Application of Trigonometry

APPLICATION OF TRIGONOMETRY CONTENTS Sol. Let AB be the building and AC be its shadow.  Angle of Elevation  Angle of Depression B 60º  ANGLE OF ELEVATION C The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e. the case when we raise our head to look at the object. (see fig.) Then, AC = 20 m and ACB = 60º. Let AB = h. Then, A  O  A B  Ex.2 3 3 )m = (20 × 1.732) m If a vertical pole 6m high has a shadow of length 2 3 metres, find the angle of elevation of the sun. Sol. Let AB be the vertical pole and AC be its shadow. Let the angle of elevation be . Then, B Angle of depression C EXAMPLES  Ex.1 h = 20 3 = 34.64 m. B The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e. the case when we lower our head to look at the point being viewed. (See fig.) AB = tan 60º = AC  h = (20 × Angle of elevation  ANGLE OF DEPRESSION A The shadow of a building is 20 m long when the angle of elevation of the sun is 60º. Find the height of the building.  C A AB = 6 m, AC = 2 3 m and ACB = . Now, tan  = 6 AB = = AC 2 3 3 = tan 60º.   = 60º. Ex.3 A ladder against a vertical wall makes an angle of 45º with the ground. The foot of the ladder is 3m from the wall. Find the length of the ladder. Sol. Let AB be the wall and CB, the ladder.   200  3   m = 173.2 m.  AB =    2   Ex.5 The pilot of a helicopter, at an altitude of 1200m finds that the two ships are sailing towards it in the same direction. The angle of depression of the ships as observed from the helicopter are 60º and 45º respectively. Find the distance between the two ships. Sol. Let B the position of the helicopter and let C, D be the ships. Let AB be the vertical height. B B 60º 1200 m 45º C A D Then, AC = 3m and ACB = 45º CB = sec 45º = Now, AC  CB = 2  3 Sol. 2 ACB = 60º and ADB = 45º. Then,  Length of the ladder = CB = 3 2  A balloon is connected to a meteorological station by a cable of length 200 m, inclined at 60º to the horizontal. Find the height of the balloon from the ground. Assume that there is no slack in the cable.   AB 3 = 200 2 1200 3 = 400 3 m. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30º and that of the top of the flagstaff is 45º. Find the height of the tower. Sol. Let AB be the tower and BC be the flagstaff. A 3 AB = sin 60º = BC 2 1 AC = 1200 3 Ex.6 60º Then, 1 AC = cot 60º = AB 3  AC = B Then, BC = 200 m and ACB = 60º AD = cot 45º = 1 AB AD = 1  AD = 1200 m 1200 And, Let B be the balloon and AB be the vertical height. Let C be the meteorological station and CB be the cable. C 45º 60º C A Then, AB = 1200 m, = (3 × 1.41) m = 4.23 m Ex.4 45º  BE = (30 – x) m. C Now, B  30º O 45º 1 1 BE BE = tan 30º =  = DE 30 3 3 Let O be the point of observation. Then, AOB = 30º and AOC = 45º.  BE = OA = cot45º = 1 AC   OA = AC = h + 7. And,  OA = cot 30º = AB OA = h 7 3  OA = h 3 Ex.7 Sol. 3 1  3  1 7( 3  1) = = 9.562 m 2 3 1 (a) 15(1  3 ) m (b) 30( 3  1) m  1   m (c) 301  3   1   m (d) 301  3   30    CD = AE = AB – BE =  30  3  Sol. Let AB be the cliff and CD be the tower. D B  E  A C Then, AB = 25 m. From B draw BE  CD. 30º Let EBD = ACB = . 30º Now, D  45º A . From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower. Let AB be the building and CD be the tower. E 3 Ex.8 From the top of a building 30 m high, the top and bottom of a tower are observed to have angles of depression 30º and 45º respectively. The height of the tower is : B 30  1   m = 301  3  3  h+7= h 3 h= AC = 1  AC = 30 m. 30  DE = AC = 30 m. A Then, BC = 7 m. Let AB = h. Now, AC = cot 45º = 1 AB C Then, AB = 30 m. Let DC = x. Draw DE  AB. Then AE = CD = x. DE AB = tan  and = tan  BE AC DE AB = . So, DE = AB BE AC [ BE = AC]  CD = CE + DE = AB + AB = 2AB = 50m Ex.9 (A) 30 3 m (C) Sol. shadows of a pole 15 m high, made at these two positions, is– The altitude of the sun at any instant is 60º. The height of the vertical pole that will cast a shadow of 30 m is 30 3 (B) 15 m (A) 7.5 m (B) 15 m (C) 10 3 m (D) (D) 15 2 m m Let AB be the pole and AC be its shadow. Sol. Then,  = 60º and AC = 30 m. 15 3 m When AB = 15m,  = 30º, then  AC = 15 3 B m. When AB = 15m,  = 60º, then C   A AB = tan 60º   AC  AB = 30 3 3  AB = 30 3 m (B) 50 3 m m (C) 25 m Sol.  