Maharashtra State XI Mathematics and Statistics (Part I) PDF

 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and th...

 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and the National Anthem; (b) to cherish and follow the noble ideals which inspired our national struggle for freedom; (c) to uphold and protect the sovereignty, unity and integrity of India; (d) to defend the country and render national service when called upon to do so; (e) to promote harmony and the spirit of common brotherhood amongst all the people of India transcending religious, linguistic and regional or sectional diversities, to renounce practices derogatory to the dignity of women; (f) to value and preserve the rich heritage of our composite culture; (g) to protect and improve the natural environment including forests, lakes, rivers and wild life and to have compassion for living creatures; (h) to develop the scientific temper, humanism and the spirit of inquiry and reform; (i) to safeguard public property and to abjure violence; (j) to strive towards excellence in all spheres of individual and collective activity so that the nation constantly rises to higher levels of endeavour and achievement; (k) who is a parent or guardian to provide opportunities for education to his child or, as the case may be, ward between the age of six and fourteen years. The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 20.06.2019 and it has been decided to implement it from the educational year 2019-20. Mathematics and Statistics (Arts and Science) Part - I STANDARD - XI Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune - 411 004 Download DIKSHA App on your smartphone. If you scan the Q.R.Code on this page of your textbook, you will be able to access full text and the audio-visual study material relevant to each lesson provided as teaching and learning aids.  Maharashtra State Bureau of Textbook Production and Curriculum Research reserves all rights relating to the book. No part of this book should be reproduced without the written permission of the Director, Maharashtra State Bureau of Textbook Production and curriculum Research, Pune. Cover, Illustrations and Mathematics - I (Science and Arts) Computer Drawings Commitee Members Shri. Sandip Koli, Dr. Mangala Narlikar (Chairman) Artist Dr. Sharad Gore (Member) Shri. Prahallad Chippalagatti (Member) Typesetter Shri. Prasad Kunte (Member) Baladev Computers Shri. Sujit Shinde (Member) Smt. Prajakti Gokhale (Member) Co-ordinator Shri. Sandeep Panchbhai (Member) Ujjwala Shrikant Godbole Shri. Ramakant Sarode (Member) I/C Special Officer for Mathematics Smt. Pooja Jadhav (Member) Smt. Ujjwala Godbole (Member-Secretary) Production Sachin Mehta Chief Production Officer Mathematics - I (Science and Arts) Sanjay Kamble Study Group Members Production Officer Prashant Harne Asst. Production Officer Dr. Ishwar Patil Dr. Pradeep Mugale Dr. Pradnyankumar Bhojankar Shri. Milind Patil. Paper Shri. Pradipkumar Bhavsar Shri. Balkrishna Mapari 70 GSM Cream wove Shri. Prafullchandra Pawar Shri. Uday Mahindrakar Print Order No. Shri. Devanand Bagul Smt. Swati Powar N/PB/2022-23/1,40,000 Shri. Swapnil Shinde Smt. Mahadevi Mane Printer Shri. Sachin Batwal Smt. Nileema Khaldkar BHAVANI INDUSTRIES,KOLHAPUR Smt. Deepti Kharpas Shri. Amit Totade Shri. Pramod Deshpande Smt. Gauri Prachand Shri. Sharadchandra Walagade Smt. Supriya Abhyankar Publisher Shri. Dhananjay Panhalkar Shri. Dilip Panchamia Vivek Uttam Gosavi, Controller Shri. Vinayak Godbole Maharashtra State Textbook Bureau, Prabhadevi Mumbai- 400 025  PREFACE Dear Students, Welcome to the eleventh standard! You have successfully completed your secondary education and have entered the higher secondary level. You will now need to learn certain mathematical concepts and acquire some statistical skills to add more applicability to your work. Maharashtra State Bureau of Textbook Production and Curriculum Research has modified and restructured the curriculum in Mathematics in accordance with changing needs. The curriculum of Mathematics is divided in two parts. Part 1 covers topics in Trignometry, Algebra, Co-ordinate Geometry and Statistics. Part 2 covers Complex Numbers, Sets and Relations, Calculus and Combinatorics. There is a special emphasis on applications. Activities are added in every chapter for creative thinking. Some material will be made available on E-balbharati website (ebalbharati.in). It contains a list of specimen practical problems on each chapter. Students should complete the practical exercises under the guidance of their teachers. Maintain a journal and take teacher’s signature on every completed practical. You are encouraged to use modern technology in your studies. Explore the Internet for more recent information and problems on topics in the curriculum. You will enjoy learning if you study the textbook thoroughly and manage to solve problems. On the title page Q.R. code is given. It will help you to get more knowledge and clarity about the contents. This textbook is prepared by Mathematics Subject Committee and members of study group. This book has also been reviewed by senior teachers and subject experts. The Bureau is grateful to all of them and would like to thank for their contribution in the form of creative writing, constructive criticism and valuable suggestions in making this book useful to you and helpful to your teachers. The Bureau hopes that the textbook will be received well by all stakeholders in the right spirit. You are now ready to study. Best wishes for a happy learning experience. Pune Date : 20 June 2019 (Dr. Sunil Magar) Indian Solar Date : 30 Jyeshtha 1941 Director Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune.  