28. Height and Distance Solutions Part 1#

  1. The diagram is given below:

    1st problem

    Let BCBC be the tower, AA the point of observation and θ\theta as angle of elevation.

    Since the tower is vertical it forms a right-angle triangle with right angle at BB. Thus,

    tanθ=BCAB=1003100=3θ=60\tan\theta = \frac{BC}{AB} = \frac{100\sqrt{3}}{100} = \sqrt{3} \Rightarrow \theta = 60^\circ.

  2. The diagram is given below:

    2nd problem

    Let BCBC be the tower, AA the point of observation and the angle of elevation is 3030^\circ.

    Since the tower is vertical it forms a right-angle triangle with right angle at BB. Thus,

    tan30=BCABBC=30.13=103\tan30^\circ = \frac{BC}{AB} \Rightarrow BC = 30.\frac{1}{\sqrt{3}} = 10\sqrt{3} m.

  3. The diagram is given below:

    3rd problem

    Let BCBC be the height of kite, ACAC be the length of string the angle of elevation is 6060^\circ.

    Since the kite would be vertical it forms a right-angle triangle with right angle at BB. Thus,

    sin60=BCACAC=60.23=403\sin60^\circ = \frac{BC}{AC} \Rightarrow AC = \frac{60.2}{\sqrt{3}} = 40\sqrt{3} m.

  4. The diagram is given below:

    4th problem

    Let BCBC be the height of kite, ACAC be the length of string the angle of elevation is 6060^\circ.

    Since the kite would be vertical it forms a right-angle triangle with right angle at BB. Thus,

    sin60=BCACBC=10032=503\sin60^\circ = \frac{BC}{AC} \Rightarrow BC = 100\frac{\sqrt{3}}{2}= 50\sqrt{3} m.

  5. The diagram is given below:

    5th problem

    Let BCBC be the pole, AA the point where rope is tied to the ground and the angle of elevation is 3030^\circ.

    Since the pole is vertical it forms a right-angle triangle with right angle at BB. Thus,

    sin30=BCACAC=12sin30=24\sin30^\circ = \frac{BC}{AC} \Rightarrow AC = \frac{12}{\sin30^\circ} = 24 m.

    Thus, the acrobat has to climb 2424 m.

  6. The diagram is given below:

    6th problem

    Let BCBC be the pole, AA the point where rope is tied to the ground and the angle of elevation is 3030^\circ.

    Since the pole is vertical it forms a right-angle triangle with right angle at BB. Thus,

    sin30=BCACBC=20.12=10\sin30^\circ = \frac{BC}{AC} \Rightarrow BC = 20.\frac{1}{2} = 10 m.

  7. The diagram is given below:

    7th problem

    Let the shaded region represent the river and vertical lines the banks. ABAB represents the bridge, making an angle of 4545^\circ with the bank. Let BCBC represent the width of river, which clearly makes a right angle triangle with right angle at CC.

    Clearly, sin45=BCABBC=150.12=752\sin45^\circ = \frac{BC}{AB} \Rightarrow BC = 150.\frac{1}{\sqrt{2}} = 75\sqrt{2} m.

    Thus, width of the river is 74274\sqrt{2} meters.

  8. The diagram is given below:

    8th problem

    Let ABAB be the observer, 1.51.5 m tall. CECE be the tower. Draw line BDBD parallel to ACAC which will make CD=1.5CD = 1.5 m. In right angle triangle BDEBDE angle of elevation B=45\angle B = 45^\circ. Given, BD=28.5BD = 28.5 m. Thus,

    tan45=DEBDDE=28.5\tan45^\circ = \frac{DE}{BD} \Rightarrow DE = 28.5 m. CE=CD+DE=1.5+28.5=30\therefore CE = CD + DE = 1.5 + 28.5 = 30 m.

  9. The diagram is given below:

    9th problem

    BDBD is the pole and ACAC is the ladder. CC is the point which the electrician need to reach to repair the pole which is 1.31.3 m below the top of the pole. Total height of the pole is 44 m, thus, BC=41.3=2.7BC = 4 - 1.3 = 2.7 m.

