comprehensive trigonometry with challenging problems and solutions for jee main and advance pdf

Comprehensive Trigonometry 13 February 2017 02:12:47 PM ABOUT THE AUTHOR REJAUL MAKSHUD (R M) Post graduated from Calcutta University in PURE MATHEMATICS having teaching experience of 15+ years in many prestigious institution of India Presently, he trains IIT Aspirants at RACE IIT ACADEMY, Jamshedpur, playing a role of DIRECTOR cum HOD OF MATHEMATICS Rejaul Makshud McGraw Hill Education (India) Private Limited CHENNAI Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai-600116 Comprehensive Trigonometry with Challenging Problems & Solutions for Jee Main and Advanced Copyright © 2017, McGraw Hill Education (India) Private Limited No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited ISBN (13): 978-93-5260-510-1 ISBN (10): 93-5260-510-1 Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services If such services are required, the assistance of an appropriate professional should be sought Typeset at Script Makers, 19, A1-B, DDA Market, Paschim Vihar, New Delhi, India, and text and cover printed at Cover Designer: Creative Designer visit us at: www.mheducation.co.in PREFACE This text book on TRIGONOMETRY with Problems & Solutions for JEE Main and Advanced is meant for aspirants preparing for the entrance examination of different technical institutions, especially NIT/IIT/BITSAT/IISc In writing this book I have drawn heavily from my long teaching experience at National Level Institutes After many years of teaching I have realised the need of designing a book that will help the readers to build their base, improve their level of mathematical concepts and enjoy the subject This book is designed keeping in view the new pattern of questions asked in JEE Main and Advanced Exams It has eight chapters Each chapter has a large number of worked out problems and exercise based problems as given below: Level – I: Questions based on Fundamentals Level – II: Mixed Problems (Objective Type Questions) Level – III: Problems for JEE Advanced Exam (0 .9): Integer type Questions Passages: Comprehensive link passages Matching: Match Matrix Reasoning: Assertion and Reasoning Previous years papers: Questions asked in past IIT-JEE Exams become easy So please don’t jump to exercise problems before you go through the Concept Booster and the objectives Once you are arranged in a manner that they gradually require advanced thinking tackle any type of problem easily and skilfully My special thanks goes to Mr M.P Singh (IISc Bangalore), Mr Manoj Kumar (IIT, Delhi), Mr Nazre Hussain (B Tech.), Dr Syed Kashan Ali (MBBS) and Mr Shahid Iqbal, who have helped, inspired and motivated me to accomplish this task As a matter of fact, teaching being the best learning process, I must thank all my students who inspired me most for writing this book I would like to convey my affectionate thanks to my wife, who helped me immensely and my children who bore with patience my neglect during the period I remained devoted to this book I also convey my sincere thanks to Mr Biswajit of McGraw Hill Education for publishing this book in such a beautiful format and to all my learned teachers— Mr Swapan Halder, Mr Jadunandan Mishra, Mr Mahadev Roy and Mr Dilip Bhattacharya, who instilled the value of quality teaching in me I have tried my best to keep this book error-free I shall be grateful to the readers for their constructive suggestions toward the improvement of the book REJAUL MAKSHUD M Sc (Calcutta University, Kolkata) Dedicated to My Beloved Mom and Dad Contents Preface The Ratios and Identities 1.1 Introduction 1.2 Application of Trigonometry 1.3 Trigonometrical Functions 1.4 Measurement of Angles 1.5 Some Solved Examples Exercise 1.6 Trigonometrical Ratios 1.7 Limits of the Values of Trigonometrical Functions 1.8 Some Solved Examples Exercise 1.9 Measurement of the Angles of Different T-ratios 1.10 Some Solved Examples Exercise Exercise 1.11 T-ratios of Compound Angles 1.12 Some Important Deductions 1.13 Some Solved Examples Exercise 1.14 Transformation Formulae Exercise 1.15 Multiple Angles 1.16 Some Important Deductions Exercise 1.17 The