# INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS

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1 CHAPTER 8 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS (A) Min Concepts nd Results Trigonometric Rtios of the ngle A in tringle ABC right ngled t B re defined s: sine of A = sin A = side opposite to A BC hypotenuse AC cosine of A = cos A = tngent of A = tn A = side djcent to A AB = hypotenuse AC side opposite to A BC = side djcent to ngle A AB cosecnt of A = cosec A = AC = sin A BC AC secnt of A = sec A = = cos A AB cotngent of A = cot A = tn A = sin A cosa cos A, cot A = sina AB tn A BC

2 88 EXEMPLAR PROBLEMS The vlues of trigonometric rtios of n ngle do not vry with the lengths of the sides of the tringle, if the ngle remins the sme. If one trigonometric rtio of n ngle is given, the other trigonometric rtios of the ngle cn be determined. Trigonometric rtios of ngles: 0, 0, 45, 60 nd 90. A sin A 0 cos A 0 tn A 0 Not defined cosec A Not defined sec A Not defined cot A Not defined 0 The vlue of sin A or cos A never exceeds, wheres the vlue of sec A or cosec A is lwys greter thn or equl to. Trigonometric rtios of complementry ngles: sin (90 A) = cos A, cos (90 A) = sin A tn (90 A) = cot A, cot (90 A) = tn A sec (90 A) = cosec A, cosec (90 A) = sec A Trigonometric identities: cos A + sin A = + tn A = sec A cot A + = cosec A

3 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 89 The line of sight is the line from the eye of n observer to the point in the object viewed by the observer. The ngle of elevtion of n object viewed, is the ngle formed by the line of sight with the horizontl when it is bove the horizontl level. The ngle of depression of n object viewed, is the ngle formed by the line of sight with the horizontl when it is below the horizontl level. The height or length of n object or the distnce between two distinct objects cn be determined with the help of trigonometric rtios. (B) Multiple Choice Questions Choose the correct nswer from the given four options: Smple Question : The vlue of (sin0 + cos0 ) (sin60 + cos60 ) is (A) (B) 0 (C) (D) Solution : Answer (B) Smple Question : The vlue of tn0 cot 60 is (A) (B) (C) (D) Solution : Answer (D) Smple Question : The vlue of (sin 45 + cos 45 ) is (A) (B) (C) (D) Solution : Answer (B) EXERCISE 8. Choose the correct nswer from the given four options:. If cos A = 4, then the vlue of tn A is 5 (A) 5 (B) 4 (C) 4 (D) 5

4 90 EXEMPLAR PROBLEMS. If sin A =, then the vlue of cot A is (A) (B) (C) (D). The vlue of the expression [cosec (75 + θ) sec (5 θ) tn (55 + θ) + cot (5 θ)] is (A) (B) 0 (C) (D) 4. Given tht sinθ =, then cosθ is equl to b b (A) b (B) b (C) b b (D) b 5. If cos (α + β) = 0, then sin (α β) cn be reduced to (A) cos β (B) cos β (C) sin α (D) sin α 6. The vlue of (tn tn tn... tn89 ) is (A) 0 (B) (C) (D) 7. If cos 9α = sinα nd 9α < 90, then the vlue of tn5α is (A) (B) (C) (D) 0 8. If ΔABC is right ngled t C, then the vlue of cos (A+B) is (A) 0 (B) (C) 9. If sina + sin A =, then the vlue of the expression (cos A + cos 4 A) is (A) (B) (C) (D) 0. Given tht sinα = nd cosβ =, then the vlue of (α + β) is (A) 0 (B) 0 (C) 60 (D) 90 (D)

5 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 9. The vlue of the expression sin sin 68 cos + cos 68 + sin 6 cos6 sin (A) (B) (C) (D) 0 is. If 4 tnθ =, then 4sinθ cosθ 4sinθ+ cosθ is equl to (A) (B) (C). If sinθ cosθ = 0, then the vlue of (sin 4 θ + cos 4 θ) is (A) (B) 4 4. sin (45 + θ) cos (45 θ) is equl to (A) cosθ (B) 0 (C) sinθ (D) 5. A pole 6 m high csts shdow m long on the ground, then the Sun s elevtion is (A\) 60 (B) 45 (C) 0 (D) 90 (C) (C) Short Answer Questions with Resoning Write True or Flse nd justify your nswer. Smple Question : The vlue of sinθ + cosθ is lwys greter thn. Solution : Flse. The vlue of (sinθ + cosθ) for θ = 0 is. Smple Question : The vlue of tnθ (θ < 90 ) increses s θ increses. Solution : True. (D) (D) 4 4