    We are given than ladder makes an angle of 6060^\circ with the horizontal.

    sin60=BCACAC=BCsin60=2.732=3.12\therefore \sin60^\circ = \frac{BC}{AC} \Rightarrow AC = \frac{BC}{\sin60^\circ} = \frac{2.7\sqrt{3}}{2} = 3.12 m.

  10. The diagram is given below:

    10th problem

    AA is the point of observation. BB is the foot of the tower and CC is the top of the tower. CDCD is the height of the water tank above the tower. Given AB=40AB = 40 m .

    In ABC,tan30=BCABBC=AB.tan30=403=23.1\triangle ABC, \tan30^\circ = \frac{BC}{AB} \Rightarrow BC = AB.\tan30^\circ = \frac{40}{\sqrt{3}} = 23.1 m, which is height of the toweer.

    In ABD,tan45=BDABBD=AB.tan45=40.1=40\triangle ABD, \tan45\circ = \frac{BD}{AB} \Rightarrow BD = AB.\tan45^\circ = 40.1 = 40 m which is combined height of the tower and water tank. Thus, height or depth of the water tank =CD=BDBC=4023.1=16.9= CD = BD - BC = 40 - 23.1 = 16.9 m.

  11. The diagram is given below:

    11th problem

    The shaded region is the river. ABAB is the tree and CC is the initial point of the observer. DD is the final point of observation. Given, CD=20CD = 20 m. Let AB=hAB = h m and AC=xAC = x m.

    In ABC,tan60=hxh=3x\triangle ABC, \tan60^\circ = \frac{h}{x} \Rightarrow h = \sqrt{3}x

    In ABDtan30=hx+203h=x+20\triangle ABD \tan30^\circ = \frac{h}{x + 20}\Rightarrow \sqrt{3}h = x + 20

    3x=x+20x=10\Rightarrow 3x = x + 20 \Rightarrow x = 10 m. h=103\Rightarrow h = 10\sqrt{3} m.

  12. The diagram is given below:

    12th problem

    ACAC is the tree before breaking. Portion BCBC has borken and has become BDBD which makes an angle of 6060^\circ with remaining portion of tree standing. If AB=xAB = x m, then BD=12xBD = 12 - x because original height of the tree is given as 1212 m.

    In ABD,sin60=x12x32=x12xx=5.57\triangle ABD, \sin60^\circ = \frac{x}{12 - x} \Rightarrow \frac{\sqrt{3}}{2} = \frac{x}{12 - x} \Rightarrow x = 5.57

  13. The diagram is given below:

    13th problem

    ACAC is the tree before breaking. Portion BCBC has borken and has become BDBD which makes an angle of 3030^\circ with remaining portion of tree standing. If AB=xAB = x m, then BD=lxBD = l - x where ll is the original height of the tree.

    In ABD,sin30=xlx=123x=l\triangle ABD, \sin30^\circ = \frac{x}{l - x} = \frac{1}{2} \Rightarrow 3x = l

    cos30=30lx=32x=17.32l=51.96\cos30^\circ = \frac{30}{l - x} = \frac{\sqrt{3}}{2} \Rightarrow x = 17.32 \Rightarrow l = 51.96 m.

  14. The diagram is given below:

    14th problem1

    ABAB is the tower. Initial observation point is DD where angle of elevation is α\alpha such that tanα=512\tan\alpha = \frac{5}{12}. CC is the second point of observation where angle of elevation is β\beta such that tanβ=34\tan\beta = \frac{3}{4}. Given, CD=192CD = 192 meters. Let hh be the height of the tower and xx be the distance of CC from the foot of the tower i.e. AA.

    In ABC,tanβ=34=hx\triangle ABC, \tan\beta = \frac{3}{4} = \frac{h}{x}

    In ABD,tanα=512=hx+192h=180\triangle ABD, \tan\alpha = \frac{5}{12} = \frac{h}{x + 192} \Rightarrow h = 180 meters.