Maximum and Minimum Values of f (x) = a cos x + b sin x + c Exercise 1.18 Sub–Multiple Angles 1.19 Some Solved Examples Exercise 1.20 Conditional Trigonometrical Identities 1.21 Some Solved Examples Exercise 10 1.22 Trigonometrical Series 1.23 Different Types of the Summation of a Trigonometrical Series Exercise 11 Exercise 12 Exercise 13 Problems for JEE Advanced Exam Level I (Problems Based on Fundamentals) Level II (Mixed Problems) Level III (Tougher Problems for JEE Advanced) Integer Type Questions Link Comprehension Type (For JEE Advanced Exam Only) Match Matrix Assertion & Reason (Questions Asked in Past IIT-JEE Exams) Answers Hints and Solutions v 1-99 1 2 5 10 11 13 15 17 17 19 20 22 23 25 26 27 33 34 36 37 40 43 43 44 47 47 48 48 49 50 51 65 67 69 70 71 72 74 74 77 79 viii Contents Integer Type Questions Questions asked in IIT-JEE Exams Graphs of Trigonometric Functions 2.1 Introduction 2.2 Characteristics of co-sine Function 2.3 Characteristics of Tangent Function 2.4 Characteristics of co-tangent Function 2.5 Characteristics of co-secant Function 2.6 Characteristics of Secant Function: Level I (Questions Based on Fundamentals) Level II (For JEE Main Exam Only) Level III (For JEE Advanced Exam Only ) Answers The Trigonometric Equation 3.2 3.3 3.4 3.5 Solution of a Trigonometric Equation General solution of Trigonometric Equations Ranges of Trigonometric Functions Some Solved Examples Exercise 3.6 A Trigonometric Equation is of the Form Exercise 3.7 Principal Value 3.8 Method to Find Out the Principal Value Exercise 3.9 Solutions in Case of Two Equations are Given: Exercise 3.10 Some Important Remarks to Keep in Mind While Solving a Trigonometric Equation 3.11 Types of Trigonometric Equations Exercise Exercise Exercise Exercise Exercise Exercise 10 Exercise 11 Exercise 12 Exercise 13 Exercise 14 Level I (Questions Based on Fundamentals) Level II (Mixed Problesms) Level III (Problems for JEE Advanced Exam) Integer Type Questions Linked Comprehension Type (For JEE Advanced Exams Only) Match Matrix Assertion and Reason Questions Asked in Past IIT-JEE exams Answers Hints and Solutions Level III Integer Type Questions Past IIT-JEE Questions Trigonometric In-Equation 4.1 Trigonometric Inequalities Type - I: An inequation is of the form sin x > k 89 91 100-118 100 109 111 112 112 113 115 116 117 118 119-183 119 119 119 120 121 122 123 123 123 124 124 127 128 128 128 129 129 130 130 131 131 132 132 133 140 141 144 145 146 147 149 150 151 156 170 175 177 184-195 184 184 Contents Type - II: An in-equation is of the form sin x < k Type - III: An in-equation is of the form cos x > k Type - IV: An in-equation is of the form cos x < k Type - V: An in-equation is of the form tan x > k Type -VI: An inequation is of the form tan x < k 4.2 Some Solved Examples Comprehensive Link Passage Answers Logarithm 5.1 Introduction 5.2 Some Solved Examples 5.3 Logarithmic Equation 5.4 Logarithmic Inequation Problems for JEE Advanced Exam Comprehensive Link Passages Match Matrix Integer Type Questions Questions Asked in Past IIT-JEE Exams Answers Hints and Solutions Integer Type Questions Questions Asked In Past IIT-JEE Exams Inverse Trigonometric Function 6.1 Introduction to Inverse Function 6.2 Some Solved Examples Exercise 6.3 Inverse Trigonometric Functions 6.4 Graphs of Inverse Trigonometric Functions Characteristics of arc sine function Characteristics of arc cosine function: Characteristics of arc tangent function Characteristics of arc co-tangent function: Characteristics of arc co-secant function: Characteristics of arc secant function: Exercise 6.5 Constant Property Exercise 6.5 Conversion of Inverse Trigonometric Functions Exercise 6.6 Composition of Trigonometric functions and its Inverse 6.7 Composition of Inverse Trigono Metric Functions and Trigonometric Functions Exercise 6.9 Sum of Angles Exercise 6.10 Multiple Angles Exercise 6.11 More Multiple Angles Exercise Problems for JEE Main Exam Questions with Solutions of Past JEE Main Exams Problems for JEE Advanced Exam Level II (Mixed Problems) Level III (Problems for JEE Advanced) Integer Type Questions Comprehensive link passage Match-Matrix ix 185 