C

D

E30

30 24 m (pole)

15 m

F

(pole) hA B

C30

30

150 m (Tower)

Sample Content

186

186


15243

h

24 h = 15 33 3

24 5 3h 24 (5 1.732) = h 24 8.66 = h h = 15.34 m 26. Let AB and EC represent the building and the

tower respectively. EAD = 30 and DAC = 60 ABCD is a rectangle. AB = DC = 60 m and AD = BC Let the height of the tower be h m. In ADC, ADC = 90 and DAC = 60

tan 60 = DCAD

603AD

…[ DC = AB = 60]

AD = 603

= 60 33 3

= 20 3 m

In ADE, ADE = 90 and EAD = 30

tan 30 = EDAD

13

= EC DCAD

1 603 20 3

h

h 60 = 20 h = 80 m EC AB = 80 60 = 20 m 27.

seg AF || seg BE and AE intersects them FAE = AEB = 45 ...[Alternate angles] seg AF || seg CD and AD intersect them FAD = ADC = 60 ...[Alternate angles] BCDE is a rectangle. BC = ED = h and BE = CD In ABE, ABE = 90 and AEB = 45

tan 45 = ABBE

1 = AC BCBE

1 = 60BE

h

BE = 60 h ... (i) In ACD, C = 90 and ADC = 60

tan 60 = ACCD

603BE

...[ BE = CD]

60360 h

...[From (i)]

60 h = 603

60 h = 20 3 60 h = 20 1.73 h = 60 34.6 h = 25.4 m 28. Let the height of the cloud above the level of

point A be h m. C is the position of cloud, and E is its reflection in the lake.

CD = DE = (20 + h) m In ABC, ABC = 90 and CAB = 30

tan 30 = BCAB

1AB3

h

A

E

CB

60 m

3060

(Building)

h D

A

C

B

E

20 m

3060

20

h

(Cloud)

20 + h

Lake surface

(Reflection)

D

45 60

45

60

A

C D

h (tower)

E B

F

60 m (building)

Sample Content

187


AB 3 h ...(i) In ABE, ABE = 90 and BAE = 60

tan 60 = BEAB

BD DE3AB

20 203AB

h

40AB3

h

4033

hh …[From (i)]

3h = 40 + h 2h = 40 h = 20 m In ABC,

sin 30 = BCAC

1 202 AC

AC = 40 m 29. In ABC, B = 90 and C = 60

tan 60 = 50h

350

h

h = 50 3 m

30. In ABC, B = 90 and A = 60

cos 60 = ABAC

1 AB2 20

AB = 202

= 10 m 31. i.

AG = 19372

km

In AGF,

cos 30 = AGAF

32

= 19372AF

AF = 19373

AF = 1118.36 km ii. PH = 1937

2 km

In FHP,

cos 60 = PHPF

12

= 19372PF

PF = 1937 km iii. In AGF,

tan 30 = FGAG

13

= FG1937

2

FG = 19372 3

= 19373.464

= 559.18 km

AD = GI = 7816m = 7.816 km FI = FG + GI = 559.18 + 7.816 = 566.996 km iv. Height of Nanda Devi mountain (AD) = 7816 m Distance between man and Nanda Devi

mountain (CD) = 7816 m

tan = ADUD

= 78167816

= 1

tan = tan 45 = 45 v. tan 45 = PS

ST

1 = 1930ST

ST = 1930 m = 1.930 km

A

D 7816 m

U

7816 m

h (Tower)

C B

A

6050 m

C B

60

A 20 m

(Ladder) Wall

P

T

1930 m

R 45

A P

SQ R I D B

H

G60 30

FSatellite

Nanda DeviC

Mullayanagiri

Sample Content

188

188


32. i. seg QR || seg TS TSQ = SQR = 30 ...[Alternate angles] ii. In SQR,

SRtan30QR

1 40QR3

QR = 40 3 m iii. In PQR,

PQtan 60QR

340 3

h

h = 120 m iv. If the height of an object and the length of its

shadow are equal, then the angle of elevation is 45.