  15. The diagram is given below:

    15th problem

    ABAB is the tower. When the sun’s altittude is 4545^\circ the shadow reached CC. When the shadow reached the altitude of sun becomes 3030^\circ. Let hh meters be the height and xx meters be the distance of of initial point of observation from foot of the tower. Given CD=10CD = 10 meters.

    In ABC,tan45=1=hxx=h\triangle ABC, \tan45^\circ = 1 = \frac{h}{x}\Rightarrow x = h

    In ABD,tan30=13=hx+10h=1031=13.66\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{x + 10}\Rightarrow h = \frac{10}{\sqrt{3} - 1} = 13.66 meters.

  16. The diagram is given below:

    16th problem

    This problem is same as previous problem, where 1010 m is replaced by 11 km. Processing similarly, we obtain h=1.366h = 1.366 km.

  17. The diagram is given below:

    17th problem

    This problem is same as two previous problems. The height of the mountain is 5.0715.071 km.

  18. The diagram is given below:

    18th problem

    This problem is same as 1111-th. Proceeding similarly, we find width of river as 2020 m and height of the tree as 20320\sqrt{3} m.

  19. The diagram is given below:

    19th problem

    Height of the plane is 12001200 m which is ABAB. The ships are located at CC and DD. Let CD=dCD = d m and AC=xAC = x m.

    In ABC,tan60=1200xx=1200sqrt3=4003\triangle ABC, \tan60^\circ = \frac{1200}{x} \Rightarrow x = \frac{1200}{sqrt{3}} = 400\sqrt{3} m.

    In ABC,tan30=1200x+dx+d=12003d=8003\triangle ABC, \tan30^\circ = \frac{1200}{x + d} \Rightarrow x + d = 1200\sqrt{3} \Rightarrow d = 800\sqrt{3} m.

  20. The diagram is given below:

    20th problem

    Let ABAB be the flag staff having height hh and ACAC be the shadow when sun’s altitude is 6060^\circ. Let ADAD be the shadow when sun’s altitude is θ\theta^\circ. If we let AC=xAC = x m then AD=3xCD=2xAD = 3x \Rightarrow CD = 2x.

    In ABC,tan60=hxh=3x\triangle ABC, \tan60^\circ = \frac{h}{x} \Rightarrow h = \sqrt{3}x.

    In ABDtanθ=h3x=13θ=30\triangle ABD \tan\theta = \frac{h}{3x} = \frac{1}{\sqrt{3}}\Rightarrow \theta = 30^\circ.

  21. The diagram is given below:

    21st problem

    Let ABAB be the height of the plane, equal to 200200 m. Let the shaded region present the river such that width CD=xCD = x m.

    In ABD,tan45=200ADAD=200\triangle ABD, \tan45^\circ = \frac{200}{AD} \Rightarrow AD = 200 m.

    Clearly, AC=200xAC = 200 - x m. In ABC,tan60=200200xx=84.53\triangle ABC, \tan60^\circ = \frac{200}{200 - x} \Rightarrow x = 84.53 m.

  22. The diagram is given below:

    22nd problem

    Let ACAC and BDBD represent the towers having height hh. Given the distance between towers is 100100 m which is CDCD. Let the point of observation be EE which is at distance xx from CC and 100x100 - x from DD. Angle of elevations are given as 3030^\circ and 6060^\circ.

    In ACE,tan60=3=hxh=3x\triangle ACE, \tan60^\circ = \sqrt{3} = \frac{h}{x}\Rightarrow h = \sqrt{3}x.

    In BDE,tan30=13=h100xx=25,h=253\triangle BDE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{100 - x} \Rightarrow x = 25, h = 25\sqrt{3} m.

