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www.sakshieducation.com TRIGONOMETRY This chapter comes under Paper-II. From this chapter 5 Marks question -1 (1×5 = 5M), 4 Marks questions - 2 (2×4 = 8M), 2 Marks questions - 2 (2×2 = 4M), 1 Mark question - 1 (1×1 = 1M) and 6 objective bits will be given. Altogether we can score 21 marks from it. This chapter is not only very important for S.S.C exam but also to higher classes like Intermediate and Engineering etc. The important material relevant to this topic is given below for the sake of the students who are going to appear S.S.C public exams. ♦ Trigonometry is derived from three "Greek" roots ie Tri + Goma + Metron. Here Tri-means ♦ ♦ ♦ ♦ ♦ ♦ three, Gomia means angle and Metron means measurement, this Trigonometry is the study of "Three angle measurement". Trigonometry is an analytical study of a three angled Geometric figure ie "triangle". "Hipparchus" established the relationship between the sides and angles of a triangle. Radian: The angle subtended by an arc of length equals to the radius of the circle at it's centre is called 'radian'. 2πc = 360°, πc = 180° 90° = 100g = πc/2 or 180° = 200g = πc The relationship among D, G, and C is r D G C D G C or = = = = 90 100 π / 2 180 200 π 1c r r ♦ If l is the length of arc of a sector, r is the radius, θ is the angle at the centre of sector then the relationship among them is l = rθ or r = ♦ ♦ ♦ ♦ l l or θ = , θ r Here θ should be radian only. 1° = 0.01746 radian ; 1c = 58°16' 3πc/4 = 135° (degrees) 120° = 400g/3 (grades) 270° = 3πc/2 (radians) Problem: 1 (1 Mark) A minute hand of a table clock is 3 cm long. How far it's tip move in 20 minutes? Sol: The length of minutes hand r = 3 cm Angle made by minutes hand in 60 minutes ---- 360° 20 minutes ------- ?(θ) www.sakshieducation.com www.sakshieducation.com 6 θ= 20 × 360 πc 2π c = 120° = 120 × = 180 3 60 1 r = 3cm We know that l = rθ θ The distance travelled by tip of minutes hand l = rθ 2π = 3× 3 = 2× = r=? 22 7 44 cm 7 Problem: 2 (1 Mark) A wheel makes 360 revolutions in 1 minute. Through how many radians does it turn in 1 second. Sol: A wheel makes in 1 minute → 360 revolutions i.e. 60 seconds → 360 × 2πc 6 1 second → 360 × 2πc 60 1 1 second → 12πc ∴ The wheel makes 12πc in 1 second. Trigonometric Ratio's: ♦ sin θ = C side opp.to θ BC = hypotenuse AC ♦ cos θ = side adjacent to θ AB = hypotenuse AC ♦ tan θ = side opp.to θ BC = side adj.to θ AB ♦ cot θ = side adj.to θ AB = side opp.to θ BC ♦ sec θ = hyp AC = sideadj.to θ AB hyp. Opp.side θ A www.sakshieducation.com adj.side B www.sakshieducation.com ♦ co sec θ = ♦ tan θ = hyp AC = sideopp.to θ BC sin θ cos θ ;cot θ = cos θ sin θ ♦ sinθ.cosθ = 1; cosθ.secθ = 1 ; tanθ.cotθ = 1 Trigonometric Values θ ratio 0° 30° sin 0 1/2 1/√2 √3/2 1 cos 1 √3/2 1/√2 1/2 0 tan 0 1/√3 1 √3 ∞ cot ∞ √3 1 1/√3 0 sec 1 2/√3 √2 2 ∞ cosec ∞ 2 √2 2/√3 45° 60° Trigonometric Identities: I. sin2θ + cos2θ = 1 ; sin2θ = 1 – cos2θ ; cos2θ = 1 – sin2θ II. sec2θ – tan2θ = 1 ; sec2θ = 1 + tan2θ ; sec2 – 1 = tan2θ III. cosec2θ – cot2θ = 1 ; cosec2θ = 1 + cot2θ ; cosec2θ – 1 = cot2θ Problem: 3 (2 Marks) If tan (A − B ) = 1 1 , sin A = find B in circular measure. 