20.1 Heights and distances

Transcription

1 20 Heigts an istances INTROUTION In tis capter, we will learn te practical use of trigonometry in our ay-to-ay life. We will see ow trigonometry is use for fining te eigts an istances of various objects, witout actually measuring tem. 20. Heigts an istances To obtain te eigt of a mountain (pillar or minar) or te breat of a river, we require te following two efinitions: ngle of elevation an angle of epression Let O be an eye of an observer an OX be te orizontal line rawn troug O. Let be an object, ten (i) if is above OX as sown in fig. (), XO is calle angle of elevation (as observe from O). (ii) if is below OX as sown in fig. (2), XO is calle angle of epression (as observe from O). Object O angle of elevation ^ X O angle of epression X ^ Object () (2) Te line O, joining te eye to te object, is calle te line of sigt. Remarks Wile ealing wit problems on eigts an istances, usually, we fin require sie = a certain t-ratio of a known angle. known sie Remember tat 2 = 44 an 3 = 732

2 Illustrative Eamples Eample. tower stans vertically on te groun. From a point on te groun wic is 5 m away from te foot of te tower, te angle of elevation of te top of te tower is foun to be. Fin te eigt of te tower. Solution. Let M be te tower of eigt metres an O be te point on te groun 5 m away from te foot of tower. Ten, te angle of elevation = MO = (given). In ΔOM, OM = 90. From ΔOM, we get tan = M om 3 = 5 = 5 3 = = Hence, te eigt of te tower = m. Eample 2. circus artist is climbing a 20 m long rope, wic is tigtly stretce an tie from te top of a vertical pole to te groun. Fin te eigt of te pole, if te angle mae by te rope wit te groun level is. Solution. Let be te vertical pole an be te rope wic is tigtly stretce an tie from te top of pole to te groun, ten = 20 m. Let metres be te eigt of te vertical pole. Ten, angle of elevation = = (given). From rigt angle Δ, we get sin = = 2 20 = 0. Hence, te eigt of te pole is 0 m. Eample 3. kite is flying at a eigt of 60 metres from te level groun, attace to a string incline at to te orizontal. Fin te lengt of te string. Solution. Let be te kite an M = 60 metres, te eigt of te kite. Let te string be el at te point O, ten MO = (given) an O is te lengt of te string. From rigt-angle ΔOM, sin = M 3 60 = o 2 o Kite o = = 40 3 = = Hence, te lengt of te string = m. O Groun M O 5 m 20 m M 60 m Eample 4. If te lengt of a saow cast by a pole be 3 times te lengt of te pole, fin te angle of elevation of te sun. Solution. Let M be te pole, ten its saow sun = OM = 3 M (accoring to given). Let MO = θ, te angle of elevation of te sun. 504 Unerstaning ISE matematics θ O saow M pole

3 From rigt-angle OM, we get tan θ = M om but OM = 3 M M tan θ = = = tan 3 M 3 θ =. Hence, te angle of elevation of te sun =. Eample 5. n electrician as to repair an electric fault on a pole of eigt 5 m. He nees to reac a point 3 m below te top of te pole to unertake te repair work (as sown in te ajoining figure). Wat soul be te lengt of te laer tat e soul use wic, wen incline at an angle of to te orizontal, woul enable im to reac te require position? lso, ow far from te foot of te pole soul e place te foot of te laer? Take 3 = 73 How te electricity can be save for te progress of te country? (Value ase) Solution. Let l metres be te lengt of te laer an metres be te istance between te foot of te pole an te foot of te laer. = = 5 m 3 m = 3 7 m. From rigt angle Δ, we get sin = 3 37 =. 2 l l = = = lso, tan = 3 = 37. = = = = 4 27 (appro.) = 2 3 (appro.). Hence, te lengt of te laer soul be 4 27 m (appro.) an e soul place te foot of te laer at a istance of 2 3 m (appro.) from te foot of te pole. To save electricity, we soul minimise te use of electrical appliances. Flourescent bulbs an manually operate macines are also elpful in saving electricity. Eample 6. tree breaks ue to storm an te broken part bens (witout being etace) so tat te top of te tree touces te groun making an angle of wit it. Te istance between te foot of te tree to te point were te top touces te groun is 8 m. Fin te eigt of te tree. Solution. Let be te tree. Wen broken at point by te storm, let its top touc te groun so tat = an = 8 m. From rigt angle Δ, we get 5 m 3 m Laer tan = = 3 8 m = 8 3 m. lso cos = 3 2 Te eigt of tree = + = 8 m = = 6 3 m. 8 6 m + m m Heigts an istances 505