  23. The diagram is given below:

    23rd problem

    Let ABAB be the light house, CC and DD are the two locations of the ship. The height of the light house is given as 100100 m. The angle of elevations are given as 3030^\circ and 4545^\circ. Let AC=yAC = y m and CD=xCD = x m.

    In ABC,tan45=1=100yy=100\triangle ABC, \tan45^\circ = 1 = \frac{100}{y} \Rightarrow y = 100.

    In ABD,tan30=13=100x+y=100100+xx=73.2\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{100}{x + y} = \frac{100}{100 + x} \Rightarrow x = 73.2 m.

  24. The diagram is given below:

    24th problem

    The diagram represents the top PQPQ and XYXY as given in the problem. The angle of elevations are also given. Draw YZYZ parallel to ZQZQ and thus, PZ=40PZ = 40 m. Let ZQ=xZQ = x.

    In QYZ,tan45=1=ZQYZYZ=x\triangle QYZ, \tan45^\circ = 1 = \frac{ZQ}{YZ} \Rightarrow YZ = x m. Thus, PX=xPX = x m.

    In PQX,tan60=3=x+40xx=4031\triangle PQX, \tan60^\circ = \sqrt{3} = \frac{x + 40}{x} \Rightarrow x = \frac{40}{\sqrt{3} - 1} m.

    Height of toewr is x+40=40331x + 40 = \frac{40\sqrt{3}}{\sqrt{3} - 1}

    In PQX,sin60=32=PQXQXQ=8031\triangle PQX, \sin60^\circ = \frac{\sqrt{3}}{2} = \frac{PQ}{XQ} \Rightarrow XQ = \frac{80}{\sqrt{3} - 1} m.

  25. The diagram is given below:

    25th problem

    Let ABAB and CDCD are the houses. Given CD=15CD = 15 m. Let the width of the street is AC=ED=xAC = ED = x m. The angle of depression and elevation are given as 4545^\circ and 3030^\circ respectively. Draw EDACED\parallel AC.

    In ACD,tan45=1=CDACAC=15\triangle ACD, \tan45^\circ = 1 = \frac{CD}{AC} \Rightarrow AC = 15 m. Thus, EDED is also 1515 m because EDED is paralle to ACAC.

    In BED,tan30=13=BEEDBE=53\triangle BED, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BE}{ED} \Rightarrow BE = 5\sqrt{3} m.

    Thus, total height of the house =15+53=23.66= 15 + 5\sqrt{3} = 23.66 m.

  26. The diagram is given below:

    26th problem

    Let ABAB represent the building and CDCD the tower. Let CD=hCD = h m and given AB=60AB = 60 m. Also, let AC=xAC = x m. Draw DEACDE\parallel AC, thus CE=xCE = x m and AE=hAE = h m.

    The angles of depression are given which would be same as angle of elevation from top and bottom of tower.

    In ABC,tan60=3=60xx=203\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{60}{x} \Rightarrow x = 20\sqrt{3} m.

    In ADE,tan30=13=BExBE=20\triangle ADE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BE}{x}\Rightarrow BE = 20 m.

    \therefore Height of the building CD=AE=ABBE=6020=40CD = AE = AB - BE = 60 - 20 = 40 m.

  27. The diagram is given below:

    27th problem

    Let CDCD represent the deck of the ship with height 1010 m and ABAB the hill. The water level is ACAC. Draw DEACDE||AC and let AC=DE=xAC = DE = x m.

    The angle of elevation are shown as given in the question.

    In ACD,tan30=13=CDxx=103\triangle ACD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{CD}{x} \Rightarrow x = 10\sqrt{3} m.

    In BDC,tan60=3=BExBE=30\triangle BDC, \tan60^\circ = \sqrt{3} = \frac{BE}{x} \Rightarrow BE = 30 m.

    Thus, height of the hill =AE+BE=10+30=40= AE + BE = 10 + 30 = 40 m.

  28. The diagram is given below:

    28th problem

    Let CECE be the line in which plane is flying and ABDABD be the horizontal ground. Since the plane is flying at a constant height of 360033600\sqrt{3} m, we have BC=DE=36003BC = DE = 3600\sqrt{3} m. Let AB=xAB = x m and BD=yBD = y m.