3 2 Sol: tan(A–B) = 1/√3 tan (A − B ) = tan 30° A – B = 30° ____ (1) sinA = 1/√2 sin A = sin 45° A = 45° ____ (2 ) Substitute (2) in (1) 45° – B = 30° –B = 30° – 45° –B = –15° B = 15° www.sakshieducation.com 90° 1 www.sakshieducation.com 1 Now A = 45° = 45 × πc 180c πc = 4 4 1 πc πc = B = 15° = 15 × 180 12 12 Problem: 4 πc πc πc 2 3 Find the value of 32cot − 8 sec + 8cot 4 3 6 2 Sol: c c πc 2 π 3 π − 8 sec + 8cot 32cot 4 3 6 2 = 32cot 2 45° 60° 30° 1 1 1 180° 180° 180° − 8 sec 2 + 8cot 3 4 3 6 = 32 cot2 45° – 8 sec2 60° + 8 cot3 30° = 32(1)2 – 8(2)2 + 8.(√3)3 = 32 – 8(4) + 8 × 3√3 = 32 − 32 + 24 3 = 24√3 Problem: 5 If 4 cotA = 3 then find sin A + cos A sin A − cos A C Sol: 4 cot A = 3 sideadj.to θ sideopp.to θ By Pythagorean Theorem, AC2 = AB2 + BC2 AC2 = 32 + 42 AC2 = 9 + 16 AC2 = 25 ⇒ AC = 5 5 4 ⇒ cot A = 3 / 4 = sin A = θ A 4 3 , cos A = 5 5 4 3 + sin A + cos A 5 5 = sin A − cos A 4 − 3 5 5 www.sakshieducation.com 3 B www.sakshieducation.com = 7/ 5 =7 1/ 5 Problem: 6 (2 Marks) Show that 1 − cos θ = cos ecθ − cot θ 1 + cos θ Proof: L.H.S = 1 − cos θ 1 + cos θ = 1 − cos θ 1 − cos θ × 1 + cos θ 1 − cos θ = (1 − cos θ )2 1 − cos2 θ (1 − cos θ )2 1 − cos2 θ = 1 − cos θ sin2 θ = 1 − cos θ sin θ = 1 cos θ − sin θ sin θ = cosecθ – cotθ = R.H.S Problem: 7 (4 Marks) Prove that tan θ + sec θ − 1 1 + sin θ = tan θ − sec θ + 1 cos θ Proof: tan θ + sec θ − 1 tan θ − sec θ + 1 = tan θ + sec θ − sec 2 θ − tan2 θ tan θ − sec θ + 1 ∵ sec 2 θ − tan2 θ = 1 www.sakshieducation.com www.sakshieducation.com = (tan θ + sec θ ) − (sec θ − tan θ )(sec θ + tan θ ) = (tan θ + sec θ )1 − (sec θ − tan θ ) = (tan θ + sec θ )(1 − sec θ + tan θ ) tan θ − sec θ + 1 tan θ − sec θ + 1 tan θ − sec θ + 1 = tan θ + sec θ 1 = sin θ 1 + cos θ cos θ = sin θ + 1 cos θ = 1 + sin θ cos θ Problem: 8 (4 Marks) P2 − 1 If secθ + tanθ = P then show that sin θ = 2 P +1 Proof: Given that P = secθ + tanθ R.H.S= P2 − 1 (sec θ + tan θ ) − 1 = P2 + 1 (sec θ + tan θ )2 + 1 2 sec 2 θ + tan 2 θ + 2sec θ tan θ − 1 = sec 2 θ + tan 2 θ + 2sec θ tan θ + 1 = sec 2 θ − 1 + tan 2 θ + 2sec θ tan θ sec 2 θ + tan 2 θ + 1 + 2sec θ tan θ = tan2 θ + tan 2 θ + 2sec θ tan θ sec 2 θ + sec 2 θ + 2sec θ tan θ 2 tan2 θ + 2sec θ tan θ = 2sec 2 θ + 2sec θ tan θ = ( ) 2 sec θ (sec θ + tan θ ) 2 tan θ tan θ + sec θ www.sakshieducation.com www.sakshieducation.com sin θ cos θ = 1/ cos θ = sinθ = L.H.S Problem: 9 (4 Marks) Eliminate θ from x = a sinθ – b cosθ, y = a cosθ + b sinθ Sol: x = a sinθ – b cosθ, y = a cosθ + b sinθ Consider x = a sinθ – b cosθ Squaring on both sides x2 = (a