4 = m = 24 3 m = 8 = m = m 3 m Hence, te eigt of te tree is 3 86 m (appro.). Eample 7. n observer 5 m tall is 28 5 m away from a cimney. Te angle of elevation of te top of te cimney from is eye is 45. Wat is te eigt of te cimney? Solution. Let be an observer of eigt 5 m wic is 28 5 m away from a cimney of eigt metres. = 5 m an = 28 5 m. From, raw E, ten E is a rectangle. E = E = = ( 5) m an E = = 28 5 m. Given, angle of elevation E = 45. From rigt angle ΔE, we get tan 45 = e e = E = 28 5 = 30. Hence, te eigt of te cimney = 30 m. Eample 8. Te angle of elevation of te top of a ill at te foot of te tower is an te angle of elevation of te top of te tower from te foot of te ill is. If te tower is 20 m ig, fin (i) te eigt of te ill (ii) te istance between te ill an te tower. Solution. Let = metres be te eigt of te ill an be te tower, ten = 20 metres (given). Let = metres be te istance between te foot of te ill an te foot of te tower. Ten = an = (given). From rigt angle, we get 28 5 m tan = 3 = = 3 () From rigt angle, we get tan = 20 = 3 20 m = 20 3 (2) (i) Substituting te value of from (2) in (), we get = 3 (20 3 ) = 60. Te eigt of te ill = 60 metres. (ii) From (2), = 20 3 = = Te istance between te ill an te tower = metres. Eample 9. statue, 6 m tall, stans on te top of a peestal. From a point on te groun, te angle of elevation of te top of te statue is an from te same point te angle of elevation of te top of te peestal is 45. Fin te eigt of te peestal. Solution. Let be te peestal an be te statue wic stans on te top of peestal, an O be te point of observation on te groun, = 6 m. 506 Unerstaning ISE matematics

5 Let = metres an O = metres. Given O = 45 an O = From rigt angle ΔO, we get tan 45 = o = = (i) From rigt angle ΔO, we get tan = o 3 = = = 6 (using (i)) 6. ( 3 ) = 6 = = = 6. ( 3 + ) = (0 8) ( 3 + ) = 0 8 ( ) = = 2 9 (appro). O 45 metres 6 m metres Hence, te eigt of te peestal = 2 9 m (appro). Eample 0. pole 6 m ig is fie on te top of a tower. Te angle of elevation of te top of te pole observe from a point on te groun is an te angle of epression of te point from te top of te tower is 45. Fin te eigt of te tower. (Take 3 = 73) Solution. Let metres be te eigt of te tower an be te pole of eigt 6 m fie on te top of te tower. is te point of observation from te groun, ten =. From top of tower, te angle of epression of te point is 45, so = 45. Let be metres. From rigt angle Δ, we get tan 45 = = = From rigt angle Δ, we get (i) tan = 3 = = = + 6 (using (i)) ( 3 ) = 6 = = 3 + = 3( 3 + ) 3 + = 3( 73 + ) = = 8 9. Hence, te eigt of te tower is 8 9 metres. Eample. Te angles of elevation of te top of a tower from two points an Q at istances of a an b respectively from te base an in te same straigt line wit it are complementary. rove tat te eigt of te tower is ab. Solution. Let be te tower of eigt (units). = a, Q = b. s te angles of elevation are complementary, if = θ ten Q = 90 θ. From rigt angle Δ, we get tan θ = a (i) θ 45 Q θ 6 Heigts an istances 507