XI Mathematics and Statistics (Part I) for Arts and Science Sr. Area Topic Competency Statement No The student will be able to­ - understand angle of any measure. understand different systems of Angle and Its 1 Angle measurement of angle and relations measurement between them. convert an angle from one system to the other. understand definitions of trigonometric functions of angle of any measure. Trigonometric Trigonometric 2 find the values of trigonometric Functions Functions functions of certain angles. draw graphs of trigonometric functions. find the trigonometric functions of sum Trigonomet- or difference of the angles. ric Functions Trigonometric find the trigonometric functions of of compound Functions of multiple and sub-multiple angles. 3 angles and compound express the sum or difference of two fractorization angles trigonometric functions as product formulae learn some rules of trigonometric ratios of angles of a triangle. find value of a determinant. reduce determinant to simple form. Determinant solve linear equations in 2 or 3 variables Determinants find area of triangle using 4 and Matrices determinants. understand types of matrices. Matrices Perform algebraic operations of the matrices.  understand locus and its equation. find equation of a straight line in different forms. 5 Straight Line Straight Line find angle between given two straight lines and the distance of a point from given line. find equation of circle satisfying given conditions. 6 Circle Circle learn and use the properties of circle. find the equation of tangent to the circle. find the equations of conic sections Parabola, satisfying given conditions. 7 Conic Section Ellipse, learn and use the properties of conics. Hyperbola find the equation of tangent to the conic. Measures of Measures of calculate range, standard deviation 8 dispersion dispersion and variance from given data. calculate probability of an event and learn conditional probability 9 Probability Probability learn and use Baye’s theorem and its applications Note :- Extra examples for competitive section and practice are given on e-balbharti. The activities which can be conducted as a part of practicals are also mentioned in pdf form on our website.  INDEX Sr. No. Chapter Page No. 1 Angle and its Measurement 1 2 Trigonometry - I 14 3 Trigonometry - II 35 4 Determinants and Matrices 59 5 Straight Line 103 6 Circle 127 7 Conic Sections 140 8 Measures of Dispersion 179 9 Probability 193 Answers 216  1 Angle and its measurement Now we will differentiate between such Let's Study angles. Directed angle. Let's Learn Angles of different measurements Units of measure of an angle 1.1 Directed Angles: Length of an arc of a circle. Consider the Area of a sector of a circle. ray OA. Rotate it about O till it takes the position OB as shown in Let's Recall Fig. 1.3. Then angle so obtained We know how to draw the acute angles due to the rotation of different measures. is called directed In a circle we can find arc length and area angle AOB. We Fig. 1.3 of a sector in terms of the central angle define the notion and the radius. of directed angle as follows: Activity No. 1 Definition: Draw the angle ABC of measure 40° The ordered pair of rays (OA,OB ) together with the rotation of the ray OA to the position of the ray OB is called the directed angle ∠AOB. If the rotation of the initial ray is anticlockwise then the measure of directed angle is considered as positive and if it is clockwise then the measure of directed angle Fig. 1.1 is considered as negative. In the ordered pair (OA,OB ), the ray OA is called the initial arm and the ray OB is called the terminal arm. Fig. 1.2 In the Fig.s 1.1 and 1.2 both the angles are of 40°. But one is measured in anticlockwise direction and the other is measured in clockwise Fig. 1.3(a) Fig. 1.3(b) direction. 