185 186 186 187 187 192 194 196-231 196 198 201 204 207 218 219 219 220 220 222 226 228 232-329 232 233 235 235 236 236 236 236 237 237 237 240 241 242 243 245 245 247 251 252 260 261 263 263 265 266 273 275 289 292 294 295 296 434 Comprehensive Trigonometry with Challenging Problems & Solutions for Jee Main and Advanced 35 F A O I E C B fi A B C + + =p 2 fi A B C + =p 2 fi Cˆ Ê Ê A Bˆ tan Á + ˜ = tan Á p - ˜ Ë Ë 2¯ 2¯ fi Ê Bˆ Ê Aˆ tan Á ˜ + tan Á ˜ Ë 2¯ Ë 2¯ Ê Cˆ = cot Á ˜ Ë 2¯ Ê Bˆ Ê Aˆ - tan Á ˜ tan Á ˜ Ë 2¯ Ë 2¯ fi Ê Bˆ Ê Aˆ tan Á ˜ + tan Á ˜ Ë 2¯ Ë 2¯ = A B Ê Cˆ Ê ˆ Ê ˆ - tan Á ˜ tan Á ˜ tan Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ fi Ê Aˆ ÊCˆ Ê Bˆ ÊCˆ tan Á ˜ tan Á ˜ + tan Á ˜ tan Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯ D Let O be the circumcentre and OF be perpendicular to AB Let I be the in-centre and IE perpendicular to AC Then –OAF = 90∞ - C –OAI = –IAF - –OAF = A - (90∞ - C ) = A A+ B+C +C 2 = C-B Also, AI = IE r = Ê Aˆ Ê Aˆ sin Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ Ê Bˆ Ê Cˆ = R sin Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ Ê Aˆ Ê Bˆ = - tan Á ˜ tan Á ˜ Ë 2¯ Ë 2¯ fi Ê Aˆ Ê Bˆ + tan Á ˜ tan Á ˜ = Ë 2¯ Ë 2¯ Ê Aˆ Ê Bˆ Ê Cˆ Dividing both sides by tan Á ˜ tan Á ˜ tan Á ˜ , Ë 2¯ Ë 2¯ Ë 2¯ we get, fi Ê Bˆ Ê Cˆ Ê Ê B Cˆˆ = - sin Á ˜ sin Á ˜ Á cos Á + ˜ ˜ Ë 2¯ Ë 2¯Ë Ë 2 ¯¯ Ê Bˆ Ê Cˆ Ê Aˆ OI = R - sin Á ˜ sin Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Also, OI Ê Aˆ Ê Bˆ Ê Cˆ = R - R ¥ R sin Á ˜ sin Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ = R - Rr fi OI = R - Rr Hence, the result 36 We have, A + B + C = p 1 + + ÊCˆ ÊCˆ ÊCˆ tan Á ˜ tan Á ˜ tan Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ = Ê Bˆ Ê Cˆ Ê Aˆ = - sin Á ˜ sin Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ fi Ê Aˆ ÊCˆ Ê Bˆ ÊCˆ tan Á ˜ tan Á ˜ + tan Á ˜ tan Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯ fi 1 ÊCˆ ÊCˆ ÊCˆ tan Á ˜ tan Á ˜ tan Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ê Aˆ Ê Bˆ ÊCˆ cot Á ˜ + cot Á ˜ + cot Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ê Aˆ Ê Bˆ ÊCˆ = cot Á ˜ cot Á ˜ cot Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ 37 Figure Here, HE = JK = r1 But IE = r So, IH = r - r1 Ê p Aˆ In a right triangle IHJ, –JIH = Á - ˜ Ë 2¯ Properties of Triangles fi r Ê p Aˆ tan Á - ˜ = Ë 2 ¯ r - r1 fi r Ê Aˆ cot Á ˜ = Ë ¯ r - r1 = a+b 40 r r Ê Bˆ Ê Cˆ Similarly, cot Á ˜ = , cot Á ˜ = Ë ¯ r - r2 Ë ¯ r - r3 In a triangle ABC, we have Ê Aˆ Ê Bˆ Ê Cˆ Ê Aˆ Ê Bˆ Ê Cˆ cot Á ˜ + cot Á ˜ + cot Á ˜ = cot Á ˜ cot Á ˜ cot Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯ fi r r r1 r r r + + = r - r1 r - r2 r - r3 r - r1 r - r2 r - r3 fi r r1r2 r3 r1 r + + = r - r1 r - r2 r - r3 ( r - r1 ) ( r - r2 ) ( r - r3 ) Ê A- B + Cˆ 38 We have, 2ac sin Á ˜¯ Ë [IIT-JEE-2000] 41 By the sine rule, a b c = = sin A sin B sin C a b c = = sin A sin B sin (p - ( A + B )) a b c = = sin A sin B sin ( A + B ) if we know a, sinA, sin B b c and C, A, B and C by using the half angle formulae, if we know the values of a, b and c a b c = = = 2R By using, sin A sin B sin C –B, –C and the sides a, b and c, if we know a, sinB and R –B, –C and the sides b and c, Ê A + C Bˆ - ˜ = 2ac sin Á Ë 2¯ if we just know a,sin A, R , since Ê p B Bˆ = 2ac sin Á - - ˜ Ë 2 2¯ a b c = = = R gives sin A sin B sin C Êp ˆ = 2ac sin Á - B˜ Ë2 ¯ b c & from which we cannot obtain sin B sin C = 2ac cos B b, c, –B & –C Ê a + c2 - b2 ˆ = 2ac Á ˜ 2ac Ë ¯ ( = a +c -b 2 ) 39 Ans (a) As ABC be a right angled triangle, circum radius of this triangle is half of its hypotemuse a + b2 Thus, R = Ê Cˆ Êpˆ Also, r = ( s - c ) tan Á ˜ = ( s - c ) tan Á ˜ = ( s - c ) Ë 2¯ Ë 4¯ Now, ( R + r ) = 42 Let the angles of the triangle ABC are 4q , q and q Also 4q + q + q = 180∞ fi 6q = 180∞ fi q= = 2s - c = a+b+c-c 180∞ = 30∞ By sine rules, fi fi a + b + 2s - 2c = c + 2s - 2c 435 fi a b c = = sin A sin B sin C a b c = = sin(120∞) sin(30∞) sin(30∞) a b c = = sin (60∞) sin (30∞) sin (30∞) a = b c = 1 2 436 Comprehensive Trigonometry with Challenging Problems & Solutions for Jee Main and Advanced a fi = ( 3l ) cos A = b c = 1 Hence, the required ratio is = a = a+b+c 2+ A= fi + ( 2l ) - ( l ) 2 = ( 3l ).