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1 TRIGONOMETRY UNIT-6 "Te matematician is fascinated wit te marvelous beauty of te forms e constructs, and in teir beauty e finds everlasting trut.". If xcosθ ysinθ a, xsinθ + ycos θ b, prove tat x +y a +b. xcosθ - y sinθ a xsinθ + y cosθ b Squaring and adding x +y a +b.. Prove tat sec θ+cosec θ can never be less tan. S.T Sec θ + Cosec θ can never be less tan. If possible let it be less tan. + Tan θ + + Cot θ <. + Tan θ + Cot θ (Tanθ + Cotθ) <. Wic is not possible. 3. If sinϕ, sow tat 3cosϕ-4cos 3 ϕ 0. Sin ϕ ½ ϕ 30 o Substituting in place of ϕ 30 o. We get If 7sin ϕ+3cos ϕ 4, sow tat tanϕ. If 7 Sin ϕ + 3 Cos ϕ 4 S.T. Tanϕ 3 7 Sin ϕ + 3 Cos ϕ 4 (Sin ϕ + Cos ϕ) 3 Sin ϕ Cos ϕ Sin ϕ Cos ϕ 3 43

2 Tan ϕ 3 Tanϕ 3 5. If cosϕ+sinϕ cosϕ, prove tat cosϕ - sinϕ sin ϕ. Cosϕ + Sinϕ Cosϕ ( Cosϕ + Sinϕ) Cos ϕ Cos ϕ + Sin ϕ+cosϕ Sinϕ Cos ϕ Cos ϕ - Cosϕ Sinϕ+ Sin ϕ Sin ϕ Sin ϕ - Cos ϕ (Cosϕ - Sinϕ) Sin ϕ - Cos ϕ Sin ϕ & - Sin ϕ Cos ϕ or Cosϕ - Sinϕ Sinϕ. 6. If tana+sinam and tana-sinan, sow tat m -n 4 TanA + SinA m TanA SinA n. m -n 4 mn. m -n (TanA + SinA) -(TanA - SinA) 4 TanA SinA TanA + SinA ( TanA SinA RHS 4 mn 4 ( ) ) 4 Tan A Sin A 4 Sin A Sin ACos A Cos A 4 Sin A 4 Cos A Sin A 4 4 TanA SinA Cos A m n 4 mn 7. If seca, prove tat seca+tanax or. 44

3 Secϕ x + 4x Sec ϕ ( x + ) 4x Tan ϕ ( x + ) - 4x Tan ϕ ( x - ) 4x (Sec ϕ + Tan ϕ) Tanϕ + x - 4x Secϕ + Tanϕ x + x or x + x - 4x 4x 8. If A, B are acute angles and sina cosb, ten find te value of A+B. A + B 90 o 9. a)solve for ϕ, if tan5ϕ. Tan 5ϕ 45 ϕ ϕ9 o. 5 Sinϕ + Cosϕ b)solve for ϕ if Cosϕ Sinϕ Sinϕ + Cosϕ Cosϕ Sinϕ Sin ϕ + (Cosφ) Sinϕ( + Cosϕ) 4 Sin ϕ + + Cos ϕ + Cosϕ 4 Sinϕ + SinϕCosϕ + Cosϕ Sinϕ( + Cosϕ) 4 45

4 + ( + Cosϕ) 4 Sinϕ( + Cosϕ) 4 Sinϕ Sinϕ Sinϕ Sin30 ϕ 30 o 0. If Cosα m Cosβ m Cos α Cos β LHS (m +n ) Cos β Cosα n Sinβ n Cos α Cos α + Cos β Cos β Sin β Cos α Cos βsin Cos α n Sin β (m +n ) Cos α Sin Cos β Cos β n β β. If 7 cosecϕ-3cotϕ 7, prove tat 7cotϕ - 3cosecϕ 3. 7 Cosecϕ-Cotϕ7 P.T 7Cotϕ - 3 Cosecϕ3 7 Cosecϕ-3Cotϕ7 7Cosecϕ-73Cotϕ 7(Cosecϕ-)3Cotϕ 46