1 O is called the vertex as shown in fig 1.3(a) Half of one rotation angle is straight angle. and 1.3(b). Right angle: Observe Fig 1.3(b) and note that One fourth of one rotation angle is called (OA,OB ) ≠ (OB ,OA) as one right angle, it is also half of a straight ∠ AOB ≠ ∠ BOA angle. One rotation angle is four right angles. ∠ AOB ≠ ∠ BOA even though they have same amount of rotation. Zero angle: If the ray OA has zero rotation, that is it does not rotate, the initial arm itself is a terminal arm OB, the angle so formed is zero angle. Fig. 1.4 One rotation angle: Fig. 1.7 After one complete rotation if the initial Angles in Standard position : ray OA coincides with the terminal ray OB In the rectangular co-ordinate system, a then so formed angle is known as one rotation directed angle with its vertex at origin O and angle m∠ AOB = 360°. the initial ray along the positive X-axis, is called angle in standard position. Fig. 1.5 Straight angle: After the rotation, if the initial ray OA and the terminal ray OB are in opposite directions then directed angle so formed is known as straight angle (fig. 1.6). Fig. 1.8 In adjoint Fig. 1.8, ∠XOP, ∠XOQ and Fig. 1.6 ∠XOR are in standard positions. But, ∠POQ is not in standard position. Note that in this case AOB is a straight line. 2 Angle in a Quadrant: A directed angle in standard position is said to be in a particular quadrant if its terminal ray lies in that quadrant. Fig. 1.10 (b) In Fig. 1.8, directed angles ∠XOP, ∠XOQ and ∠XOR lie in first, second and In Fig. 1.10(a), the directed angles having third quadrants respectively. measure 30°, 390°, −330° have the same initial arm, ray OA and the same terminal arm, ray Quadrantal Angles: OB. Hence, these angles are co-terminal angles. A directed angle in standard position whose terminal ray lies along X-axis or Y-axis If the two directed angles are co-terminal is called a quadrantal angle. angles then difference between measures of these two directed angles is an integral multiple of 360° e.g. in figure 1.10(a), 390° − (−330)° = 720° = 2 × 360°. 1.1.1 Measures of angles: The amount of rotation from the initial ray OA to the terminal ray OB gives the measure of angle AOB. It is measured in two systems. Fig. 1.9 In Fig. 1.9, ∠XOP, ∠XOQ, ∠XOR and x ∠XOS are all quadrantal angles. Co-terminal angles: Fig. 1.11 Directed angles in standard position having the same terminal ray are called co-terminal 1) Sexagesimal system (Degree measure) angles. 2) Circular system (Radian measure) 1.1.2 Sexagesimal System (Degree Measure): In this system, the unit of measurement of angle is degree. One rotation angle is divided into 360 equal parts, the measure of each part is called as one degree. th ∴ 1 part of one complete rotation Fig. 1.10 (a) 360 is called one degree and is denoted by 1°. 3  1 th part of one degree is called one Theorem : 60 The radian so defined is independent of the minute and is denoted by 1'. radius of the circle used and πc = 1800. Proof: Let us consider a circle with 1 th part of one minute is called one centre at O and radius r. Let AB be an arc of 60 length r. Join OA and OB. Then ∠AOB = 1c. second and is denoted by 1''. Produce AO to meet the circle at C. 1° = 60' 1' = 60'' m ∠ (one rotation angle) = 360° m ∠ (straight angle) = 180° m ∠ (right angle) = 90° 1.1.3 Circular System (Radian Measure): In this system, the unit of measurement of an angle is a radian. Let r be the radius of a circle with centre Fig. 1.13 O. Let A and B be two points on circle such that the length of arc AB is r. Then the Clearly, ∠AOC = a straight angle measure of the central angle AOB is defined = 2 right angles as 1 radian. It is denoted by 1c. Since measures of the angles at the centre of a circle are proportional to the lengths of the corresponding arcs subtending them: l (arcAB) ∴ = l (arcABC) = = ∴ m∠AOB =. m∠AOC (2 right angles) ∴ 1c = m∠AOB = , π Fig. 1.12 a constant independent of r. Thus, one radian is the measure of an Hence one radian is well defined. angle subtended at the centre of a circle by Also, πc = 2 right angles = 1800. an arc whose length is equal to the radius of the circle. 4 Activity 2 : Verify the above result by taking v) Relation between angle and time in a the circles having different radii. clock. (R is rotation.) Let an angle have its measure r in radian Min Hand Hr Hand and θ in degrees. Then its proportion with the 1R = 360° 1R = 360° straight angle is the same in either measure. 1R : 60 min 1R : 12 Hrs r θ p 60 min : 360° 12 Hrs : 360° \ = \ rc = θ° × p 180 180 1min : 6° rotation 1 Hr : 30° We use this relation to convert radian 60 min : 30° measure into degree and vice-versa. 1° 1 min : Notes: 2 The word ‘minute’ is used for time i) To convert degree measure into radian measurement as well as 60th part of degree p of angle. measure, multiply degree measure by. 180 vi) Please note that “minute” in time and ii) To convert radian measure into degree “minute” as a fraction of degree angle are 180 different. measure, multiply radian measure by. p iii) Taking π = 3.14, SOLVED EXAMPLES  180 ° we have 1 =  c  Ex. 1) Convert the following degree measures  π  in the radian measures. = 57.3248°  1 ° i) 70° ii) -120° iii)   Here fractional degree is given in decimal 4 fraction. It can be converted into minutes π c and seconds as follows Solution : We know that θ° = θ × 180 0.3248° = (0.3248 × 60)' π c i) 70° = 70 × = 19.488' 180 c = 19' + (.488 × 60)'' 7π ∴ 70° = 18 ≈ 19' 29'' c π Thus, 1 = 57° 19' 29'' c ii) -120° = – 120 × 180 iv) In the table given below, certain degree 2π c measures are expressed in terms of radians. ∴ -120° = - 3 c Degree 15 30 45 60 90 120 180 270 360 1 0 1 π iii) = × π π π π π 2π 3π 4 4 180 Radian 12 6 