6 9 EXEMPLAR PROBLEMS In Fig. 8., B is moved closer to C long BC. It is observed tht (i) θ increses (s θ > θ, θ > θ,...) nd (ii) BC decreses (B C < BC, B C < B C,...) Thus the perpendiculr AC remins fixed nd the bse BC decreses. Hence tnθ increses s θ increses. Smple Question : tnθ increses fster thn sinθ s θ increses. Solution : True We know tht sinθ increses s θ increses but cosθ decreses s θ increses. We hve sin θ tn θ= cosθ Now s θ increses, sinθ increses but cosθ decreses. Therefore, in cse of tnθ, the numertor increses nd the denomintor decreses. But in cse of sinθ which cn be seen s sin θ, only the numertor increses but the denomintor remins fixed t. Hence tnθ increses fster thn sinθ s θ increses. Smple Question 4 : The vlue of sinθ is, where is positive number. Solution : Flse. We know tht 0 or, but sinθ is not greter thn. Alterntively, there exists the following three posibilities : Cse. If <, then Cse. If =, then Cse. If >, then However, sin θ cnnot be greter thn.

7 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 9 EXERCISE 8. Write True or Flse nd justify your nswer in ech of the following:. tn 47 = cot 4. The vlue of the expression (cos sin 67 ) is positive.. The vlue of the expression (sin 80 cos 80 ) is negtive. 4. ( cos θ) sec θ = tn θ 5. If cosa + cos A =, then sin A + sin 4 A =. 6. (tn θ + ) ( tn θ + ) = 5 tn θ + sec θ. 7. If the length of the shdow of tower is incresing, then the ngle of elevtion of the sun is lso incresing. 8. If mn stnding on pltform metres bove the surfce of lke observes cloud nd its reflection in the lke, then the ngle of elevtion of the cloud is equl to the ngle of depression of its reflection. 9. The vlue of sinθ cn be, where is positive number, nd. b 0. cos θ =, where nd b re two distinct numbers such tht b > 0. b. The ngle of elevtion of the top of tower is 0. If the height of the tower is doubled, then the ngle of elevtion of its top will lso be doubled.. If the height of tower nd the distnce of the point of observtion from its foot, both, re incresed by 0%, then the ngle of elevtion of its top remins unchnged. (D) Short Answer Questions Smple Question : Prove tht sin 6 θ + cos 6 θ + sin θ cos θ = Solution : We know tht sin θ + cos θ = Therefore, (sin θ + cos θ) = or, (sin θ) + (cos θ) + sin θ cos θ (sin θ + cos θ) = or, sin 6 θ + cos 6 θ + sin θ cos θ = Smple Question : Prove tht (sin 4 θ cos 4 θ +) cosec θ =

8 94 EXEMPLAR PROBLEMS Solution : L.H.S. = (sin 4 θ cos 4 θ +) cosec θ = [(sin θ cos θ) (sin θ + cos θ) + ] cosec θ = (sin θ cos θ + ) cosec θ [Becuse sin θ + cos θ =] = sin θ cosec θ [Becuse cos θ = sin θ ] = = RHS Smple Question : Given tht α + β = 90, show tht Solution : cosαcosec β cosαsinβ = sin α cosαcosec β cosαsinβ = cosα cosec(90 α) cosα sin (90 α) = cosαsec α cosαcosα [Given α + β = 90 ] = cos = sin α Smple Question 4 : If sin θ + cos θ =, then prove tht tn θ + cot θ = Solution : sin θ + cos θ = (Given) or (sin θ + cos θ) = or sin θ + cos θ + sinθ cosθ = sinθ cosθ = [sin θ + cos θ = ] or sin θ cos θ = = sin θ + cos θ or sin cos sin cos Therefore, tnθ + cotθ =

9 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 95 Prove the following (from Q. to Q.7): EXERCISE 8.. sin θ + cosθ + = cosecθ + cosθ sinθ. tna tna seca seca cosec A. If tn A = 4, then sina cosa = 5 4. (sin α + cos α) (tn α + cot α) = sec α + cosec α 5. ( cot 0 ) = tn 60 sin cot α + = cosecα + cosecα 7. tn θ + tn (90 θ) = sec θ sec (90 θ) 8. Find the ngle of elevtion of the sun when the shdow of pole h metres high is h metres long. 9. If tn θ =, then find the vlue of sin θ cos θ. 0. A ldder 5 metres long just reches the top of verticl wll. If the ldder mkes n ngle of 60 with the wll, find the height of the wll.. Simplify ( + tn θ) ( sinθ) ( + sinθ). If sin θ cos θ =, then find the vlue of θ.. Show tht cos (45 +θ ) + cos (45 θ) = tn (60 +θ ) tn (0 θ) 4. An observer.5 metres tll is 0.5 metres wy from tower metres high. Determine the ngle of elevtion of the top of the tower from the eye of the observer. 5. Show tht tn 4 θ + tn θ = sec 4 θ sec θ.