5 7(Cosecϕ-) (Cosecϕ+)3Cotϕ(Cosecϕ+) 7(Cosec ϕ-)3cotϕ(cosecϕ+) 7Cot ϕ3 Cotϕ (Cosecϕ+) 7Cotϕ 3(Cosecϕ+) 7Cotϕ-3 Cosecϕ3. (sin 6 ϕ+cos 6 ϕ) 3(sin 4 ϕ+cos 4 ϕ)+ 0 (Sin ϕ) 3 + (Cos ϕ) 3-3 (Sin 4 ϕ+(cos 4 ϕ)+0 Consider (Sin ϕ) 3 +(Cos ϕ) 3 (Sin ϕ+cos ϕ) 3-3 Sin ϕcos ϕ( Sin ϕ+cos ϕ) - 3Sin ϕ Cos ϕ Sin 4 ϕ+cos 4 ϕ(sin ϕ) +(Cos ϕ) (Sin ϕ+cos ϕ) - Sin ϕ Cos ϕ - Sin ϕ Cos ϕ (Sin 6 ϕ+cos 6 ϕ)-3(sin 4 ϕ+cos 4 ϕ) + (-3 Sin ϕ Cos ϕ)-3 (- Sin ϕ+cos ϕ)+ 3. 5(sin 8 A- cos 8 A) (sin A ) (- sin A cos A) Proceed as in Question No. 4. If tanθ 6 5 & θ +φ 90 o wat is te value of cotφ. Tanθ 5 i.e. Cotφ 6 5. Wat is te value of tanϕ in terms of sinϕ. Sinϕ Tan ϕ Cosϕ Tan ϕ Sinϕ Sin ϕ 6. If Secϕ+Tanϕ4 find sin ϕ, cosϕ Sec ϕ + Tan ϕ 4 5 Since ϕ + θ 90 6 o. + Cosϕ Sinϕ 4 Cosϕ + Sinϕ 4 Cosϕ 47

6 ( + Sinϕ) 6 Cos ϕ apply (C & D) ( + Sinφ) ( + Sinφ) + Cos φ 6 + Cos φ 6 ( + Sinφ) Sinφ( + Sinφ) 7 Sinφ 5 5 Sinϕ 7 Cosϕ Sin ϕ p 7. Secϕ+Tanϕp, prove tat sinϕ p P Secϕ + Tanϕ P. P.T Sinϕ P Proceed as in Question No Prove geometrically te value of Sin 60 o Exercise for practice. 9. tan θ 3 sin θ If,sow tat + tan θ 3 + cos θ Exercise for practice. 0. If xsecθ and tanθ,ten find te value of x x. x () Exercise for practice. 48

7 HEIGHTS AND DISTANCES. If te angle of elevation of cloud from a point meters above a lake is and te angle of depression of its reflection in te lake is cloud is., prove tat te eigt of te Ans : If te angle of elevation of cloud from a point n meters above a lake is and te angle of depression of its reflection in te lake is β, prove tat te eigt of te tan β + tanα cloud is tan β tanα Let AB be te surface of te lake and Let p be an point of observation suc tat AP meters. Let c be te position of te cloud and c be its reflection in te lake. Ten CPM and MPC β. Let CM x. Ten, CB CM + MB CM + PA x + CM In CPM, we ave tan PM x tan AB [ PM AB] AB x cot In PMC, we ave C' M tanβ PM.. x + tanβ [Θ C MC B+BM x + + n] AB AB (x + ) cot β From & x cot (x + ) cot β 49

8 x (cot - cot β) cot β (on equating te values of AB) tan β tanα x x tanα tan β tan β tanα + tan β tan β tanα x tan β tanα Hence, te eigt CB of te cloud is given by CB is given by CB x + tanα CB + tan β tanα tanα + tan β tanα (tanα + tan β ) CB- tan β tanα tan β tanα. From an aero plane vertically above a straigt orizontal road, te angles of depression of two consecutive milestones on opposite sides of te aero plane are observed to be α and β. Sow tat te eigt of te aero plane above te road is Let P Q be QB be x Given : AB mile QB x AQ (- x) mile in PAQ PQ Tan α AQ Tan α x. x Tanα. In PQB Tan β x x Tanβ Substitute for x in equation () + Tanβ Tanα 50