sinθ – b cosθ)2 x2 = a2 sin2θ + b2 cos2θ – 2ab sinθcosθ ____ (1) Consider y = a cosθ + b sinθ Squaring on both sides y2 = (a cosθ + b sinθ)2 y2 = a2 cos2θ + b2 sin2θ + 2ab cosθ sinθ ____ (2) (1) + (2) x 2 + y2 = a2 sin2 θ + b2 cos2 θ − 2ab sin θ cos θ + a2 cos2 θ + b2 sin2 θ + 2ab cos θ .sin θ ⇒ x2 + y2 = a2(sin2θ + cos2θ) + b2 (cos2θ + sin2θ) ⇒ x2 + y2 = a2(1) + b2(1) ⇒ x2 + y2 = a2 + b2 Problem: 10 (1 Mark) If x = secθ + tanθ, y = secθ – tanθ then find the relationship between x and y. Sol: x = secθ + tanθ y = secθ – tanθ Consider xy = (secθ + tanθ) (secθ – tanθ) = sec2θ – tan2θ = 1 ♦ sin(90–θ) = cosθ, tan(90–θ) = cotθ, sec(90–θ) = cosecθ, ♦ sin(–θ) = –sinθ, tan(–θ) = –tanθ, sec(–θ) = secθ, cos(90–θ) = sinθ, cot(90–θ) = tanθ cosec (90–θ) = secθ cos(–θ) = cosθ cot(–θ) = –cotθ cosec(–θ) = –cosecθ www.sakshieducation.com www.sakshieducation.com Problem: 11 (4 Marks) Show that sec(270–θ) = –cosecθ Sol: L.H.S sec(270 – θ) = sec[90 + 180 – θ) Let 180 – θ= x = sec[90 + x] = sec[90–(–x)] = cosec(–x) = –cosec x = –cosec(180–θ) = –cosec[90 + 90 – θ] Let 90 – θ=y = –cosec [90 + y] = –cosec [90–(–y)] = –sec(–y) = –secy = –sec(90–θ) = –cosecθ = R.H.S ♦ ♦ ♦ ♦ ♦ ♦ sin(180 + θ) = –sinθ cos(270 + θ) = –sinθ tan(90 + θ) = –cotθ sec(270 – θ) = –cosecθ cot(360 + θ) = cotθ cosec (360 – θ) = –cosecθ Angle of Elevation: If the object is at a heigher level than the eye then the angle between the horizontal line drawn through the observers eye and the line joining the eye to an object is called "Angle of Elevation". t igh e lin s of Angle of elevation Eye Horizontal line Angle of Depression: If the object is at a lower level than the eye, then the angle between the horizontal line drawn through the observers eye and the line joining the eye to any object is called "Angle of Depression". Angle of depression ht ig fs eo lin object www.sakshieducation.com Horizontal line www.sakshieducation.com Problem: 12 (2 Marks) Find the height of mountain cliff, if the angle of elevation of it's top, from a point 300m from it's foot is found to be 60°. Sol: Let AB is the mountain cliff. AC is the distance between cliff and observing point. B ∠BCA = 60° In ∆ABC, tan60 ° = AB AC AB 300 ⇒ AB = 300√3 m ⇒ AB = 300 × 1.732 ⇒ AB = 519.6m ∴ Height of mountain cliff = 519.6m 3= Mountain Cliff θ = 60° A 300 m C Problem: 13 (4 Marks) The upper part of a tree, broken by wind into two parts, makes an angle of 30° with the ground. The top of the tree touches the ground at a distance of 20m from the foot of the tree. Find the height of the tree before it was broken. Sol: Let AB is the tree before it was broken. C is broken point. CB' is broken part. B ∠CB'A = 30° y AB' = 20m Let AC = x, BC = y = B'C AC C In ∆CB'A, tan30 ° = AB ' 1 x = 3 20 30° 20 3 ⇒x= × 3 3 ⇒x= B' 20 3 m ____ (1) 3 In ∆CB'A, cos30 ° = x y AB ' CB ' www.sakshieducation.com 20 m A www.sakshieducation.com 