6 From rigt angle ΔQ, we get tan (90 θ ) = b cot θ = Multiplying (i) an (ii), we get tan θ cot θ = a = 2 b ab b (ii) 2 = ab = ab ( > 0) Hence, te eigt of te tower is ab units. Eample 2. person staning on te bank of a river observes tat te angle of elevation of te top of a tree staning on te opposite bank is. Wen e moves 50 m away from te bank e fins tat te angle of elevation to be. alculate : (i) te wit of te river an (ii) te eigt of te tree. (2003) Solution. Let = metres be te eigt of te tree an = metres be te breat of te river. Let be te first position of te man, an be te position after moving 50 metres away from te bank, ten = 50 m, = an =. From rigt-angle, we get tan = 3 = = 3 () From rigt-angle, we get tan = = m 3 = 50 + (2) (i) Substituting te value of from () in (2), we get 3 ( 3 ) = = = 50 = 25. Te breat of te river = 25 metres. (ii) From (), we get = 3 25 m = m = 43 3 m. Te eigt of te tree = 43 3 metres. Eample 3. n aeroplane at an altitue of 500 metres fins tat two sips are sailing towars it in te same irection. Te angles of epression as observe from te aeroplane are 45 an respectively. Fin te istance between te two sips. (206) Solution. Let be te position of te Horizontal aeroplane at an altitue of 500 m, ten M = 500 m. Let an be te positions of te two sips wose angles of epression as observe from are an 45 respectively, ten M = an M = From rigt angle ΔM, 45 M tan 45 = M M = 500 M M = 500. tan = M M 3 = 500 M = M 508 Unerstaning ISE matematics From rigt angle ΔM, 500 m

7 From figure, = M M = = 500( 3 ) = 500 ( 732 ) = = 098. Hence, te istance between te sips = 098 metres. Eample 4. Te saow of a tower staning on a level groun is foun to be 40 m longer wen te sun s altitue is tan wen it is. Fin te eigt of te tower. Solution. Let te eigt of te tower be metres an te lengt of its saow be metres wen te sun s altitue i.e. elevation is. Wen te sun s altitue is, ten te lengt of saow of te tower is 40 m longer i.e. = metres, = metres an = 40 metres. From rigt angle Δ, we get tan = = 3 (i) 40 m From rigt angle Δ, we get tan = = = = 40 = 20. From (i), = 20 3 = = Hence, te eigt of te tower is metres. (using (i)) Eample 5. Two persons staning on te same sie of a tower in a straigt line wit it, measure te angles of elevation of te top of te tower as 25 an 50 respectively. If te eigt of te tower is 70 m, fin te istance between te two persons. (2004) Solution. Let, be te positions of te two persons an be te tower, ten = 70 m (given) = 25 an = 50. From rigt-angle, we get cot 25 = = 70 = 70 cot 25 () 25 From rigt-angle, we get cot 50 = = 70 = 70 cot 50 (2) From figure, = = 70 cot cot 50 (Using () an (2)) = 70 (cot 25 cot 50 ) = 70 [cot (90 65 ) cot (90 40 )] = 70 (tan 65 tan 40 ) = 70 ( ) (From trigonometric tables) = = Te istance between te two persons = 9 4 metres (approimately). Eample 6. From te top of a cliff 50 m ig, te angles of epression of two boats are an. Fin te istance between te boats, if te boats are (i) on te same sie of te cliff (ii) on te opposite sies of te cliff. 50 Heigts an istances 509

8 Solution. Let = 50 m be te eigt of te cliff an let, be te two boats, ten = an =. Horizontal oats on same sie cliff cliff Horizontal oats on opposite sies (i) (ii) In bot cases (boats on same sie of te cliff or boats on opposite sies of te cliff), from rigt-angle, we get tan = = 50 3 = 50 From rigt-angle, we get 3 = 50 3 () tan = 50 = 3 = 50 3 (2) (i) oats on te same sie of te cliff. istance between te boats = = ( ) m = 00 3 m = m = 73 2 m. (ii) oats on te opposite sies of te cliff. istance between te boats = + = ( ) m = m = m = m. Eample 7. From te top of a ligt ouse 00 m ig te angles of epression of two sips on opposite sies of it are 48 an 36 respectively. Fin te istance between te two sips to te nearest metre. (200) Solution. = 48 an = 36. From rigt angle, cot 48 = = cot 48. From rigt angle, cot 36 = = cot 36 Te istance between te two sips = + = cot 48 + cot 36 = (cot (90 42 ) + cot (90 54 )) = (tan 42 + tan 54 ) 50 Unerstaning ISE matematics (From figure (i)) (From figure (ii)) 48 36