Total Marks: 15 1. 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°. The height of the tower is [1] (A) 5 3 m (B) 10 3 m (C) 15 3 m (D) 20 3 m 2. A pole 20 m high casts a shadow 20 3 m long on the ground, then the Sun’s elevation is [1] (A) 30 (B) 45 (C) 60 (D) 90 3. From a point P on the ground the angle of elevation of the top of a tower is 30 and that of the top of a

flagpole fixed on the top of the tower, is 60. If the length of the flagpole is 5m, then the height of the tower is [1]

(A) 1.5 m (B) 2.5 m (C) 3 m (D) 3.5 m 4. If an observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high, then the angle of elevation of the

top of the tower from the eye of the observer is [1] (A) 30 (B) 45 (C) 60 (D) 90 5. In the given figure, if AD = 7 3 m, then the value of x + y is [1] (A) 10 m (B) 14 m (C) 21 m (D) 28 m 6. 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°. Assuming that there is no slack in the string, the length of the string is [1]

(A) 10 3 m (B) 20 3 m (C) 40 3 m (D) 60 3 m 7. From a window, 15 m high above the ground, the angles of elevation and depression of the top and the foot

of a house on the opposite side of the street are 30 and 45 respectively. The height of the opposite house is (Take 3 = 1.732) [1] (A) 15.66 m (B) 23.66 m (C) 28.66 m (D) 32.66 m 8. The top of two poles of height 16 m and 10 m are connected by a wire of length l metre. If the wire makes an

angle 30 with the horizontal, then l is [1] (A) 10 m (B) 12 m (C) 18 m (D) 24 m

Topic Test

R Q 30 60

P

T S h 30

40 m (Chimney)

(Tower)

A

CDB 30

7 3m

60

x ySample Content

189


9. The angle of depression of two ships on either side of a lighthouse are 45 and 30 respectively. If the height of the lighthouse is 30 m, then the distance between the two ships is (Take 3 = 1.73) [1]

(A) 61.9 m (B) 68.25 m (C) 81.9 m (D) 95.25 m 10. Assertion : If a pole 30 m high casts a shadow 10 3 m long on the ground, then the angle of elevation of the

sun is 30.

Reason : In a right angled triangle, cos = side adjacent to hypotenuse

. [1]

(A) Assertion is true, Reason is true. Reason is a correct explanation for Assertion. (B) Assertion is true, Reason is true. Reason is not a correct explanation for Assertion.

(C) Assertion is true, Reason is false. (D) Assertion is false, Reason is true. 11. Assertion : If the angle of elevation of the top of a tower from a point on the ground, which is 100 m away

from the foot of the tower, is 30°, then the height of the tower is 100 3 m.

Reason : In a right angled triangle, tan = side adjacent to hypotenuse

. [1]

(A) Assertion is true, Reason is true. Reason is a correct explanation for Assertion. (B) Assertion is true, Reason is true. Reason is not a correct explanation for Assertion.

(C) Assertion is true, Reason is false. (D) Assertion is false, Reason is true. 12. Attempt any four sub-parts of the given question. [4] A group of students of class X visited India Gate on an education trip. The teacher and students had interest

in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919.The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kingsway), is about 138 feet (42 metres) in height.

i. What is the angle of elevation if they are standing at a distance of 42m away from the monument? (A) 30 (B) 45 (C) 60 (D) 0 ii. They want to see the tower at an angle of 60. So, they want to know the distance where they should stand

and hence find the distance. (A) 7 3 m (B) 14 3 m (C) 21 3 m (D) 28 3 m

iii. If the altitude of the Sun is at 60, then the height of the vertical tower that will cast a shadow of length 20 m is

(A) 20 3 m (B) 203

m (C) 153

m (D) 15 3 m iv. The ratio of the height of the tower and the length of its shadow on the ground is 1 : 1. The angle of

elevation of the Sun is (A) 30 (B) 45 (C) 60 (D) 90 v. The angle formed by the line of sight with the horizontal when the object viewed is below the horizontal

level is (A) corresponding angle (B) angle of elevation (C) angle of depression (D) complete angle