(2l ) 6l 3l = p Similarly, B = Ê 2p ˆ I n = n (OA1 ) (OA1 ) sin Á ˜ Ë n¯ = p p ,C = A : B : C = 30∞ : 60∞ : 90∞ = : : 45 p Ê 2p ˆ sin Á ˜ Ë n¯ A Êpˆ B1 B2 = ( B1 L ) = (OL ) tan Á ˜ Ë n¯ Êpˆ = 2.1.tan Á ˜ Ë n¯ P Êpˆ = tan Á ˜ Ë n¯ In On (n / 2) sin (2q ) p = , where q = n tan q n Now, = + tan q tan q ( ) = cos q = (1 + cos 2q ) = 1Ê Ê 2I ˆ Á1 + - Á n ˜ Ë n ¯ ÁË In = 2ˆ ˜ ˜¯ 44 Ans (d) Let the sides of the triangle be l , 3l , 2l E C BD = cot (30∞) = PD We have, BD = PD DE = PQ = PR + RQ = BC = BD + DE + EC = + + = = ¥ ( BC ) = ¥4 ( ) +1 ( + 1) = (3 + + ) = (4 + ) = (6 + ) = 2ˆ On Ê Ê 2I ˆ Á1 + - Á n ˜ ˜ Ë n ¯ ˜ ÁË ¯ By the cosine rule, D Area of DABC ) = cos q ( 30° B Ê1 ˆ Êpˆ Thus, On = n Á ( B1 B2 ) (OL )˜ = n tan Á ˜ Ë2 ¯ Ë n¯ tan q Q R 46 We have, = b-c a sin B - sin C sin A ( ) +1 Properties of Triangles Ê B + Cˆ Ê B - Cˆ cos Á sin Á Ë ˜¯ Ë ˜¯ = Ê Aˆ Ê Aˆ sin Á ˜ cos Á ˜ Ë 2¯ Ë 2¯ Ê B - Cˆ Ê p Aˆ cos Á - ˜ sin Á Ë ˜¯ Ë 2¯ = Ê Aˆ Ê Aˆ sin Á ˜ cos Á ˜ Ë 2¯ Ë 2¯ Ê Aˆ Ê B - Cˆ Ê B - Cˆ sin Á ˜ sin Á sin Á Ë 2¯ Ë ˜¯ Ë ˜¯ = = Ê Aˆ Ê Aˆ Ê Aˆ cos Á ˜ sin Á ˜ cos Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ = ¥ l2 = ¥4 2+ = 7+4 Aˆ Ê B - Cˆ ˜ ˜ = a sin ÁË 2¯ ¯ a fi a fi Now, r = b c = 1 2 = = b c = = l ( say ) 1 bc sin A D bc sin A = = s 2s (a + b + c ) Ê 3ˆ Á ˜l Ë ¯ fi 3= fi Ê 3ˆ 2+ =Á ˜l Ë ¯ fi ( ( ) +1+1 l ) ( l = 2+ ) Thus, the area of the DABC = bc.sin A = l.l.sin(120∞) ) ) A E B D C F Let AD = p ar ( DABC ) = ar ( DABD ) + ( DADC ) 1 Ê Aˆ Ê Aˆ bc sin A = bp sin Á ˜ + cp sin Á ˜ Ë 2¯ Ë 2¯ 2 a b c = = sin A sin B sin C fi ( ) 48 Ans (a, b, c) 47 We have, from sine rule, a b c = = sin(120∞) sin(30∞) sin(30∞) ( = 12 + Ê B - Cˆ sin Á Ë ˜¯ b-c = Thus, a Ê Aˆ cos Á ˜ Ë 2¯ (b - c ) cos ÊÁË ( fi Ê Aˆ bc sin A = p (b + c ) sin Á ˜ Ë 2¯ fi Ê Aˆ Ê Aˆ Ê Aˆ bc sin Á ˜ cos Á ˜ = p (b + c ) sin Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ fi p= bc Ê Aˆ cos Á ˜ (b + c ) Ë ¯ AD = Also, bc Ê Aˆ cos Á ˜ (b + c ) Ë ¯ AD Ê Aˆ = cos Á ˜ Ë 2¯ AE Ê A ˆ 2bc AE = AD sec Á ˜ = = Ë 2¯ b + c 1 + b c Thus, AE is the H.M of b and c DE Ê A ˆ FD Ê Aˆ = sin Á ˜ , = sin Á ˜ Again, Ë ¯ Ë 2¯ AD AD Ê Aˆ EF = DE + FD = AD sin Á ˜ Ë 2¯ = 4bc Ê Aˆ Ê Aˆ cos Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ b+c 437 438 Comprehensive Trigonometry with Challenging Problems & Solutions for Jee Main and Advanced 2bc ¥ sin A b+c = Also, 49 Ans (b) Given AB CD , CD = AB fi Let AB = a, CD = 2a and the radius of the circle be r Let the circle touches at P, BC at Q, AD at R and CD at S Then AR = AP = r, BP = BQ = a – r, DR = DS = r and CQ = CS = 2a – r 2 In triangle BEC, BC = BE + EC fi fi fi fi ( a - r + a - r ) = ( r )2 + a (3a - 2r )2 = (2r )2 + a 9a + 4r - 12ar = 4r + a a= r Also, ar (Quad ABCD ) = 18 fi fi ar (Quad ABED ) + ar ( DBCE ) = 18 a.2r + a.2r = 18 fi 3ar = 18 fi 3¥ fi r =4 fi r=2 ¥ r = 18 fi 50 Ans (b, d) R T Ê Aˆ Ê B - Cˆ Ê Aˆ = sin Á ˜ sin Á ˜ cos Á ˜ Ë 2¯ Ë ¯ Ë 2¯ fi Ê B - Cˆ Ê Aˆ = sin Á ˜ cos Á Ë ˜¯ Ë 2¯ fi Ê B - Cˆ Êp B + Cˆ cos Á = sin Á ˜ Ë ˜¯ Ë2 ¯ fi Ê B - Cˆ Ê B + Cˆ = cos Á cos Á Ë ˜¯ Ë ˜¯ fi Ê B - Cˆ cos Á Ë ˜¯ = Ê B + Cˆ cos Á Ë ˜¯ fi Ê B + Cˆ Ê B - Cˆ + cos Á cos Á Ë ˜¯ + Ë ˜¯ = Ê B + Cˆ -1 Ê B - Cˆ cos Á cos ˜ ÁË Ë ˜¯ ¯ fi Ê Cˆ Ê Bˆ cos Á ˜ cos Á ˜ Ë 2¯ Ë 2¯ =3 Ê Cˆ Ê Bˆ sin Á ˜ sin Á ˜ Ë 2¯ Ë 2¯ fi Ê Bˆ Ê Cˆ cot Á ˜ cot Á ˜ = Ë 2¯ Ë 2¯ fi fi 1 PS ST 1 + > PS ST QS ¥ SR 1 + > PS ST QR fi Here, PS ¥ ST = QS ¥ SR Now, A.M > G.M 1 + PS ST > fi 2 QR Ê B - Cˆ Ê Aˆ Ê p Aˆ = sin Á ˜ cos Á - ˜ cos Á ˜ Ë ¯ Ë 2¯ Ë 2¯ S Q QS ¥ SR > fi P O Ê Aˆ 51 We have, cos B + cos C = sin Á ˜ Ë 2¯ Ê B + Cˆ Ê B - Cˆ Ê Aˆ = sin Á ˜ cos Á cos Á fi Ë ˜¯ Ë ˜¯ Ë 2¯ Thus, the radius is r = QS + SR > QS ¥ SR s (s - b) (s - a) (s - c) fi s =3 s-a fi 3s - 3a = s fi 2s = 3a fi a + b + c = 3a s (s - c) =3 (s - a ) (s - b) Properties of Triangles fi b + c = 2a = 100 + 36 + 60 Hence, the result Êpˆ Êpˆ 52 We have cos Á ˜ + cos Á ˜ = + Ë 2k ¯ Ë k¯ +1 Êpˆ Êpˆ cos Á ˜ + cos Á ˜ = fi Ë 2k ¯ Ë k¯ = 196 fi C = 14 Now, 2s = a + b + c = + 10 + 14 = 30 fi s = 15 fi p +1 Êqˆ =q cos Á ˜ + cos (q ) = , where Ë 2¯ k Therefore, r = fi +1 Êqˆ Êqˆ cos Á ˜ - + cos Á ˜ = Ë 2¯ Ë 2¯ fi fi +1 3+3 Êqˆ Êqˆ +1 = cos Á ˜ cos Á ˜ = Ë 2¯ Ë 2¯ 2 fi 2t + t - fi t= fi t= 439 ( 2 We have a + b - c ( ) = -1 ± (2 -2 - 3 , ) ( 2 Êpˆ a +b -c Now, cos Á ˜ = Ë 6¯ 2ab fi fi Êqˆ Êp ˆ ÁË ˜¯ = ÁË ˜¯ fi fi Êpˆ q =Á ˜ Ë 3¯ fi p Êpˆ =Á ˜ fi k = k Ë 3¯ fi fi sin C = Ê 2p ˆ = sin Á ˜ , Ë 3¯ fi Thus, ( )( ) ) x2 - x2 + x - = 2 x2 + x + x2 - ( )( ( x + x + 1) = (2 x + x - 1) (2 - ) x + (2 - ) x - (1 + ) = x = - ( + ) , ( + 1) x = ( + 1) 2 54 Ans (c) We have sin P - sin P sin P + sin P = sin P - sin P cos P sin P + sin P cos P = - cos P + cos P since C is obtuse Ê 2p ˆ C =Á ˜ fi Ë 3¯ = 2 Also, c = a + b - 2ab cos C Ê Pˆ = tan Á ˜ Ë 2¯ Ê 2p ˆ 2 = 10 + - 2.10.6.cos Á ˜ Ë 3¯ ) = ( x - 1) ( x + 1) x + x - fi 15 = 6.10.sin C 2 = x + x - 3x - x + Êqˆ Êpˆ = cos Á ˜ cos Á ˜ = Ë 2¯ Ë 6¯ fi -4 x - x - +1 t= ab sin C ) 4 = x + x + + 2x + 2x + + x - 2x + fi 53 We have, ar ( DABC ) = ) ( 2 = x + x + + x - - ( x + 1) 4 r2 = 53 Ans (b) 3+3 Êqˆ = , where t = cos Á ˜ Ë 2¯ -1 ± + + D 15 = = s 15 sin ( P / 2) cos ( P / 2) 440 Comprehensive Trigonometry with Challenging Problems & Solutions for Jee Main and Advanced = (s - b) (s - c) s (s - a) = (s - b) (s - c) s (s - a ) (s - b) ( s - c) fi ((s - b) (s - c))2 fi = = = fi fi D2 ÊÊ 7ˆ Ê 5ˆ ˆ ÁË ÁË - ˜¯ ÁË - ˜¯ ˜¯ D2 Ê Ê 3ˆ ˆ ÁË ÁË ˜¯ ˜¯ D2 Ê ˆ =Á Ë D ˜¯ 57 Ans (b) Given a + b = x , ab = y Also, x2 - c2 = y fi (a + b)2 - c = ab Now, , where s = Thus, a + b - c = - ab a + b2 - c2 ab ==2ab 2ab cos C = C = 120∞ abc D ,r = R= 4D s r 4D = R s ( abc ) ˆ Ê1 Á ab sin 120∞˜ ¯ Ë2 = x+c y.c Ê1 3ˆ Á y ˜ ¯ Ë2 = x+c y.c = 3y ( x + c) c 2 CHAPTER The Heights and Distances 8.1 INTRODUCTION 8.3 BEARING OF A LINE In this chapter we shall study how to measure the height of the object and distance the points with the help of trigonometric relations 8.2 ANGLE OF ELEVATION, ANGLE OF DEPRESSION AND THE LINE OF SIGHT The bearing of a horizontal line, i.e a line in a horizontal plane is the positive acute angle made by this line with the north - south line in the same horizontal plane If a line is said to bear 20∞ west of north, we mean that it is inclined to the north direction at angle of 20∞, this angle is measured from the north towards the west N D 15° 40° P f eo ht sig W Lin Angle of elevation O A Let the point O is the observer and the point P is the object under observation The line OP is called the line of sight of the point P Let OA be the horizontal line in the same vertical plane with OP The acute angle –AOP, between the line of sight and the horizontal line known as the angle of elevation O Angle of depression Lin fs E O C 70° 35° B S Here, the bearings of the lines OA, OB, OC and OD are, respectively, N 40∞ E , S 35∞, S 75∞W and N15∞W N P (NE) 45° A W eo A Q (ENE) 45° 22° O E igh t P The acute angle –AOP, between the line of sight and the horizontal line known as the angle of depression, where the object P below the horizontal line OA S North East means equally inclined to north and east South - East means equally inclined to south and east E-N-E means equally inclined to east and north east 442 Comprehensive Trigonometry with Challenging Problems & Solutions for Jee Main and Advanced 8.4 SOME SOLVED EXAMPLES Ex-1 A chimney of 20 m high standing vertically on the -1 Ê ˆ top of a building, subtends an angle of tan Á ˜ Ë 6¯ at a distance of 70 m from the foot of the building Find the height of the building Soln Let QR be the chimney, PQ be the vertical tower and O be the point of observation We have OP = 70, QR = 20 and tanq = R q a P Let –POQ = a and –QOR = q and PQ = h h + 20 70 Then tan (q + a ) = fi tan q + tan a h + 20 = - tan q tan a 70 fi fi So, fi fi = Êpˆ tan q = tan Á ˜ Ë 6¯ tanq = q= p Hence, the angle of elevation is 30∞ Ex-3 The angle of elevation of the top of a tower at any p and after moving 20 m point on the ground is p towards the tower it becomes Q O Let the angle of elevation of the sun be q Then it shadow on the ground OA = BQ Given OA = = BQ , PQ = Find the height of the tower Soln Let PQ be the vertical tower of height h and A and B are the point of observations We have AB = 20 m P h + 70 = h + 20 h 70 1- 70 70 + 6h h + 20 = 420 - h 70 h p /6 A p /3 B Let BQ = x m fi 4900 + 420h = 420h - h + 8400 - 20h Êpˆ h In DPBQ, tan Á ˜ = Ë 3¯ x fi h + 20h = 3500 fi (h + 10)2 = 3500 + 100 = 3600 fi fi h + 10 = 60 fi h = 50 Hence, the height of the building is 50 m Ex-2 If a flag stuff on m high, placed on the top of a q O (i) h Êpˆ In DPAQ, tan Á ˜ = Ë ¯ 20 + x fi = h 20 + x 20 + x = h fi 20 + x = x R fi fi fi 20 + x = 3x x = 20 x = 10 Q Hence, the height of the vertical tower = 10 q A h=x fi tower costs a shadow of m along the ground, then find the angle of elevation of the sun Soln Let PQ be the flag staff placed on the top of a vertcal tower QR B Q R ( ) Ex-4 When the suns altitude increases from 30∞ to 60∞ , the length of the shadow of a tower decrease by m Find the height of the tower The Heights and Distances Soln Let PR be the height of the tower when the angle of elevation is (x – 5), the length of the shadow is QR = x (say) P We have QR =10 and –MRQ = Ê p ˆ QM In DMRQ, tan Á ˜ = Ë 4¯ 10 QM 1= fi 10 30° Q Also, 60° x- S fi R When the angle of elevation is 60°, then the length of the shadow is (x – 5) h In DPRQ, tan (30∞) = x h = fi x fi x=h fi 3= fi 3= h x-5 h x-5 h 3-5 3h - = h fi 2h = fi h= fi RM = 10 fi Thus, PM = 10 PQ = QM + PM = 10 + 10 = 10 ) +1 P 5 Ex-5 A tree is broken by wind, its upper part touches the ground at a point 10 m from the foot of the tree and p with the ground Find the whole makes an angle length of the tree Soln Let PQ be the whole length of the tree and PM its broken part, which touches the ground at R Hence, the height of the tower is P M p /3 p /6 M 40 m p /4 10 Q Q N Let the breadth of the river be QN = x PQ h Êpˆ = In DPQM, tan Á ˜ = Ë ¯ x + 40 x + 40 fi fi h= = h x + 40 x + 40 Êpˆ h In DPQN, tan Á ˜ = Ë 3¯ x R ( Soln Let PQ be the tree of height h and M, N be the point of observations We have MN = 40 m h fi Ê p ˆ 10 cos Á ˜ = Ë ¯ RM 10 = RM Ex-6 A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank p is , when he retreats back 40 m from the bank he p find that the angle to be Then find the breadth of the river (i) In DPRS, tan (60∞) = p QM = 10 fi h 443 fi 3= h x (i) 444 Comprehensive Trigonometry with Challenging Problems & Solutions for Jee Main and Advanced Hence, the height of the hill = ( x + h) h=x fi .(ii) From (i) and (ii), we get, x + 40 x 3= = 3x = x + 40 x = 40 x = 20 fi fi fi = Hence, the breadth of the river is 20 m Ex-7 Let the angle of elevation of the top of a hill observed from the top and bottom of a building of height h are a and b, respectively Then find the height of the hill Soln Let AB be the hill, B being its top and CD be the building of height h Thus, CD = h and consider BM = x h cot b +h cot a - cot b h cot a (cot a - cot b ) Ex-8 A ladder rests against a vertical wall an angle a to the horizontal Its foot is pulled away from the wall through a distance ‘a’ so that it slides a distance ‘b’ down the wall making an angle b with the horizontal Êa + bˆ Prove that a = b tan Á Ë ˜¯ Soln Let AC represents the ladder of length l After being pulled through a distance ‘a’ away from the wall, the new position of the ladder is A’C ‘ A B x a D h C h b DM = fi fi AC = DM = x+h AC B – ACB = a and –A ' C ' B = b Thus, a = BC '- BC = l cos b - l cos a = l (cos b - cos a ) and b = AB - A ' B ' = l sin a - l sin b = l (sin a - sin b ) x+h tan (b ) x+h tan (b ) C¢ C Clearly, AC = A ' C ' = l and CC ' = a , AA ' = b x = x cot (a ) .(i) tan (a ) In DBCA, tan (b ) = a b A x In DBDM, tan (a ) = DM fi A¢ M (ii) From (i) and (ii), we get, Therefore, fi a (cos b - cos a ) = b (sin a - sin b ) fi a = b fi a Êa + bˆ = tan Á Ë ˜¯ b fi Êa + bˆ a = b tan Á Ë ˜¯ x x+h = tan (a ) tan (b ) fi x x h = tan (a ) tan b tan ( b ) fi Ê 1 ˆ h xÁ = ˜ Ë tan (a ) tan b ¯ tan ( b ) fi x (cot a - cot b ) = h cot b fi x= h cot b (cot a - cot b ) a l (cos b - cos a ) = b l (sin a - sin b ) Ê b +aˆ Êa - bˆ sin Á sin Á Ë ˜¯ Ë ˜¯ Êa - bˆ Êa + bˆ sin Á cos Á Ë ˜¯ Ë ˜¯ Hence, the result The Heights and Distances Ex-9 At the foot of a mountain the elevation of its summit is 45°, after ascending km towards the mountain upon an incline of 30°, the elevation changes to 60° Find the height of the mountain Soln Let C be the foot and A be the top of the mountain and AB = h cot (q ) = 445 b cot a - a cot b b-a Soln Let A and B be two stations due south of the tower OP which leans towards north P A 60° D mk h E b 30° 45° S C F B and AE = AB – BE = AB – DF = h - fi b cot (a ) - a cot ( b ) = ) ( ) -1 = ( ) +1 N Q From (i) and (ii), we get, AE From DAED, tan (60∞) = DE h2 3= fi h2 3h- = hfi 2 3 -1 h = - = fi 2 h= x C a+x .