9 = 00 ( ) metres = metres = metres = 228 metres (to te nearest metre). Eample 8. Two poles of equal eigts are staning opposite to eac oter on eiter sie of te roa wic is 80 m wie. From a point between tem on te roa, te angles of elevation of te top of te poles are an, respectively. Fin te eigt of te poles an te istances of te point from te poles. Solution. Let, be poles of eigt metres (eac) an = metres, ten = (80 ) metres. Given, = an =. From rigt angle Δ, we get 80 tan = = 3 (i) From rigt angle Δ, we get tan = 3 = = = 80 4 = 80 = 20. From (i), = 20 3 = = Heigts an istances (using (i)) lso = (80 20) m = 60 m. Hence, te eigt of eac pole is m; istances are : 20 m from te pole wose elevation is an 60 m from te pole wose elevation is. Eample 9. boy of eigt 7 m is staning 20 m away from a flagstaff on te same level groun. He observes tat te angle of elevation of te top of te flagstaff is 24. alculate te 24 eigt of te flagstaff. E Horizontal Solution. Let be te boy of eigt 7 m an be 20 m te flagstaff of eigt metres, ten = 20 m (given). Troug (eye of te boy) raw orizontal line E to meet at E, ten E = 24 (given) an E = = 20 m. ( E is a rectangle) lso E = E = = ( 7) metres. From rigt-angle E, we get e tan 24 = e tan 24 = = 20 tan 24 7 = (using tables of natural tangents) 7 = = = Te eigt of te flagstaff = metres. Eample 20. girl 5 m tall is staning at some istance from a 30 m ig tower. Te angle of elevation from er eye to te top of te tower increase from to as se walks towars te tower. Fin te istance se walke towars te tower. boy flag staff 5

10 Solution. Let be te tower an GE represent te girl wen te angle of elevation is ; an F be te position of girl wen te angle of elevation is. Let G = metres an = y metres = = GE = (30 5) m = 28 5 m. From, tan = 3 = (i) y From G, tan = g 3 = (ii) + y From (i) an (ii), we get y + y = y = y 3y = + y = 2y = G.5 m E F (using (i)) = 57 3 = 9 3 = = istance walke by te girl towars tower = 32 9 metres (appro.). Eample 2. In te figure given alongsie, from te top of a builing, 60 metres ig, te angles of epression of te top an bottom of a vertical lamp post are observe to be an respectively. Fin : (i) te orizontal istance between an. (ii) te eigt of te lamp post. (203) Solution. (i) Let metres be te orizontal istance between an. s te angle of epression of te bottom of te lamp post wen observe from te top of builing is, Horizontal = From rigt angle Δ, 60 E tan = 3 = = 20 3 (i) = = Hence, te orizontal istance between an = metres. (ii) From, raw E. Ten e = = metres an E = E = = (60 ) metres. s te angle of epression of te top of te lamp post wen observe from is, so E = From rigt angle Δ E, e 60 tan = = e 3 3 = = 60 = 40 Hence, te eigt of te lamp post is 40 metres. 52 Unerstaning ISE matematics (using (i))