    In ABC,tan60=3=36003xx=3600\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{3600\sqrt{3}}{x} \Rightarrow x = 3600 m.

    In ADE,tan40=13=36003x+yy=7200\triangle ADE, \tan40^\circ = \frac{1}{\sqrt{3}} = \frac{3600\sqrt{3}}{x + y} \Rightarrow y = 7200 m.

    Thus, the plane flies 72007200 m in 3030 s. Speed of plane =720030.36001000=284= \frac{7200}{30}.\frac{3600}{1000} = 284 km/hr.

  29. The diagram is given below:

    29th problem

    Let ACAC be the river and BDBD be the tree on the island in the river. Given wdith of the river ACAC as 100100 m. Let BC=xBC = x m AB=100x\Rightarrow AB = 100 - x m. The angles of elevation are shown as given in the question. Let BD=hBD = h m be the height of the tower.

    In BCD,tan45=1=hxh=x\triangle BCD, \tan45^\circ = 1 = \frac{h}{x} \Rightarrow h = x m.

    In ABC,tan30=13=h100xx=1003+1=h\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{100 - x} \Rightarrow x = \frac{100}{\sqrt{3} + 1} = h m.

  30. The diagram is given below:

    30th problem

    Let ABAB be the first tower and CDCD be the second tower. Given AC=140AC = 140 m and CD=40CD = 40 m. Let ACAC be the horizontal plane. Draw DEACDE=140DE\parallel AC \Rightarrow DE = 140 m and AE=60AE = 60 m. Angle of elevation is shown as given in the question from top of second tower to top of first tower to be 3030^\circ.

    In BDE,tan30=13=BE140BE=1403\triangle BDE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BE}{140} \Rightarrow BE = \frac{140}{\sqrt{3}} m.

    Thus, total height of first tower is 1403+60\frac{140}{\sqrt{3}} + 60 m.

  31. The diagram is given below:

    31st problem

    Let ADAD be the horizontal ground. Let ABAB and ACAC be the heights at which planes are flying. Given AC=4000AC = 4000 m. Also, given are angles of elevation of the two aeroplanes. Let point of observation be DD and AD=bAD = b m.

    In ACD,tan60=3=ACADb=40003\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{AC}{AD}\Rightarrow b = \frac{4000}{\sqrt{3}} m.

    In ABD,tan45=1=ABADAB=b=40003\triangle ABD, \tan45^\circ = 1 = \frac{AB}{AD} \Rightarrow AB = b = \frac{4000}{\sqrt{3}} m.

    Therefore, distance between heights of two planes =4000.313= 4000.\frac{\sqrt{3} - 1}{\sqrt{3}} m.

  32. The diagram is given below:

    32nd problem

    Let BCBC be the tower where BB is the foot of the toewr. Let AA be the point of observation. Given BAC=60\angle BAC = 60^\circ.

    In ABC,tan60=3=BCABBC=203\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{BC}{AB} \Rightarrow BC = 20\sqrt{3} m.

  33. The diagram is given below:

    33rd problem

    Let BCBC be the wall and ACAC the ladder. Given distance of the foot of the ladder is 9.59.5 m away from the wall i.e. AB=9.5AB = 9.5 m. The angle of elevation is given as BAC=60\angle BAC = 60^\circ.

    In ABC,cos60=12=ABACAC=19\triangle ABC, \cos60^\circ = \frac{1}{2} = \frac{AB}{AC} \Rightarrow AC = 19 m.

  34. The diagram is given below:

    34th problem

    Let BCBC be the wall and ACAC the ladder. Given distance of the foot of the ladder is 22 m away from the wall i.e. AB=2AB = 2 m. The angle of elevation is given as BAC=60\angle BAC = 60^\circ.

    In ABC,tan60=3=BCACBC=23\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{BC}{AC} \Rightarrow BC = 2\sqrt{3} m.