Diff. in lengths of shadows  15   = 15 3  3  Ex.10 When the sun is 30º above the horizontal, the length of shadow cast by a building 50m high is50 = Let AB be the building and AC be its shadow. Then, AB = 50 m and  = 30º. 30 3 = 10 3 m . Ex.12 The heights of two poles are 80 m and 62.5 m. If the line joining their tops makes an angle of 45º with the horizontal, then the distance between the poles, is - (D) 25 3 m Sol. (A) 17.5 m (B) 56.4 m (C) 12.33 m (D) 44 m Let AB and CD be the poles such that AB = 80 m and CD = 62.5 m. B C   B D 30º AC = cot 30º = AB AC = tan60º AB  AC = 15 3 m 60º  (A) AC = tan30º AB 45º E A 3 AC  3 50  AC = 50 3 cm. Ex.11 If the elevation of the sun changed from 30º to 60º, then the difference between the lengths of C A Draw DE  AB. Then,  EDB = 45º Now, BE = AB – AE = AB – CD = 17.5 DE = cot 45º = 1 BE  DE = BE = 17.5 m. Ex.13 If the angle of elevation of cloud from a point 200 m above a lake is 30º and the angle of depression of its reflection in the lake is 60º, then the height of the cloud above the lake, is Sol. (A) 200 m (B) 500 m (C) 30 m (D) None of these Sol.  (A)  cos sec  2 (B)  cos  sec  (C)  sin cos ec 2 (D)  sin  cos ec Clearly, right angled triangles OAC and OBC are congruent. C 30º 60º B C A B   S L O C' Let CS = CS = x. 1 BC = tan 30º = AB 3 Also, 3  3  x + 200 = (AB) 3 .  3 ( x  200) = x  200 3 OC  = cosec CA 2  OC =  cosec AB BC' Also, = tan 60º = AB or x = 400.  CS = 400 m. Ex.14 A balloon of radius  makes an angle  at the eye of an observer and the angle of elevation of its centre is . The height of its centre from the ground level is given by :  . 2 Let CN  OX. Now,  x – 200 = X N  AOC = BOC = Now,  2 Let C be the centre of the balloon and O be the position of the observer at the horizontal line OX. Let OA and OB be the tangents to the balloon so that AOB = , XOC =  and CA = CB = . Let C be the cloud and C be its reflection in the lake. A  2   2 ....(i) CN = sin  OC  CN = OC sin =  cosec  sin [Using(i)] 2 Ex.15 The banks of a river are parallel. A swimmer starts from a point on one of the banks and swims in a straight line inclined to the bank at 45º and reaches the opposite bank at a point 20 m from the point opposite to the starting point. The breadth of the river is - Sol. (A) 20 m (B) 28.28 m (C) 14.14 m (D) 40 m Let A be the starting point and B, the end point of the swimmer. Then AB = 20 m and BAC = 45º. Let u m/min be the uniform speed of the trawler. A 45º Distance covered in 6 min = 6u metres. 1 BC = sin45º = AB 2 Now, 3 = 3 3u 3  h= 3 3 u = 3u. Time taken by trawler to reach A = 20  2 = 14.14 m.  BC = 2 Ex.16 A man on a cliff observes a fishing trawler at an angle of depression of 30º which is approaching the shore to the point immediately beneath the observer with a uniform speed. 6 minutes later, the angle of depression of the trawler is found to be 60º. The time taken by the trawler to reach the shore is - Sol. = 6u 3 h Now, AD = 1 BC = 20 2  2h  CD = 6u  B C (A) 3 3 min (B) (C) 1.5 min (D) 3 min 3 min distance AD 3u  A = = 3 min speed u Ex.17 A boat is being rowed away from a cliff 150m high. At the top of the cliff the angle of depression of the boat changes from 60º to 45º in 2 minutes. The speed of the boat is – Sol. (A) 2 km/hr (B) 1.9 km/hr (C) 2.4 km/hr (D) 3 km/hr Let AB be the cliff and C and D be the two positions of the ship. Then, AB = 150 m, ACB = 60º and ADB = 45º. Let AB be the cliff and C and D be the two positions of the fishing trawler. B B D 30º 60º A D C Then, ACB = 30º and ADB = 60º Let AB = h. Now,   AD = cot 60º = AB  AD = And, h 3 3   AD = cot 45º = 1 AB AD =1 150 3 3h  h  2h  = CD = AC – AD =  3 h  3 3   AD = 150 m. 1 1 AC AC = cot 60º = =  AB 150 3 3  AC = . AC = cot 30º = AB  AC = 1 Now, 45º 60º C A 150 3 = 50 3 = 86.6 m.  