9 + Tanβ Tanα Tan β + Tanα TanβTanα 3. Two stations due sout of a tower, wic leans towards nort are at distances a and b from its foot. If α and β be te elevations of te top of te tower from te situation, prove tat its inclination θ to te orizontal given by Let AB be te leaning tower and C and D be te given stations. Draw BL DA produced. Ten, BAL 0, BCA α, BDC a and DA b. Let AL x and BL In rigt ALB, we ave : AL x Cot θ Cot θ BL x Cot θ x cot θ In rigt BCL, we ave : CL Cot α a + x cot α BL a (cot α - cot θ) a (cotα cotθ )..(i)...(ii) In rigt BDL, we ave : DL DA + cot β cot β BL BLAL b + x cot β b + x b cot β b ((cot β - cot θ) b (cot β cotθ ) [using (i)]..(iii) equating te value of in (ii) and (iii), we get: a b (cotα cotθ ) (cot β cotθ ) 5

10 a cot β - a cot θ b cot α - b cot θ (b a) cot θ b cot α - a cotβ b cotα a cot βθ cot θ ( b a) 4. Te angle of elevation of te top of a tower from a point on te same level as te foot of te tower is α. On advancing p meters towards te foot of te tower, te angle of elevation becomes β. sow tat te eigt of te tower is given by 5. A boy standing on a orizontal plane finds a bird flying at a distance of 00m from im at an elevation of A girl standing on te roof of 0 meter ig building finds te angle of elevation of te same bird to be Bot te boy and te girl are on opposite sides of te bird. Find te distance of te bird from te girl. ( 4.4m) In rigt ACB AC Sin 30 AB AC 00 AC 00 AC 50m AF (50 0) 30m In rigt AFE AF Sin 45 AE 30 AE AE x m 6. From a window x meters ig above te ground in a street, te angles of elevation and depression of te top and te foot of te oter ouse on te opposite side of te street are α and β respectively. Sow tat te eigt of te opposite ouse is Meters. Let AB be te ouse and P be te window Let BQ x PC x Let AC a 5

11 PQ In PQB, tan θ or tan θ QB x x cot θ tanθ AC a In PAC, tan θ or tan θ PC x a x tan θ > ( cot θ) tan θ tan θ cot θ. te eigt of te tower AB AC + BC a + tan θ cot θ + (tanθ cot θ + ) 7. Two sips are sailing in te sea on eiter side of a ligtouse; te angles of depression of two sips as observed from te top of te ligtouse are 60 0 and 45 0 respectively. If te distance between te sips is eigt of te ligtouse. (00m) In rigt ABC Tan 60 BC 3 BC H 3 BC In rigt ABD Tan 45 BD BD BC + BD 00 BC + 3 BC 00 BC 00( ( + 3) BC m eigt of ligt ouse 00m 3 meters, find te 53

8. A round balloon of radius a subtends an angle θ at te eye of te observer wile te angle of elevation of its centre is Φ. Prove tat te eigt of te center of te balloon is a sin θ cosec Φ /.

12 8. A round balloon of radius a subtends an angle θ at te eye of te observer wile te angle of elevation of its centre is Φ. Prove tat te eigt of te center of te balloon is a sin θ cosec Φ /. Let θ be te centre of te ballon of radius r and p te eye of te observer. Let PA, PB be tangents from P to ballong. Ten APB θ. APO BPO θ Let OL be perpendicular from O on te orizontal PX. We are given tat te angle of te elevation of te centre of te ballon is φ i.e., OPL φ θ OA In OAP, we ave sin OP θ a sin OP θ OP a cosec OL In OP L, we ave sinφ OP OL OP sin φ a cosec φ sin θ. Hence, te eigt of te center of te balloon is a sin θ cosec Φ /. 9. Te angle of elevation of a jet figter from a point A on te ground is After a fligt of 5 seconds, te angle of elevation canges to If te jet is flying at a speed of 70 km/r, find te constant eigt at wic te jet is flying.(use 3.73 ( 598m) 36 km / r 0m / sec 70 km / 0 x Speed 00 m/s Distance of jet from AE speed x time 00 x m tan 60 o AC oppositeside BC adjacentside AC 3 BC BC 3 AC 54