(i) h h tan (b ) = b+x b+x cot (b ) = .(ii) h and Now, DE = BF = BC – CF = h fi a cot a = fi BC = h ( A q Let AC = a, BC = b and CQ = x h Clearly, tan a = a+x 3 CF = CD cos (30∞) = = km 2 1 DF = CD sin (30∞) = = km 2 AB h = In DABC, tan (45∞) = BC BC fi b B a ( ) Ê +1 ˆ ˜ km Hence, the height of the mountain is Á ˜ ÁË ¯ Ex-10 Two stations due south of a tower, which leans towards north are at a distances a and b from its foot If a and b are the elevations of the top of the tower from these stations, prove that its inclination q to the horizontal is given by ab + bx ab + ax (b - a ) x = h h h fi x b cot a - a cot b = h (b - a ) fi cot (q ) = b cot a - a cot b (b - a ) Hence, the result Ex-11 From an aeroplane vertically over a straight road the angles of depression of two consecutive kilometer stones on the same side are 45° and 60° Find the height of the aeroplane Soln Let B and C be two consecutive kilometer stones Then BC = km and A be the position of the plane at a certain time A X 60° 45° h 60° 45° B C D 446 Comprehensive Trigonometry with Challenging Problems & Solutions for Jee Main and Advanced Let AD = h and CD = x h In DABD, tan(45∞) = 1+ x = = 20 x +1 = h x = h -1 fi fi (i) ( fi h= ) -1 h = ( ) -1 = ( ) +1 Ê + 3ˆ Hence, the height of the plane is Á ˜ km Ë ¯ Ex-12 A bird is perched on the top of a tree 20 m high and its elevation from a point on the ground is 45° It flies off horizontally straight away from the observer and in one second the elevation of the bird is reduced to 30° Find its speed Soln Let the bird alight at B, the top of the tree BD and O be the observer, where BD = 20 m M B 20 m 30° 45° O N D Thus, BD = MN = 20 m 20 In DBOD, tan(45∞) = OD fi OD = 20 In DMON, tan(30∞) = MN 20 = ON 20 + DN fi 20 = 20 + DN fi 20 + DN = 20 fi DN = 20 Thus, speed = ( ) -1 distance time ( ( ) m / sec -1 ) - m / sec LEVEL I (PROBLEMS BASED ON FUNDAMENTALS) h In DACD, tan(60∞) = x h = fi x h = fi h -1 fi 20 A tower subtends an angle q at a point A in the plane of its base The angle of depression of the foot of the tower at a point ‘h’ m just above ‘A’ is a Find the height of the tower The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of 100 m from p its base is If the angle of elevation of the top of p the complete pillar at the same point is to be such that the height of the pillar is increased by h m, then find h A person walking along a straight road observes that at two points 1000 m apart, then angle of elevation of p 5p a vertical tower in front of him are and Find 12 the height of the tower A man in a boat rowing uniformly away from a cliff 150 m high takes minutes to change the angle of p p to Find the elevation of the top of the hill from speed of the boat A flagstaff of m high stands on a building of 25 m high The flagstaff and the building subtends equal angles at a point P, 30 m high above the ground Find the distance of P from the top of flagstaff AB is a vertical tower ‘A’ being its foot standing on a horizontal ground ‘C’ is the mid-point of AB Portion CB subtends an angle q at the point P on the ground If AP = 2AB, then find tan(q) At the foot of a mountain the elevation of its peak is p found to be , after ascending 10 m toward the p mountain up a slope of inclination, the elevation p is found to be Find the height of the mountain A man finds that at a point due south of a vertical p tower the angle of elevation of the tower is He then walks due west 10 m on the horizontal plane p and find the angle of elevation of the tower to be Find the original distance of the man from the tower The angle of elevation of the top of vertical tower from a point A on the horizontal ground is found to The Heights and Distances p From ‘A’ a man walks 10 m up a path sloping at an angle p/6 After this the slope becomes steeper and after walking up another 10 m, the man reaches the top of