CD = AD – AC = (150 – 86.6) m = 63.4 m Thus, distance covered in 2 min. = 63.4 m  Speed of the boat  63.4 60  =   km / hr. = 1.9 km/hr. 1000   2 Ex.18 A tower is 100 3 metres high. Find the angle of elevation of its top from a point 100 metres away from its foot. Sol. In ABD, we have tan 45º =  1 = Let AB be the tower of height 100 3 metres, and let C be a point at a distance of 100 metres from the foot of the tower. tan 30º = B  h =  tan  = 3 40 3 m = 23.1 m 3 m= 23.1 + h1 = 40  h1 = (40 – 23.1)m = 16.9 m A Ex.20 Two stations due south of a leaning tower which leans towards the north are at distance a and b from its foot. If , be the elevations of the top of the tower from these stations, prove that its inclination  to the horizontal is given by In CAB, we have AB AC tan = 40 1 BC h =  AB 40 3 Substituting the value of h in (i), we get 100 3m C h  h1  h + h1 = 40 m ….(i) 40 In ABC, we have Let be the angle of elevation of the top of the tower from point C.  100 m 100 3 = 100 3 cot =   = 60º Sol. Hence, the angle of elevation of the top of the tower from a point 100 metres away from its foot is 60º. b cot   a cot  ba Let AB be the leaning tower and let C and D be two given stations at distances a and b respectively from the foot A of the tower. B Ex.19 From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30º. The angle of elevation of the top of a water tank (on the top of the tower) is 45º. Find the (i) height of the tower (ii) the depth of the tank. Sol. Let BC be the tower of height h metre and CD be the water tank of height h1 metre. Let A be a point on the ground at a distance of 40 m away from the foot B of the tower. D 30º A BD AB   a x D b Let AE = x and BE = h In ABE, we have h1 C tan = h  x = h cot  45º  BE AE In CBE, we have B tan = BE CE  tan = h x ....(i)  tan =   reflection in the lake. Then, CB = C´B. Let PM be perpendicular from P on CB. Then, CPM = and MPC´ = Let CM = x. h ax  a + x = h cot   x = h cot – a C ....(ii) x In DBE, we have  tan =   M  h h bx h B` A` x+h`  b + x = h cot   x = h cot – b ....(iii) In CPM, we have h cot = h cot – a  h (cot – cot ) = a  h = a cot   cot  tan = CM PM  tan = x AB ....(iv) From equation (i) and (iii), we get h cot = h cot – b  h (cot – cot ) = b  h = C` Then, CB = CM + MB = CM + PA = x + h. From equations (i) and (ii), we have b cot   cot  Equating the values of h from equations (iv) and (v), we get b a = cot   cot  cot   cot   a(cot – cot ) = b(cot – cot )  (b – a) cot = b cot – a cot  b cot   a cot   cot = ba Ex.21 If the angle of elevation of a cloud from a point h metres above a lake is  and the angle of depression of its reflection in the lake is , prove that the height of the cloud is h (tan   tan ) . tan   tan  Sol.  P BE tan = DE Let AB be the surface of the lake and let P be a point of observation such that AP = h metres. Let C be the position of the cloud and C´ be its [ PM = AB]  AB = x cot  ....(i) In PMC´, we have tan = C´M PM tan = x  2h AB [C´M = C´B + BM = x + h + h]  AB = (x + 2h) cot  From (i) and (ii), we have x cot = (x + 2h) cot   x(cot – cot ) = 2h cot   1 2h 1   =  x   tan   tan  tan    tan   tan   2h  =  x  tan  tan  tan     x = Hence, 2h tan  tan   tan  ....(ii) found that the angle of elevation is . Show that the height of the cliff is Height of the cloud = x + h = 2h tan  +h tan   tan  = 2h tan   h tan   h tan  tan   tan  = h (tan   tan  ) tan   tan  k (cos   sin  cot ) metres cot   cot  Sol. Ex.22 There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and Q are points directly opposite to each other on two banks, and in line with the tree. If the angles of elevation of the top of the tree from P and Q are respectively 30º and 45º, find the height of the tree. Sol. Let AB be the cliff and O be the fixed point such that the angle of elevation of the cliff from O is  i.e. AOB = . Let AOC = and OC = k metres. From C draw CD and CE perpendiculars on AB and OA respectively. Then, DCB = . Let h be the height of the cliff AB. Let OA be the tree of height h metre. In triangles POA and QOA, we have tan 30º = OA OA and tan 45º = OP OQ In OCE, we have sin = CE OC  sin = CE k  CE = k sin  ...