11 Eample 22. man staning on te eck of a sip, wic is 0 m above te water level, observes te angle of elevation of te top of a ill as an te angle of epression of te base of te ill as. alculate te istance of te ill from te sip an te eigt of te ill. Solution. Let be te eck of te sip an te man be staning at, ten = 0 metres. Let = metres be te eigt of te ill an = metres be te istance between te ill an te sip. From, raw N, ten N = = metres, N = = 0 metres an N = ( 0) metres. Given N = an N =. From rigt-angle N, we get tan = n 0 = n 3 = 0 3 = = 7 32 Te istance between te ill an te sip = 7 32 metres. From rigt-angle N, we get 0 0 tan = 3 = n 0 = 3 0 = 3 (0 3 ) = = 40. Te eigt of te ill = 40 metres. Eample 23. Te orizontal istance between two towers is 20 m. Te angle of elevation of te top an angle of epression of te bottom of te first tower as observe from te secon tower is an 24 respectively. Fin te eigt of te two towers. Give your answer E correct to 3 significant figures. (205) Solution. Let te eigt of te two towers an be metres an y metres respectively. Given = 20 metres e = 20 metres. From figure, E = = y metres an E = E = = ( y) metres. From rigt angle ΔE, e y tan 24 = = e 20 y = From rigt angle ΔE, tan = e e = y 20 y = = = y = = Hence, te eigt of te tower correct to 3 significant figures is 23 metres an te eigt of te tower correct to 3 significant figures is 53 4 metres. E 0 N Heigts an istances 53

12 Eample 24. Te lower winow of a ouse is at a eigt of 2 m above te groun an its upper winow is 4 m vertically above te lower winow. t certain instant te angle of elevation of a balloon from tese winows are observe to be an, respectively. Fin te eigt of te balloon above te groun. Solution. Let be te lower winow an be te upper winow of a ouse, ten = 2 m an = 4 m. Let be te position of te balloon at eigt metres at E tat particular instant. From, raw M an from, raw E M, 4 m ten = M M = M = ( 2) metres, E = M M E = ( 2 4) metres = ( 6) metres. 2 m Let M = metres, ten = E = M = metres. M From rigt angle Δ, we get tan = 3 = 2 (i) From rigt angle ΔE, we get tan = e e 3 iviing (i) by (ii), we get = 6 = 6 3 = = 2 2 = 6 = 8. Hence, te eigt of te balloon is 8 m. (ii) Eample 25. boy staning on te groun fins a bir flying at a istance of 00 m from im at an elevation of. girl staning on te roof of 20 m ig builing fins te angle of elevation of te same bir to be 45. Te boy an te girl are on opposite sies of te bir. Fin te istance of te bir from te girl, correct to nearest cm. Solution. Let te bir be at te point. Te boy is at te point on te groun an te girl is at te point G on te roof, 20 m above te groun. Given = 00 m an G = 20 m. From rigt-angle N, sin = n 2 = n 00 N = 50 m = N N = N G = 50 m 20 m = 30 m. From rigt-angle G, sin 45 = g 00 m N 45 G 20 m 30 g 2 = 54 Unerstaning ISE matematics

13 g = 30 2 m = (30 442) m = m = m (nearly) Hence, te bir is m away from te girl. Eample 26. Te angle of elevation of a bir from a point 50 metres above a lake is an te angle of epression of its reflection in te lake is. Fin te eigt of te bir. Solution. Let O be te point of observation 50 metres above lake level an let te bir be at te point. If is te reflection of te bir in te lake, ten M = M. Let te eigt of te bir above te lake O N 50 m level be metres, ten M N = ( 50) metres an N = ( + 50) metres. Lake ON = an ON =. From te rigt-angle ON, tan = n 50 = on 3 on ON = 3 ( 50) (i) From rigt-angle ON, tan = n = on on 3 ON = ( 50) = + 50 (using (i)) 3 50 = = 200 = 00. Te eigt of te bir (above lake level) = 00 metres. Eample 27. straigt igway leas to te foot of a tower. man staning at te top of te tower observes a car at an angle of epression of, wic is approacing te foot of te tower wit a uniform spee. Si secons later, te angle of epression of te car is foun to be. Fin te time taken by te car to reac te foot of te tower from tis point. Solution. Let M be te tower of eigt metres. Let be te position of te car wen its angle of epression (as seen from ) is an 6 secons later, let te car be at te point wen its angle of epression is. Let be metres an M be y metres. From M, tan = M M = (i) 3 + y From M, tan = M M 3 = y On iviing (ii) by (i), we get y + y = 3= y y 3y = + y y = 2. (ii) s te car takes 6 secons in moving from to i.e. to cover metres, time taken by te car in moving from to M i.e. 2 metres y M Heigts an istances 55