the tower Find the distance of ‘A’ from the foot of the tower 10 A vertical tower erected at the focus of the parabola p y = 40 x subtends an angle at the vertex of the p parabola If the tower subtends an angle at a point P lying on the parabola Then find the possible co-ordinates of point P be LEVEL II (MIXED PROBLEMS) On level ground the angle of elevation of the top of the tower is 30° On moving 20 meters near then the angle of elevation is 60° The height of the tower is (b) 10 m (a) 20 m (c) 10 ( ) -1 m (d) None From the top of a light house 60 meters high with its base at sea level, the angle of depression is 15° The distance of the boat from the foot of the light house is Ê - 1ˆ Ê + 1ˆ (a) 60 ¥ Á (b) 60 ¥ Á ˜m ˜m Ë + 1¯ Ë - 1¯ Ê + 1ˆ (c) Á ˜m Ë - 1¯ (d) None Three poles whose feet A, B, C lie on a circle subtend angles a, b, g and respectively, at the centre of the circle If the height of the poles are in A.P, then cot (a ) , cot (b ) , cot (g ) are in (a) A.P (b) G.P (c) H.P (d) None A ladder 20 ft long reaches a point 20 ft below the top of a flag The angle of elevation of the top of the flag at the foot of the ladder is 60° Then the height of the flag is (a) 25 ft (b) 30 ft (c) 35 ft (d) 40 ft From an aeroplane vertically over a straight horizontal road, the angles of depression of two consecutive milestones on opposite sides of the aeroplane are observed to 45° and 60° Then the height in miles of aeroplane above the road is 3 (a) (b) +1 -1 (c) +1 -1 (d) 447 -1 +1 LEVEL III (PROBLEMS FOR JEE MAIN) A man on a cliff observes a ship at an angle of depression 30° approaching the shore just beneath him Three minutes later the angle of depression of the ship is 60° How soon will it reach the shore? A vertical tower subtends an angle of 60° at a point on the same level as the foot of the tower On moving 100 m further from the first point in line with the tower, it subtends an angle of 30° at the point Find the height of the tower Find the height of a tower when it is found that on walking 80 m towards it along a horizontal line through its base, the angular elevation of its top changes from 30° and 60° A vertical pole on one side of a street subtends a right angle at a window exactly on the opposite side If the angle of elevation of the window from the base of the pole be 60° and the width of the street be 30 m, find the heights of the window and top of the pole An object is observed from three points A, B, C lying in a horizontal straight line which passes directly underneath the object The angular elevation at B is twice that at A and at C three times at A If AB = a, BC = b, find the height of the object A man notices two objects in a striaght line due west of him After walking a distance c due north he observes that the objects subtend an angle a at his eye and after walking a further distance c due north, an angle b Find the distance between the objects The angle of elevation of an aeroplane from a point 200 meters above a lake is 45° and the angle of depression of its replection is 75° Find the height of the aeroplane above the surface of the lake From the bottom of a pole of height h, the angle of elevation of the top of a tower is a The pole subtends an angle b at the top of the tower Find the height of the tower The angle of elevation of a cloud from a point x feet above a lake is q and the angle of depression of its reflection in the lake is j Find its height 10 A train is moving at a constant speed at an angle q East of North Observations of the train are made from a fixed point It is due north at some instant Ten minutes earlier its bearing was a West of North, where as 10 minutes afterwards its bearing is b East of North Find tan(q) 448 Comprehensive Trigonometry with Challenging Problems & Solutions for Jee Main and Advanced ANSWERS LEVEL III LEVEL I h tan q cot a h = 100 250 ( ( ( ) -1 m ) 50 m +1 ) 40 m 25 - m/ 5 ¥ 30 m , 40 m 2/9 ( ) +1 10 ( ( a (a + b) (3b - a ) 2b 3c cot b - cot a 200 m m ) h sin a cos (a - b ) cos ec b +1 20 , 20 ) LEVEL II (b) (b) (c) (b) (a) x sin (j + q ) sin (j - q ) 10 2sin a sin b sin (a - b ) ... 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