4 3 2 3 π 2 2π 0 c 1 π ∴ = 4 720 5 Ex. 2) Convert the following radian measures = 74°52'+(0.2×60)'' in the degree measures. = 74°52'12'' c 7π c -π c 4 i) ii) iii) 7 ii) −30.6947° = −[30°+0.6947°] 3 18 = −[30°+(0.6947×60)'] 180 ° Solution : We know that θ = θ × c = −[30°+41.682'] π = −[30°+41'(0.682×60)''] 180 ° c 7π 7π i) = × = −[30°41'40.92''] 3 3 π c = −30°41'41'' approximately 7π ∴ = 420° 3 Ex. 4) The measures of the angles of the triangle are in A. P. The smallest angle is 40. Find the angles of the triangle in -π 180 ° c -π degree and in radian. ii) = × 18 18 π Solution : Let the angles of the triangle be a − d, a, a + d in degrees. -π c ∴ = −10° ∴ a − d + a + a + d = 180° 18 ∴ 3a = 180° Note that, ∴ a = 60° 180° = πc Also, smallest angle Hence, c = 40°  π   180  ° 1° =   , 1c =   ∴ a - d = 40°  180   π  ∴ 60° - d = 40° 4 c 4 180 ° ∴ 60° - 40° = d iii) = × π 7 7 ∴ d = 20° 4 c 720 ° 360 ° Now, a + d = 60° + 20° = 80° ∴ = 7π = 7 11 Hence the angles are 400, 600, 800 Ex. 3) Express the following angles in degree, 40 minute and second. if they are q1c, q2c, q3c, 400 = q1c, then = 180 i) 74.87° ii) −30.6947° q1 2πc 60 πc π so that q = q = × π = Solution: 1 9 2 180 3 80 4 i) 74.87° = 74°+0.87° = π = 9 πc 180 = 74°+(0.87×60)' 2πc πc 4πc Hence the angles are , and. = 74°+(52.2)' 9 3 9 = 74°52'+0.2' 6  The angles of a triangle in degree are 40°, ∴ Sum of the remainging three angles is c 2π π c 4π c 360° − 40° = 320° 60° and 80° and in radian , and Since these threee angles are in the ratio 9 3 9 3:5:8. Ex. 5) The difference between two acute angles ∴ Degree measures of these angles are 3k, 7π c 5k, 8k, where k is constant. of a right angled triangle is. ∴ 3k + 5k + 8k = 320° 30 ∴ 16k = 320° Find the angles of the triangle in degree. ∴ k = 20° Solution : Let x and y be the acute angles of a ∴ The measures of three angles are triangle in degree. (3k)° = (3 × 20)° = 60° c 7π 7π 180 ° (5k)° = (5 × 20)° = 100° Here, x - y = = × 30 30 π and (8k)° = (8 × 20)° = 160° = 42° Ex. 7) Find the number of sides of a regular ∴ x - y = 42°.......... (I) polygon if each of its interior angle is The triangle is right angled. 4π c. ∴ x + y = 90°.......... (II) 5 adding, (I) + (II), Solution: we get x - y + x + y = 42° + 90° Let the number of sides be ‘n’. 4π c ∴ 2x = 132° each interior angle = 5 ∴ x = 66° 180 ° 4π c = × = 144° Put in (I) 5 π 66° - y = 42° ∴ 66° - 42° = y Exterior angle = 180° − 144° = 36° ∴ y = 24° ° ∴ 360 = 36° ∴ The angles of a triangle are 66°, 90° n and 24°. 360 ∴ n = 36 2π Ex. 6) One angle of a quadrilateral is ∴ n = 10 9 radian and the measures of the other ∴ number of sides of the regular polygon three angles are in the ratio 3:5:8, find is 10. their measures in degree. Ex. 8) Find the angle between hour hand and Solution : The sun of angles of a quadrilateral minute hand of a clock at is 360°. i) Quarter past five One of the angles is given to be ii) Quarter to twelve 2π c 0 Solution : 2π 180 = × = 40° 1) When a hour hand moves from one clock 9 9 π mark to the next one, it turns through an 360° angle of = 30º. 12 7  At quarter past 5, B) Draw the angles of the following measures miniute hand is pointing and determine their quadrants. to 3. Hour hand has i) –140° ii) 250° iii) 420° iv) 750° gone past 5. So the angle between them is v) 945° vi) 1120° vii) –80° viii) –330° more than 60º. In one ix) –500° x) –820° minute hour hand turns Q.2 Convert the following angles in to radian. ο Fig. 1.14 through 1 hence in 2 i) 85° ii) 250° 15 minute it has turned iii) −132° iv) 65°30′ ο  15  v) 75°30′ vi) 40°48′ through   = 7.5º.  2 Q.3 Convert the following angles in degree. Thus the angle between 7π c -5π c the hands is equal to i) ii) 3 iii) 5c 12 Fig. 1.15 60º + 7.5º = 67.5º. c 11π c -1 ii) At quarter to twelve, minute hand is pointing iv) v) 18 4 to 9, hour hand is between 11 and 12 though it is nearer 12. It will take 15 minute i.e. 7.5º Q.4 Express the following angles in degree, to reach 12. minute and second. 1 c ∴ the angle between the hands is equal to i) (183.7)° ii) (245.33)0 iii) 5 90º − 7.5º = 82.5º. Q.5 In ∆ ABC, if m∠ A = 7π c , 36 Note: m∠ B = 120°, find m∠ C in degree and In radian. 0° 30° 45° 60° 90° 180° 270° 360 degrees 5π c In πc πc π π 3π Q.6 Two angles of a triangle are and 0c πc 2πc 9 radians 6 4 3 2 2 5π c. Find the degree and radian measure 18 of third angle. EXERCISE 1.1 Q.7 In a right angled triangle, the acute angles are in the ratio 4:5. Find the angles of the triangle in degree and radian. Q.1 A) Determine which of the following pairs of angles are co-terminal. Q.8 The sum of two angles is 5πc and their i) 210°, −150° ii) 360°, −30° difference is 60°. Find their measures in degree. iii) −180°, 540° iv) −405°, 675° Q.9 The measures of the angles of a triangle v) 860°, 580° vi) 900°, −900° are in the ratio 3:7:8. Find their measures in degree and radian. 8 Q.10 The measures of the angles of a The area A of a sector is in the proportion triangle are in A.P. and the greatest is of its central angle θ. 5 times the smallest (least). Find the If the central angle θ is in radian, angles in degree and radian. θ A Q.11 In a cyclic quadrilateral two adjacent = 2π Area of the circle angles are 40° and π. Find the angles c 3 θ A ∴ = 2 of the quadrilateral in degree. 