  35. The diagram is given below:

    35th problem

    Let BCBC be the electric pole, having a height of 1010 m. Let ACAC be the length of wire. The angle of elevation is given as BAC=45\angle BAC=45^\circ.

    In ABC,sin45=12=BCACAC=102\triangle ABC, \sin45^\circ = \frac{1}{\sqrt{2}} = \frac{BC}{AC} \Rightarrow AC = 10\sqrt{2} m.

  36. The diagram is given below:

    36th problem

    Let BCBC represent the height of kite. Given BC=75BC = 75 m. Let ACAC represent the length of the string. The angle of elevation is given as 6060^\circ.

    In ABC,sin60=32=BCACAC=503\triangle ABC, \sin60^\circ = \frac{\sqrt{3}}{2} = \frac{BC}{AC} \Rightarrow AC = 50\sqrt{3} m.

  37. The diagram is given below:

    37th problem

    Let BCBC represent the wall and ACAC the ladder. Given that the length of ladder is 1515 m. The angle of elevation of the wall from foot of the tower is given as 60BAC=6060^\circ \Rightarrow \angle BAC = 60^\circ.

    In ABC,sin60=32=BCACBC=1532\triangle ABC, \sin60^\circ = \frac{\sqrt{3}}{2} = \frac{BC}{AC} \Rightarrow BC = \frac{15\sqrt{3}}{2} m.

  38. The diagram is given below:

    38th problem

    Let BCBC be the tower and CDCD be the flag staff, the heights of which are to be found. Let AA be the point of obsevation. Given that AB=70AB = 70 m. The angle of elevation of the foot and the top of flag staff are given as 4545^\circ and 6060^\circ i.e. BAC=45\angle BAC = 45^\circ and BAD=60\angle BAD = 60^\circ.

    In ABC,tan45=1=BCABBC=70\triangle ABC, \tan45^\circ = 1 = \frac{BC}{AB} \Rightarrow BC = 70 m, which is height of the tower.

    In ABD,tan60=3=BDABBD=703\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{BD}{AB} \Rightarrow BD = 70\sqrt{3} m, which is combined height of tower and flag staff. Thus, CD=70(31)CD = 70(\sqrt{3} - 1) m, which is height of flag staff.

  39. This problem is same as 12. Put 1515 instead of 1212.

  40. The diagram is given below:

    40th problem

    Let ABAB be the tower and BCBC the flag staff, whose height is 55 m. Let DD be the point of observation. Given that angle of elevation of the foot of the flag staff is 3030^\circ and that of top is 6060^\circ i.e. ADB=30\angle ADB = 30^\circ and ADC=60\angle ADC = 60^\circ. Let AB=hAB = h m and AD=xAD = x m.

    In ABD,tan30=13=hxx=3h\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{x} \Rightarrow x = \sqrt{3}h m.

    In ACD,tan60=3=h+5xh=2.5 m,x=2.53\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{h + 5}{x} \Rightarrow h = 2.5~{\rm m}, x = 2.5\sqrt{3} m.

  41. This problem is same as 15. Put 5050 m instead of 1010 m and 6060^\circ instead of 4545^\circ.

  42. This problem is similar to 15. Put 4545^\circ instead of 3030^\circ and 6060^\circ instead of 3030^\circ.

  43. The diagram is given below:

    43rd problem

    Let ABAB be the current height of the skydiver as hh m. CC and DD are two points observed at angle of depression 4545^\circ and 6060^\circ which woule be equal to angle of elevation from these points. Given that CD=100CD = 100 m. Let AC=xAC = x m.

    In ABD,tan45=1=ABAD=hx+100h=x+100\triangle ABD, \tan45^\circ = 1 = \frac{AB}{AD} = \frac{h}{x + 100} \Rightarrow h = x + 100 m.