(i) [CE = AD]  AD = k sin   1 3 =  OP = h h and 1 = OQ OP And, cos = 3 h and OQ = h cos =  OP + OQ = 3 h + h PQ = (  100 = ( 3 + 1)h  h = 100 3 1 m h = 3 + 1)h [ PQ = 100 m] 100( 3  1) m 2  h = 50(1.732 – 1) m = 36.6 m Hence, the height of the tree is 36.6 m Ex.23 The angle of elevation of a cliff from a fixed point is . After going up a distance of k metres towards the top of cliff at an angle of , it is OE OC OE k OE = k cos   ....(ii) In OAB, we have tan = AB OA  tan = h OA  OA = h cot  ....(iii)  CD = EA = OA – OE = h cot – k cos  .....(iv) [Using eqs.(ii) and (iii)] and, BD = AB – AD = AB – CE = (h – k sin ) ....(v) [Using equation (i)] 1   MS = OS – OM =  h   km 2  In BCD, we have Also, cos 30º = h  k sin  BD tan =  tan = h cot   k cos  CD h  k sin  1 = h cot   k cos  cot  Now,  h cot – k sin cot = h cot – k cos       3  PM =  h  km  2   In PSM, we have Let F be the foot and S be the summit of the mountain FOS. Then, OFS = 45º and therefore OSF = 45º. Consequently, tan 60º = 1  3     h   =  h  2   2   Let FP = 1000 m = 1 km be the slope so that  OFP = 30º. Draw PM OS and PL OF. In FPL, We have  h – S F   h = M  h = 45º 30º L O PL sin 30º = PF 1  1  PL = PF sin 30º = 1   km. = km. 2 2   1  OM = PL = km 2 3 1 3 =h 3 – 2 2  h( 3 – 1) = 1  60º …..(ii) [Using equations (i) and (ii)] Join PS. It is given that MPS = 60º. P SM PM  SM = PM. tan 60º OF = OS = h km(say). 1000m 3 2  3  km  OL =  h   2   k (cos   sin  cot  )  h = cot   cot  Sol. h = OS = OF = OL + LF  h = OL +  h(cot – cot ) = k(cos – sin cot ) Ex.24 At the foot of a mountain the elevation of its summit is 45º; after ascending 1000 m towards the mountain up a slope of 30º inclination is found to be 60º. Find the height of the mountain. FL PF  3  3 FL = PF cos 30º = 1  km = km   2  2  [Using equations (iv) and (v)]  ….(i) = = 1 3 1 3 1 ( 3  1) ( 3  1) 3 1 2 2.732 = 1.336 km 2 Hence, the height of the mountain is 1.366 km. Ex.25 The angle of elevation of the top of a tower from a point A due south of the tower is and from B due east of the tower is . If AB = d, show that the height of the tower is d . 2 cot   cot 2  Sol. In right angled triangles OAP and OBP, we have P A Let OP be the tower and let A and B be two points due south and east respectively of the tower such that OAP = and OBP = . Let OP = h. In OAP, we have tan =  tan = ....(i) h h and tan = OA OB In OAB, we have OB2 = OA2 + AB2 N P  AB2 = OB2 – OA2 h  AB2 = h2 cot2 – h2 cot2   O B  AB2 = h2 [cot2 – cot2 ] E  AB2 = h2[(cosec2 – 1) – (cosec2 – 1)] A  AB2 = h2(cosec2 – cosec2) S  sin 2   sin 2     AB2 = h2  2 2   sin  sin   tan = h OB  OB = h cot .  h = ....(ii) Since OAB is a right angled triangle. Therefore, AB2 = OA2 + OB2  d2 = h2 cot2 + h2 cot2  d 2 cot   cot 2  [Using (i) and (ii)] Ex.26 The elevation of a tower at a station A due north of it is and at a station B due west of A is . Prove that the height of the tower is AB sin  sin  . sin 2   sin 2  Sol. O  OA = h cot and OB = h cot . In OBP, we have  h = h B h OA  OA = h cot  W  Let OP be the tower and let A be a point due north of the tower OP and let B be the point due west of A. Such that OAP = and OBP = Let h be the height of the tower. AB sin  sin  sin 2   sin 2  Ex.27 An aeroplane when flying at a height of 4000m from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60º and 45º respectively. Find the vertical distance between the aeroplanes at that instant. Sol. Let P and Q be the positions of two aeroplanes when Q is vertically below P and OP = 4000 m. Let the angles of elevation of P and Q at a point A on the ground be 60º and 45º respectively. In triangles AOP and AOQ, we have tan 60º =  3 = OP OQ and tan 45º = OA OA 4000 OQ and 1 = OA OA  OA =  OQ = 4000 and OQ = OA 3 4000 3 m P Q 45º 4000 m 60º A O  Vertical distance between the aeroplanes = PQ = OP – OQ  ( 3  1) 4000   m = 4000 =  4000  m 3 3   = 1690.53 m