13 AC ED (constant eigt) BC 3 ED. tan 30 o ED oppositeside BC + CD adjacentside ED 3 BC BC ED 3 BC BC 3 (from equation ) 3 BC BC 3BC BC 3000 BC 3000 BC 3000 BC 500 m ED BC 3 (from equation ) x.73 ED 598m Te eigt of te jet figter is 598m. 0. A vertical post stands on a orizontal plane. Te angle of elevation of te top is 60 o and tat of a point x metre be te eigt of te post, ten prove tat Self Practice x. 3. A fire in a building B is reported on telepone to two fire stations P and Q, 0km apart from eac oter on a straigt road. P observes tat te fire is at an angle of 60 o to te road and Q observes tat it is an angle of 45 o to te road. Wic station sould send its team and ow muc will tis team ave to travel? (7.3km) Self Practice. A ladder sets against a wall at an angle α to te orizontal. If te foot is pulled away from te wall troug a distance of a, so tat is slides a distance b down te wall cosα cos β a making an angle β wit te orizontal. Sow tat. sin β sinα b Let CB x m. Lengt of ladder remains same 55

14 CB Cos α CA x Cos α x cos α ED AC Let Ed be ED AC () DC + CB cos β ED a + x cos β a + x cos β x cos β a () from () & () cos α cos β - a cos α - cos β - a -a (cosα - cosβ)...(3) AE + EB Sin α AC b + EB Sin α Sin α b EB EB Sin α b...(4) EB Sin β DE EB Sin β EB Sin β From (4) & (5) (5) Sin β Sin α b b sin α Sin β -b (Sin β - Sin α)...(6) Divide equation (3) wit equation (6) a (cosα cos β ) b (sin β Sinα) a b Cosα Cosβ Sinβ Sinα 56

15 3. Two stations due sout of a leaning tower wic leans towards te nort are at distances a and b from its foot. If α, β be te elevations of te top of te tower bcotα acot β from tese stations, prove tat its inclination ϕ is given by cotϕ. b a Let AE x, BE BE Tan φ AE x x x tanφ x cot φ BE tan α CE a + x a + x cot α x cot α - a BE tan β DE b + x b+x cot β x cot β - b from and cot φ cot α - a ( cot φ + cot α ) a a cot φ + cotα from and 3 cot φ cot β - b ( cot φ - cot β) b b cot φ + cot β from 4 and 5 a b cot φ + cotα cot φ + cot β a (cot β - cot φ ) b ( cot α - cot φ ) - a cot φ + b cot φ b cot α - a cot β (b a) cot φ b cot α - a cot β 57

16 b cot α - a cot β cot φ b a 4. In Figure, wat are te angles of depression from te observing positions O and O of te object at A? Self Practice ( 30 o,45 o ) 5. Te angle of elevation of te top of a tower standing on a orizontal plane from a point A is α. After walking a distance d towards te foot of te tower te angle d of elevation is found to be β. Find te eigt of te tower. ( ) cotα cot β Let BC x AB tan β CB tan β x x tan β x cot β () AB tan α DC + CB tan α d + x d + x cotα tan α x cot α - d () from and cot β cot α - d (cot α - cot β ) d d cot α cot β 58

17 6. A man on a top of a tower observes a truck at an angle of depression α were tanα and sees tat it is moving towards te base of te tower. Ten minutes 5 later, te angle of depression of te truck is found to be β were tan β 5, if te truck is moving at a uniform speed, determine ow muc more time it will take to reac te base of te tower... 0 minutes600sec A Let te speed of te truck be x m/sec CDBC-BD In rigt triangle ABC tanα BC BC 5. In rigt triangle ABD tanβ BD tanα 5 5BD ( tan β 5 ) CDBC-BD ( CD600x ) 600x 5BD-BD BD50x 50x Time taken x 50 seconds Time taken by te truck to reac te tower is 50 sec. C α D β B 59

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Downloaded from HEIGHTS AND DISTANCES 1. If te angle of elevation of cloud from a point meters aove a lake is and te angle of depression of its reflection in te lake is cloud is., prove tat te eigt of te Ans : If te angle