2π π r Q12 One angle of a quadrilateral has measure θ r2 1 ∴ A= = r 2θ 2π c 2 2 and the measures of other three 5 The arc length S of a sector is in the angles are in the ratio 2:3:4. Find their proporation of its central angle. If the central measures in degree and radian. angle is θ radians. Q.13 Find the degree and radian measure of θ S exterior and interior angle of a regular = 2π circumference of the circle i) Pentagon ii) Hexagon iii) Heptagon iv) Octagon θ S ∴ = 2π 2π r Q.14 Find the angle between hour-hand and ∴ S = rθ. minute-hand in a clock at i) ten past eleven ii) twenty past seven SOLVED EXAMPLES iii) thirty five past one iv) quarter to six Ex. 1) The diameter of a circle is 14 cm. Find v) 2:20 vi) 10:10 the length of the arc, subtending an angle of Let’s Understand 54° at the centre. Solution : Here diameter = 14 cm 1.2 ARC LENGTH AND AREA OF A ∴ Radius = r = 7 cm SECTOR:- c c π 3π θ = 54 × 180 c = 10 To find s, we know that s = rθ 3π 7 × 3 22 66 = 7 × = × = 10 10 7 10 ∴ arc length = 6.6 cm Ex. 2) In a circle of radius 12 cms, an arc PQ subtends an angle of 30° at the centre. Find the Fig. 1.16 area between the arc PQ and chord PQ. 9 Solution : To find s and A. 1 2 We know that s = rθ and A = r θ r = 12cms, 2 2π ∴ s = 15 × = 10π and θc = 30° 3 1 2π c A = × 15 × 15 × = 75π =  30 × π  2 3  180  ∴ s = 10π cm and A = 75π sq.cm. c π θ = Fig. 1.17 Ex. 4) The perimeter of a sector is equal to 6 half of the circumference of a circle. Find the 1 2 Area of sector OPQ = rθ measure of the angle of the sector at the centre 2 in radian. 1 π = ×12 ×12 × Solution : Let r be the radius of a circle. 2 6 Perimeter of a sector = half of the = 12π sq.cm....... (1) circumference 1 QR ∴ l(OA) + l(OB) + l(arc APB) = (2πr) Draw QR ⊥ OP, ∴ sin 30° = 2 12 1 ∴ r + r + r.θ ∴ QR = 12 × = 6 cms = πr 2 = Height of ∆ OPQ 2r + r.θ = πr 1 Area of ∆ OPQ = × base × height ∴ 2 + θ = π 2 ∴ θ = (π − 2)c 1 = × 12 × 6 2 Ex. 5) A pendulum of Fig. 1.18 length 21cm oscillates = 36 sq.cm........(2) through an angle of By (1) and (2), 36°. Find the length of its path. Required Area = A(Sector OPQ)−A (∆OPQ) Solution : Here r = 21 cm = (12π − 36)sq.cm. p c π c = 12(π − 3)sq.cm θ = 36° = 36 × 180 = 5 Length of its path = Ex. 3) The area of a circle is 225π sq. cm. Find the length of its arc subtending an angle l(arc AXB) of 120° at the centre. Also find the area of = s = rθ the corresponding sector. π = 21 × Solution : Let ‘r’ be the radius of a circle 5 whose 21 22 66 area is 225π sq. cm. = × = 5 7 5 ∴ πr2 = 225π Length of path = Fig. 1.19 ∴ r2 = 225 13.2 cm ∴ r = 15 cm. Ex. 6) ABCDEFGH is a regular octagon  π  c 2π c inscribed in a circle of radius 9cm. Find the θ = 120° = 120 × c  = 3  180  length of minor arc AB. 10 Solution : Here r = 9cm (7) OAB is a sector of the circle having 360 o πc centre at O and radius 12 cm. If θ = = 45° = m∠AOB = 45o, find the difference 8 4 between the area of sector OAB and l(minor arc AB) triangle AOB. = S (8) OPQ is the sector of a circle having = rθ centre at O and radius 15 cm. If   m∠POQ = 30o, find the area enclosed π  by arc PQ and chord PQ. = 9   cm 4 (9) The perimeter of a sector of the circle 9π of area 25π sq.cm is 20 cm. Find the = cm 4 area of the sector. Fig. 1.20 (10) The perimeter of the sector of the circle of area 64π sq.cm is 56 cm. Find the area of the sector. EXERCISE 1.2 (1) Find the length of an arc of a circle Let's Remember which subtends an angle of 108o at the centre, if the radius of the circle is 15 If an angle is r radian and also θ degree cm. r θ then = (2) The radius of a circle is 9 cm. Find the π 180° length of an arc of this circle which  c π  , 1° = (0.01745)c o θ = θ ×  cuts off a chord of length, equal to  180  length of radius. o (3) Find the angle in degree subtended at  180  , 1c = 57°17′48″ θ c = θ ×  the centre of a circle by an arc whose  π  length is 15 cm, if the radius of the Arc length = s = rθ. θ is in radians. circle is 25 cm. 1 2 (4) A pendulum of length 14 cm oscillates Area of a sector A = r θ , where θ is 2 through an angle of 18o. Find the length in radians. of its path. Two angles are co-terminal if and only (5) Two arcs of the same lengths subtend if the difference of their measures is an angles of 60o and 75o at the centres of integral multiple of 360. two circles. What is the ratio of radii of two circles ? Exterior angle of a regular polygon of n o (6) The area of a circle is 25π sq.cm. Find 360  sides =   the length of its arc subtending an angle  n  of 144o at the centre. Also find the area In one hour, hour’s hand covers 30o and of the corresponding sector. a minutes hand covers 360o. 11  In 1 minute, hour hand turns through 7) If the angles of a triangle are in the ratio 1 o 1:2:3, then the smallest angle in radian is   and minute hand turns through 6°. π π π π 2 A) B) C) D) 3 6 2 9 MISCELLANEOUS EXERCISE - 1 8) A semicircle is divided into two sectors whose angles are in the ratio 4:5. Find I Select the correct option from the given the ratio of their areas? alternatives. A) 5:1 B) 4:5 C) 5:4 D) 3:4  22π  c 1)   is equal to 9) Find the measure of the angle between  15  hour-hand and the minute hand of a clock A) 246o B) 264o C) 224o D) 426o at twenty minute past two. o A) 50o B) 60o C) 54o D) 65o 2) 156 is equal to 17π  c 13π  c 11π  c 10) The central angle of a sector of circle of A)   B)   C)   area 9π sq.cm is 600, the perimeter of the  15   15   15  sector is c D)  7π  A) π B) 3+π C) 6+π D) 6    15  II Answer the following. 