3 20 = 2 y y= 20 × 2 3 y= 40 3 3× 3 40 3 ____ (2 ) 3 The height of tree before it was broken = x + y y= = 20 3 40 3 + 3 3 20 = 60 3 3 = 20√3 = 20 × 1.732 = 34.64m Problem: 14 There are two temples, one on each bank of a river, just opposite to each other. One of the temple A is 40m high. As observed from the top of this temple 'A', the angles of depression of the top and foot of the other temple 'B' are 12°30' and 21°48' respectively. Find the width of the river and height of the temple "B". Sol: x P 12°30' 40 - h 21°48' 12°30' Q R d 40 m h h 21°48' A d www.sakshieducation.com B www.sakshieducation.com AP & BQ are two temples. AP = 40m AB is the width of river. ∠QPX = 12°30' ⇒ ∠RQP = 12°30', ∠BPX = 21°48' ⇒ ∠ABP = 21°48' Let AB = d ⇒ RQ = d Let BQ = h ⇒ AR = h ⇒ PR = 40–h 40 − h Consider ∆PQR, tan12°30' = d 0.2217 = 40 − h ____ (1) d In ∆PBA, tan 21°48' = 40/d 0.4000 = 40/d 4000 d× = 40 10000 1 4 0× 10, 000 d= 4 000 1 d = 100m Substitute 'd' value in (1) 40 − h 0.2217 = 100 2217 × 100 = 40 − h 10, 000 2217 = 40 − h 100 22.17 = 40–h h = 40 – 22.17 h = 17.83m ∴ Width of the river (d) = 100m. The height of temple B = 17.83m Objective Bits: 1. If sinθ = cosθ then θ = _______ 2. If tanθ – secθ = 1 then tanθ + secθ = _______ 3. cosec2 9°– cot2 9° = _______ 4. sin 420° = _______ 5. sin29° + sin281° = _______ 6. If tanθ = a/b then sinθ = _______ 7. If sin2θ = cos3θ then (i) tan 5θ = (ii) cot 5θ = www.sakshieducation.com www.sakshieducation.com sin18 ° = _______ cos72 ° 9. sin15° cos15° = 10. 1° = ______ radians 11. x = secθ – tanθ, y = secθ + tanθ then the relationship between x and y is ____ 12. cos 240° = _____ 13. The relationship among D, G, and C is 14. sin258° + cos258° = 15. An angle made by a side of a 24 sided regular polygon at it's centre is ____ 8. Answers: πc 1. θ = or 45° 4 2. 3. 4. 5. 6. –1 1 √3/2 1 a a 2 + b2 7. i) ∞ ii) 0 8. 1 9. 1/4 10. 0.01746 11. xy = 1 12. –1/2 D G C D G C or = = = = 13. 90 100 π / 2 180 200 π 14. 1 15. 15° Assignment: (2 Marks) 1. The angles of a triangle are in A.P and the greatest is three times the least. Then find the angles of triangle in circular measure. 2. Prove that 4(sin430° + cos460°) –3 (cos245° – sin290°) = 2 (1 3. 4. (4 Mark) Find the value of cos0° + sin90° + √2.sin45° Show that tan2θ + tan4θ = sec4θ – sec2θ Marks) cos θ cos θ + =4 5. Solve 1 − sin θ 1 + sin θ www.sakshieducation.com www.sakshieducation.com (2 Marks) 6. Show that cos(270 + θ) = sinθ (4 Marks) 7. A ladder 20m long is placed against a vertical wall of height 10m. Find the distance between the foot of the ladder and the wall, also find the inclination of the ladder with the horizontal. (4 Marks) 8. An aeroplane at an altitude of 2500 mts observes the angles of depression of opposite points on the two banks of a river to be 41°20' and 52°10'. Find the width of the river in metres. www.sakshieducation.com