Sample Content

190

190


1. (B) 2. (A) 3. (B) 4. (B) 5. (D) 6. (C) 7. (B) 8. (B) 9. (C) 10. (D) 11. (C) 12. i. (B) ii. (B) iii. (A) iv. (B) v. (C)

1. In ABC, B = 90 and C = 30

tan 30 = ABBC

1 AB303

AB = 303

= 30 33 3

AB 10 3 m 2. Let the angle of elevation be . In ABC, B = 90 and C =

ABtanBC

= 20 120 3 3

tan = tan30 = 30 3. In PQR, Q = 90 and P = 30

tan 30 = QRPQ

1PQ3h

PQ = 3 h ...(i)

In PQS, Q = 90 and P = 60

tan 60 = SQPQ

SR QR3PQ

53PQ

h

PQ = 53h

533hh

...[From (i)]

3h = 5 + h 2h = 5 h = 2.5 m 4. ABCD is a rectangle. AD = BC = 20.5 m and DC = AB = 1.5 m ED = EC DC = 22 1.5 ED = 20.5 m In ADE, ADE = 90 and EAD =

tan = EDAD

= 20.520.5

tan = 1 tan = tan 45 ...[tan 45 = 1] = 45o

5. In ADB, ADB = 90 and B = 30

ADtan30BD

Answers

Hints

A

B C

20 m

20 3

R

P Q

S

30 60

h (tower)

5 m (flagpole)

E

D

C

A

B

(observer)1.5 m

20.5 m

22 m (Tower)

C

A

B30 m

30Tower

Sample Content

191


D C

45E Bx

A

30

15 m 15 m

h

1 7 33

x

x = 21 m In ADC, ADC = 90 and C = 60

ADtan60DC

7 33 y

y = 7 m x + y = 21 + 7 = 28 m 6. In DEF, E = 90 and F = 60

sin 60 = DEDF

3 602 DF

DF = 60 23

= 120 33 3

DF 40 3 m 7. Let E be the window and AC be the house on

the opposite side. Let the top (A) of the house be at height h from window level.

In ABE, ABE = 90 and AEB = 30

tan 30 = hx

13

hx

h = 3

x … (i)

In CBE, CBE = 90 and CEB = 45

tan 45 = 15x

1 = 15x

x = 15

h = 153

… [From (i)]

= 15 3 5 33 3

height of the house opposite to the window = 5 3 15 = 5 (1.732) + 15 = 8.66 + 15 = 23.66 m 8.

BCDE is a rectangle. In ABE, ABE = 90 and AEB = 30

ABsin30AE

12

= AC BCAE

12

= 16 10l …[ BC = ED = 10]

1 62

l

l = 12 m 9. In ACD, ACD = 90 and D = 45

ACtan 45CD

1 = 30CD

CD = 30 m In ACB, ACB = 90 and B = 30

ACtan 30BC

60 m

D

F E

String

60

(Kite)

(Point on ground) (Pole) 10m

A

B

C D

E30(Pole)16m

l (wire)

A

C D (ship)(ship) B 30 45

30 m Sample Content

192

192


1 30BC3

BC 30 3 m BD = BC + CD = 30 3 30 = 30(1.73) + 30 = 51.9 + 30 = 81.9 m 10. In ABC, B = 90 and C =

tan = AB 30BC 10 3

tan = 3 33

tan = tan 60 = 60 11. In PQR, Q = 90 and R = 30

PQtan30QR

1 100QR3

QR 100 3 m 12. i. Height of India gate (AB) = 42 m

tan = ABBC

= 4242

= 1 tan = tan 45 = 45 ii. tan = AB

BC

tan 60 = 42x

3 = 42x

x = 423

x = 42 33 3

= 14 3 m

iii. tan = ABBC

tan 60 = 20h

3 = 20h

h = 20 3 m iv. If the height of an object and the length of its

shadow are equal, then the angle of elevation is 45.

30 m (Pole)

A

C B

10 3 m (Shadow)

P

RQ 30

(Tower)

100m

A

B 20 m

C

h

60

A

B 42 m C

42 m

A

B x C

42 m

60

Sample Content

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