3) A horse is tied to a post by a rope. If the horse moves along a circular path, always 1) Find the number of sides of a regular keeping the rope tight and describes 88 polygon if each of its interior angle is meter when it traces the angle of 72o at 3π c the centre, then the length of the rope is. 4 A) 70 m. B) 55 m. C) 40 m. D) 35 m. 2) Two circles, each of radius 7 cm, 4) If a 14cm long pendulum oscillates intersect each other. The distance through an angle of 12o, then find the between their centres is 7 2 cm. Find length of its path. the area of the portion common to both 13π 14π 15π 14π the circles. A) B) C) D) 14 13 14 15 3) ∆ PQR is an equilateral triangle 5) Angle between hands of a clock when it with side 18 cm. A circle is drawn shows the time 9.45 is on the segment QR as diameter. Find A) (7.5)o B) (12.5)o C) (17.5)o D) (22.5)o the length of the arc of this circle within the triangle. 6) 20 meter of wire is available for fancing off a flower-bed in the form of a circular 4) Find the radius of the circle in which sector of radius 5 meters, then the central angle of 60o intercepts an arc of maximum area (in sq. m.) of the flower- length 37.4 cm. bed is A) 15 B) 20 C) 25 D) 30 12 5) A wire of length 10 cm is bent so as to 9) A train is running on a circular track of form an arc of a circle of radius 4 cm. radius 1 km at the rate of 36 km per What is the angle subtended at the hour. Find the angle to the nearest centre in degrees ? minute, through which it will turn in 30 seconds. 6) If two arcs of the same length in two circles subtend angles 65o and 110o at 10) In a circle of diameter 40 cm, the length the centre. Find the ratio of their radii. of a chord is 20 cm. Find the length of minor arc of the chord. 7) The area of a circle is 81π sq.cm. 11) The angles of a quadrilateral are in A.P. Find the length of the arc subtending an and the greatest angle is double the angle of 300o at the centre and also the least. Find angles of the quadrilateral in area of the corresponding sector. radian. 8) Show that minute hand of a clock gains 5o30' on the hour hand in one minute. 13  2 Trigonometry - 1 shall now extend the definitions of trigonometric Let's Study ratios to angles of any measure in terms of co- ordinates of points on the standard circle. Trigonometric functions with the help of unit circle Let's Recall Extensions of trigonometric functions to any angle We have studied that, in a right angled Range and Signs of trigonometric functions triangle if measure of an acute angle is 'θ', then in different quadrants opposite side adjacent side Fundamental Identities and Periodicity of sinθ = , cosθ = , hypoteneous hypoteneous trigonometric functions opposite side Domain, Range and Graph of each tanθ =. (see fig 2.1 (a)) adjacent side trigonometric function 1 1 Polar Co-ordinates Also, cosecθ = , secθ = , sinθ cosθ 1 2.1 Introduction cotθ =. tanθ Trigonometry is a branch of Mathematics that deals with the relation between sides and angles of triangles. The word ‘trigonometry’ is derived Let's Learn from the Greek words 'trigonon' and ‘metron’. It means measuring the sides of triangles. Greek 2.1.1 Trigonometric functions with the help of Mathematicians used trigonometric ratios to a circle: determine unknown distances. The Egyptians Trigonometric ratios of any angle used a primitive form of trigonometry for building We have studied that in right angled ∆ ABC, pyramids in the second millennium BC. Greek 'q' is an acute angle astronomer Hipparches (190-120 BC) formulated the general principles of trigonometry and he is known as the founder of the trigonometry. We are familiar with trigonometric ratios of acute angles a in right angled triangle. We have introduced the concept of directed angle having Fig. 2.1(a) Fig. 2.1(b) any measure, in the previous chapter. We 14  adjacent side BC We consider the circle with center at origin cosq = = and radius r. Let P (x,y) be the point on the circle hypoteneous AC with m∠MOP = θ opposite side AB sinq = = Since P lies on the circle, OP = r hypoteneous AC We will now extend this definition to any ∴ x2 + y 2 = r ∴ r2=x2+y2 angle q, consider q as directed angle, The pair (x,y) of co-ordinates of P is uniquely Let 'q' be an acute angle. [See fig. 2.1 (b)] determined by θ. Thus x = rcosθ, y = rsinθ are functions of θ. consider a circle of radius 'r' with centre at origin 'O' of the co-ordinate system. OA is the initial ray of angle q, Note : OB is its terminal ray. 1) If P (x, y) lies on the unit circle P(x,y) is