10 96 EXEMPLAR PROBLEMS (E) Long Answer Questions Smple Question : A sphericl blloon of rdius r subtends n ngle θ t the eye of n observer. If the ngle of elevtion of its centre is φ, find the height of the centre of the blloon. Solution : In Fig. 8., O is the centre of blloon, whose rdius OP = r nd PAQ = θ. Also, OAB = φ. Let the height of the centre of the blloon be h. Thus OB = h. Now, from ΔOAP, sin = r, where OA = d () d Also from ΔOAB, sin = h d. () From () nd (), we get or h = r sin φ cosec θ. h sin φ h = d = θ r sin r d Smple Question : From blloon verticlly bove stright rod, the ngles of depression of two crs t n instnt re found to be 45 nd 60. If the crs re 00 m prt, find the height of the blloon.

11 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 97 Solution : Let the height of the blloon t P be h meters (see Fig. 8.4). Let A nd B be the two crs. Thus AB = 00 m. From ΔPAQ, AQ = PQ = h Now from ΔPBQ, PQ BQ = tn 60 = h or = h 00 or h = (h 00) Therefore, h = 00 = 50 ( + ) i.e., the height of the blloon is 50 ( + ) m. Smple Question : The ngle of elevtion of cloud from point h metres bove the surfce of lke is θ nd the ngle of depression of its reflection in the lke is φ. tn tn Prove tht the height of the cloud bove the lke is h φ+ θ tn φ tnθ. Solution : Let P be the cloud nd Q be its reflection in the lke (see Fig. 8.5). Let A be the point of observtion such tht AB = h.

12 98 EXEMPLAR PROBLEMS Let the height of the cloud bove the lke be x. Let AL = d. Now from ΔPAL, x h = tn θ () d From ΔQAL, x + h = tnφ () d From () nd (), we get x+ h tn φ = x h tnθ x tnφ+ tnθ or = h tnφ tnθ Therefore, x = h tn φ+ tn θ. tn φ tn θ

13 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 99 EXERCISE 8.4. If cosecθ + cotθ = p, then prove tht cosθ = p p +.. Prove tht sec θ+ cosec θ = tn θ + cot θ. The ngle of elevtion of the top of tower from certin point is 0. If the observer moves 0 metres towrds the tower, the ngle of elevtion of the top increses by 5. Find the height of the tower. 4. If + sin θ = sinθ cosθ, then prove tht tnθ = or. 5. Given tht sinθ + cosθ =, then prove tht sinθ cosθ =. 6. The ngle of elevtion of the top of tower from two points distnt s nd t from its foot re complementry. Prove tht the height of the tower is st. 7. The shdow of tower stnding on level plne is found to be 50 m longer when Sun s elevtion is 0 thn when it is 60. Find the height of the tower. 8. A verticl tower stnds on horizontl plne nd is surmounted by verticl flg stff of height h. At point on the plne, the ngles of elevtion of the bottom nd the top of the flg stff re α nd β, respectively. Prove tht the height of the h tn α tower is tnβ tn α. l + 9. If tnθ + secθ = l, then prove tht secθ =. l 0. If sinθ + cosθ = p nd secθ + cosecθ = q, then prove tht q (p ) = p.. If sinθ + b cosθ = c, then prove tht cosθ b sinθ = + b c.. Prove tht + sec θ tn θ + secθ+ tnθ = sin θ cosθ. The ngle of elevtion of the top of tower 0 m high from the foot of nother tower in the sme plne is 60 nd the ngle of elevtion of the top of the second tower from the foot of the first tower is 0. Find the distnce between the two towers nd lso the height of the other tower.

14 00 EXEMPLAR PROBLEMS 4. From the top of tower h m high, the ngles of depression of two objects, which re in line with the foot of the tower re α nd β (β > α). Find the distnce between the two objects. 5. A ldder rests ginst verticl wll t n inclintion α to the horizontl. Its foot is pulled wy from the wll through distnce p so tht its upper end slides distnce q down the wll nd then the ldder mkes n ngle β to the horizontl. Show tht p q cos β cosα = sin α sinβ. 6. The ngle of elevtion of the top of verticl tower from point on the ground is 60 o. From nother point 0 m verticlly bove the first, its ngle of elevtion is 45 o. Find the height of the tower. 7. A window of house is h metres bove the ground. From the window, the ngles of elevtion nd depression of the top nd the bottom of nother house situted on the opposite side of the lne re found to be α nd β, respectively. Prove tht the height of the other house is h ( + tn α cot β ) metres. 8. The lower window of house is t height of m bove the ground nd its upper window is 4 m verticlly bove the lower window. At certin instnt the ngles of elevtion of blloon from these windows re observed to be 60 o nd 0 o respectively. Find the height of the blloon bove the ground.

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