14 = 6 2 secons = 3 secons. Hence, te time taken by te car to reac te foot of tower = 3 secons. Eample 28. Te angle of elevation of a jet plane from a point on te groun is. fter a fligt of 5 secons, te angle of elevation canges to. If te jet plane is flying orizontally at a constant eigt of metres, fin te spee of te jet plane. Solution. Let be te position of te jet plane wen its elevation from a point on te groun is an Q be its position wen te angle of elevation is. Given M = NQ = m. From rigt angle M, tan = m am M = 500 m. From rigt angle NQ, m 3 = am nq m tan = = an 3 an N = 4500 m Q = MN = N M = ( ) m = 3000 m. M N Q Te spee of te plane = = 200 m/s = 200 m/s 8 5 km/ = 720 km/. Eample m tall girl spots a balloon moving wit te win in a orizontal line at a eigt of 9 4 m from te groun. Te angle of elevation of te balloon from te eyes of te girl at tat instant is. fter some time, te angle of elevation reuces to. Fin te istance travelle by te balloon uring tat interval. Take 3 = 732. Solution. Let represent te girl 4 m tall an, Q be te two positions of te balloon wen te angles of elevation observe from te point are Q an respectively. Troug, raw a orizontal line X. Y represents groun. From, raw M Y to meet X at M; an from M N X Q, raw QN Y to meet X at N. M = M MM = M = 9 4 m 4 m = 90 m, M Groun N Y ten NQ = M = 90 m ( MNQ is a rectangle) Let Q be metres, ten MN = Q = metres Let M be metres, ten N = M + MN = ( + ) metres. In rigt angle M, M = tan = m 90 3 = (i) am In rigt angle NQ, QN = nq 90 tan = = an 3 + (ii) 56 Unerstaning ISE matematics

15 iviing (i) by (ii), we get = = 3 = + = 2 (iii) From (i), = 90 = From (iii), = = 60 3 = = Hence, te istance travelle by te balloon uring te interval = metres. Eample 30. sperical balloon of raius r subtens an angle θ at te eye of an observer. If te angle of elevation of its centre is ϕ, fin te eigt of te centre of te balloon. Solution. Let be te centre of a sperical balloon of raius r an te eye of te observer be at te point O. Let O an O be tangents rawn from O to te spere. Since te balloon subtens an angle θ is an eye of te observer, so O = θ. From geometry, we know tat O bisects O, so O = O = θ. 2 From rigt angle ΔO, we ave sin θ = sin θ 2 o = r (i) 2 o Let te eigt of te centre of te balloon be i.e. M =. From rigt angle ΔOM, we ave sin ϕ = M o sin ϕ = iviing (ii) by (i), we get o (ii) o = sin φ = r sin ϕ cosec θ o r θ. sin 2 2 Hence, te eigt of te centre of balloon is r sin ϕ cosec θ. 2 O θ ϕ M Eercise 20. n electric pole is 0 metres ig. If its saow is 0 3 metres in lengt, fin te elevation of te sun. 2. Te angle of elevation of te top of a tower, from a point on te groun an at a istance of 50 m from its foot, is. Fin te eigt of te tower correct to one place of ecimal. 3. laer is place against a wall suc tat it just reaces te top of te wall. Te foot of te laer is 5 metres away from te wall an te laer is incline at an angle of wit te groun. Fin te eigt of te wall. 4. Wat is te angle of elevation of te sun wen te lengt of saow of a vertical pole is equal to its eigt? 5. river is 60 m wie. tree of unknown eigt is on one bank. Te angle of elevation of te top of te tree from te point eactly opposite to te foot of te tree, on te oter bank, is. Fin te eigt of te tree. 6. From a point on level groun, te angle of elevation of te top of a tower is. If te tower is 00 m ig, ow far is from te foot of te tower? Heigts an istances 57