a point on the circle and on ray OB. then cosθ = x Draw PM ⊥ to OA. and sinθ = y. ∴ ∴ OM = x, PM = y and OP = r. P (x, y) ≡ P(cosθ, using ∆ PMO we get, sinθ) 2) The trigonometric OM x PM y functions do not cos θ = = , sin θ = = , OP r OP r depend on the position of the Fig. 2.2 Now we define, for any θ ∈ R point on the terminal arm but they depend x x − co-ordinate of P on measure of the angle. cos θ = = r Distance of P from origin Point P(x,y) is y y − co-ordinate of P on the circle sin θ = = r Distance of P from origin of radius r and Q (x',y') is on the y , x unit circle. tan θ = cot θ = x y Considering r r results on similar cosec θ = , sec θ = y x triangles. y y' Fig. 2.3 For every angle 'q', there is corresponding sinθ = r = 1 , unique point P(x,y) on the circle, which is on the terminal ray of 'q', so trignometric ratios of q are ∴ y = r sinθ also trignometric functions of 'q'. y1 = sinθ and Note that : 1) Trignometric ratios / functions are x x' independent of radius 'r'. cosθ = r = 1 , x = r cosθ x1 = cosθ 2) Trignometric ratios of coterminal angles are same. 15 2.1.2 Signs of trigonometric functions in 3π 3) In the third quadrant (π < θ < ), both x different quadrants : 2 and y are negative, hence Trigonometric functions have positive or negative values depending on the quadrant cosθ = y is negative in which the point sinθ = x is negative P(x, y) lies. Let us find signs of tanθ = y is positive x trigonometric ratios in different Hence only tanθ is quadrants. If the positive, sinθ and cosθ terminal arm of an are negative for θ in the angle θ intersects Fig. 2.7 third quadrant. the unit circle in the point P(x, y), Fig. 2.4 3π 4) In the fourth quadrant ( < θ < 2π), x is then cosθ = x, 2 y positive and y is negative, hence sinθ =y and tanθ =. The values of x and x sinθ = y is negative y are positive or negative depending on the quadrant in which P lies. cosθ = x is positive π tanθ = y is negative 1) In the first quadrant (0 < θ < ), both x and x 2 y are positive, hence Hence only cosθ is cosθ = x is positive positive; sinθ and tanθ sinθ = y is positive are negative for θ in the Fig. 2.8 tanθ = y is positive fourth quadrant. x Hence all You can check sinθ & cosecθ, have the trigonometric functions same sign, cosθ & secθ have the same sign and simillarly tanθ & cotθ have the same sign, when of θ are positive in the they exists. first quadrant. Fig. 2.5 Remark: Signs of cosecθ, secθ and cotθ are π same as signs of sinθ, cosθ and tanθ respectively. 2) In the second quadrant ( < θ < π), y is 2 positive and x is negative, hence 2.1.3 Range of cosθ and sinθ : P(x, y) is point cosθ = y is positive on the unit circle. m∠AOB = θ. OP = 1 sinθ = x is is negative ∴ x2 + y2 = 1 tanθ = y is negative ∴ x2 ≤ 1 and y2 ≤ 1 x ∴ -1 ≤ x ≤1 and -1 ≤ y ≤ 1 Hence only sinθ is ∴ -1 ≤ cosθ ≤ 1 and -1 ≤ sinθ ≤ 1 positive, cosθ and tanθ are negative for θ in the second quadrant. Fig. 2.6 16  SOLVED EXAMPLE π c 2) Angle of measure 90° or : Let 2 m ∠ XOP = 90°. Its terminal arm intersects Ex.1. Find the signs of the following : unit circle in P(0,1). i) sin 300° ii) cos 400° iii) cot (−206°) Hence x = 0 and y = 1 Solution: ∴ sin90° = y = 1 (For given θ, we need to find coterminal angle which lies between 0° and 360°) cos90° = x = 0 i) 270° < 300° < 360° tan90° is not defined ∴ 300° angle lies in the fourth quadrant. as cos90° = 0 ∴ sin 300° is negative. 1 1 cosec90° = y = 1 ii) 400° = 360° + 40° =1 Fig. 2.10 ∴ 400° and 40° are co-terminal angles sec90° is not defined as x = 0 (hence their trigonometric ratios are same) x 0 cot90° = y = 1 = 0 Since 40° lies in the first quadrant, 400° also lies in the first quadrant. (Activity) : ∴ cos 400° is positive. Find trigonometric functions of angles 180°, iii) −206° = −360° + 154° 270°. 154° and −206° are coterminal angles. Since 154° lies in the second quadrant, therefore 3) Angle of measure 360° or (2π)c : Since 360° cot (−206°) is negative. and 0° are co-terminal angles, trigonometric functions of 360° are same as those of 0°. 2.1.4 Trigonometric Functions of specific angles c 4) Angle of measure 120° or 2π : 1) Angle of measure 0° : Let m∠XOP = 0°. 3 Its terminal arm intersects unit circle in Let m ∠ XOP = 120°. Its terminal arm intersects P(1,0). Hence x = 1 and y = 0. unit circle in P(x, y). We have defined, Draw PQ perpendicular to the X-axis sinθ = y, cosθ = x ∴ ∆ OPQ is 30° - 60° - 90° triangle. y and tanθ = x 1 ∴ OQ = and 2 ∴ sin0° = 0, cos0° = 1, 3 PQ = and OP = 1 0 2 and tan0° = 1 = 0 Fig. 2.9 As P lies in the second cosec0° is not defined as y = 0, sec0° = 1 and cot0° 1 quadrant, x = - 2 is not defined as y = 0 and y = 3 2 Fig. 2.11 17  3 1 ∴ sin120° = y = cosec225° = y = – 2 2 1 1 sec225° = cos120° = x=- 2 x =– 2 3 1 y 2 2 tan120° = = =– 3 cot225° = =1 x 1 1 2 2 1 2 cosec120° = = 2.1.3 Trigonometric functions of negative y 3 angles: 1 sec120° = =–2 Let P(x, y) be x 1 any point on the unit x circle with center at 2 1 cot120° = y = =– the origin such that 3 √3 θ ∠AOP = θ. 2 If ∠AOQ = – θ. then 5) Angle of measure 225° or 5π c the co-ordinates of Q 4 Let m ∠XOP = 225°. Its will be (x, –y). terminal arm intersects By definition Fig. 2.13 unit circle in P(x,y). Draw sin θ = y and sin (–θ) Q 225 PQ perpendicular to the = –y X-axis at Q. cos θ = x and cos (– θ) = x ∴ ∆ OPQ 45° – 45° – 90° Therefore sin(–θ) = – sinθ and cos(–θ) = cosθ triangle. Fig. 2.12 1 1 sin(−θ ) − sin θ ∴ OQ = and PQ = and OP = 1 tan(−θ ) = = − tanθ = 2 2 cos(−θ ) cosθ 1 As P lies in the third quadrant, x = – and 1 2 cos(−θ ) cosθ y=– cot(−θ ) = = − cotθ = 2 sin(−θ ) − sin θ 1 ∴ sin225° = y = – 1 1 2 sec(−θ=) = = secθ cos(−θ ) cosθ 1 cos225° = x=– 2 1 1 cosec(−θ ) = = − cosecθ = y sin(−θ ) − sin θ tan225° = =1 x 18  π 6) Angle of measure – 60° or – : Trig. Fun. 3 sinθ cosθ Angles Let m ∠ XOP = –60°. 0° = 0c 0 1 Its terminal arm πc 1 3 intersects unit circle 30° = 2 6 2 in P (x, y). - 1 1 πc Draw PQ perpendicular 45° = 2 2 4 to the X-axis. ∴ ∆ OPQ is 30° – 60° πc 3 1 Fig. 2.14 60° = 2 2 – 90° triangle. 3 1 3 πc OQ = and PQ = and OP = 1 1 0 2 2 90° = 2 1 As P lies in the fourth quadrant, x = and 2 180° = π 0 −1 3 y=– 2 3π −1 0 270° = 3 2 ∴ sin(–60°) = y=– 2 (Activity) : 1 cos(–60°) = x= 2 Find trigonometric functions of angles 150°, y - 3 210°, 330°, – 45°, – 120°, – 3π and complete the tan(–60°) = = 2 =– 3 4 x table. 1 2 Trig. 1 2 Fun. cosec(–60°) = =– y 3 sin θ cos θ tan θ cosecθ sec θ cot θ θ 1 Angle sec(–60°) = x =2 1 150° x 2 1 210° cot(–60°) = y = =– 3 330° 3 2 –45° Note : Angles –60° and 300° are co-terminal –120° angles therefore values of their trigonometric 3π functions are same. – 4 The trigonometric functions of 0°, 30°, 45°, 60°, 90° are tabulated in the following table. 19  iv) sin π + 2 cos π + 3 sin 3π + 4 cos 3π SOLVED EXAMPLES 2 2 – 5 sec π – 6 cosec 3π Ex.1 For θ = 30°, Verify that sin2θ = 2sinθ cosθ 2 = 0 + 2(–1) + 3(–1) + 4(0) – 5(–1) – 6(–1) Solution: Given θ = 30° ∴ 2θ = 600 =0–2–3+0+5+6=6 1 sinθ = sin30° = 2 Ex.3 Find all trigonometric functions of the cosθ = cos30° = 3 angle made by OP with X-axis where P is 2 (−5, 12). sin2θ = sin60° = 3 Solution: Let θ be the measure of the angle in 2 1 standard position whose terminal arm passes L.H.S. = 2sinθ cosθ = 2 × 2 × 3 through P(−5,12). 2 = 3 = sin2θ = R.H.S. r = OP = (−5) 2 + 122 = 13 2 P(x, y) = (–5, 12) ∴ x = −5 , y = 12 Ex.2 Evaluate the following : y 12 13 i) cos30° × cos60° + sin30° × sin60° sinθ = = cosecθ = r = r 13 y 12 ii) 4cos345° − 3cos45° + sin45° 13 cosθ = x = 5 secθ = r = π π π r 13 x 12 iii) sin20 + sin2 + sin2 + sin2 6 3 2 y 12 5 tanθ = = cot θ = xy = iv) sin π + 2 cos π + 3 sin 3π + 4 cos 3π x 5 12 2 2 3π 3π – 5 sec π – 6 cosec Ex.4 secθ = – 3 and π < θ < then find the 2 2 Solution : values of other trigonometric functions. i) cos30° × cos60° + sin30° × sin60° Solution : Given sec θ = – 3 = 3 1 1 ×2 + 2× 3 = 3 ∴ cosθ = – 1 2 2 2 3 ii) 4cos345° − 3cos45° + sin45° We have tan2 θ = sec2 θ – 1 3 ∴ tan2 θ = 9 – 1 = 8 = 4  1   −3 1 + 1  2 2 2 3π ∴ tan2 θ = 8 and π < θ < , the third 2 =4 1 − 2 2 2 2 quadrant. 2 ∴ tan θ = 2 2 Hence cotθ = 1 = 2 − 2 =0 2 2 2 2 π π π Also we have, sin θ = tan θ cos θ iii) sin2 0 + sin2 + sin2 + sin2 6 3 2 2 2 2 1 2 2 1 2 =2 2 3 =– 3 = (0)2 + + 23 + (1)2 2 2 ∴ cosecθ = – 32 1 3 2 2 =0+ + +1=2 4 4 20 Ex.5 If secx = 13 , x lies in the fourth quadrant, ∴ tan2 θ = 1 and 3π < θ < 2π (the fourth 5 2 quadrant) find the values of other trigonometric functions. ∴ tanθ = –1. Hence cotθ = – 1 Solution : Since secx = 13 , we have cosx = 5 Now sinθ = tanθ cosθ = (–1) 1 =– 1 5 13 2 2 Now tan x = sec x – 1 2 2 Hence cosec θ = – 2 13 2 169 144 ∴ tan2 x = –1= –1= 5 25 25 1 + tan θ + cosecθ 1 + (−1) + (− 2 ) = = −1 ∴ tan2 x = 144 and x lies in the fourth 1 + cos θ − cosecθ 1 + (−1) − (− 2 ) 25 quadrant. 3 –12 5 Ex.8 If sinθ = − and 180° < θ < 270° then find ∴ tan x = cot x = – 5 5 12 all trigonometric functions of θ. Further we have, sin x = tan x × cos x 12 5 12 Solution : Since 180° < θ < 270°, θ lies in the =– × =– third quadrant. 5 13 13 1 3 5 13 Since, sinθ = − ∴ cosecθ = − And cosec x = =– sin x 12 5 3 Now cos θ = 1 − sin θ 2 2 4 9 16 Ex.6 If tanA = , find the value of ∴ cos2θ =1 − = 3 25 25 2 sin A – 3 cos A 4 2 sin A + 3 cos A ∴ cosθ = − ∴ secθ = − 5 5 4 Solution : Given expression sinθ Now tanθ = cosθ 2 sin A – 3 cos A sin A - 3 cos A 2 cos A cos A 2 sin A + 3 cos A = ∴ tanθ = 3 ∴ cotθ = 4 sin A + 3 cos A 2 cos A cos A 4 3 = 2 tan A – 3 2 tan A + 3 EXERCISE 2.1 4 2   − 3 3 1 =   = – 1) Find the trigonometric functions of 4 17 2   + 3 3 0°, 30°, 45°, 60°, 150°, 180°, 210°, 300°, 330°, −30°, −45°, −60°, −90°, −120°, −225°, 2 3π −240°, −270°, −315° Ex.7 If sec θ = 2 , < θ < 2π then find the 2 1 + tan θ + cosecθ 2) State the signs of value of. i) tan380° ii) cot230° iii) sec468° 1 + cot θ − cosecθ 2 1 Solution : Given secθ = 2 ∴ cosθ = 2 3) State the signs of cos 4c and cos4°. Which of these two is greater ? Now tan2 θ = sec2 θ – 1 = 2 – 1 = 1 21 4) State the quadrant in which θ lies if 1 values in the domain. For example cosecθ = sinθ i) sinθ < 0 and tanθ >0 is true for all admissible values of θ. Hence this ii) cosθ < 0 and tanθ >0 is an identity. Identities enable us to simplify 5) Evaluate each of the fol