    In ABC,tan60=3=ABAC=hxh=3x\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} = \frac{h}{x} \Rightarrow h = \sqrt{3}x m.

    x=10031,h=100331\Rightarrow x = \frac{100}{\sqrt{3} - 1}, h = \frac{100\sqrt{3}}{\sqrt{3} - 1} m.

  44. The diagram is given below:

    44th problem

    Let ABAB be the tower having a height of 150150 m. Let CC and DD be the points observed such that ADB=45\angle ADB = 45^\circ and ACB=60\angle ACB = 60^\circ. Let AC=yAC = y m and CD=xCD = x m. We have to find xx.

    In ABC,tan60=3=ABAC=150yy=503\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} = \frac{150}{y} \Rightarrow y = 50\sqrt{3} m.

    In ABD,tan45=1=ABAD=150x+yx=150503\triangle ABD, \tan45^\circ = 1 = \frac{AB}{AD} = \frac{150}{x + y}\Rightarrow x = 150 - 50\sqrt{3} m.

  45. The diagram is given below:

    45th problem

    Let ABAB be the towerr having a height of hh m. Let CC and DD be the points observed such that ADB=30\angle ADB = 30^\circ and ACB=60\angle ACB = 60^\circ. Let AC=xAC = x m. Given CD=150CD = 150 m.

    In ABC,tan60=3=ABBC=hxh=3x\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{BC} = \frac{h}{x} \Rightarrow h = \sqrt{3}x m.

    In ABD,tan30=13=ABAD=150x+150h=753\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{150}{x + 150} \Rightarrow h = 75\sqrt{3} m.

  46. The diagram is given below:

    46th problem

    Let ABAB be the towerr having a height of hh m. Let CC and DD be the points observed such that ADB=30\angle ADB = 30^\circ and ACB=60\angle ACB = 60^\circ. Let AC=xAC = x m. Given CD=100CD = 100 m.

    In ABC,tan60=3=ABBC=hxh=3x\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{BC} = \frac{h}{x} \Rightarrow h = \sqrt{3}x m.

    In ABD,tan30=13=ABAD=hx+100x=50\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{h}{x + 100} \Rightarrow x = 50 m.

    Thus, h=503h = 50\sqrt{3} m. Distance of initial point =x+100=150= x + 100 = 150 m.

  47. The diagram is given below:

    47th problem

    Let ABAB be the tower and CDCD be the building. Given CD=15CD = 15 m. ACAC is the horizontal plane joining foot of the building and foot of the tower having width xx m. Draw DEACDE||AC then DE=xDE = x m and AE=15AE = 15 m.

    In BDE,tan30=hxx=3h\triangle BDE, \tan30^\circ = \frac{h}{x}\Rightarrow x = \sqrt{3}h m.

    In ABC,tan60=hx+15h=7.5\triangle ABC, \tan60^\circ = \frac{h}{x + 15}\Rightarrow h = 7.5 m and x=7.53x = 7.5\sqrt{3} m.

  48. The diagram is given below:

    48th problem

    Let ABAB be the tower and BCBC be the flag staff having heights xx and yy m respectively. The distance of foot of tower from the point of observation 99 m. The angles of elevation of the foot and the top of the flag staff are 3030^\circ and 6060^\circ as given in the question.

    In ABD,tan30=13=x9x=33\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{x}{9}\Rightarrow x = 3\sqrt{3} m.

    In ACD,tan60=3=x+y9y=63\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{x + y}{9} \Rightarrow y = 6\sqrt{3} m.

  49. The diagram is given below:

    49th problem

    Let ACAC be the full tree and BCBC is the portion which has fallen. BCBC becomes BDBD after falling and angle of elevation is 3030^\circ. Let the height of remaining portion of tree be AB=xAB = x m. Given, AD=8AD = 8 m and BC=BDBC = BD, which is broken part of tree.

    In ABC,tan30=13=ABADAB=83\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} \Rightarrow AB = \frac{8}{\sqrt{3}} m.

    Also, cos30=32=ADBDBD=43\cos30^\circ = \frac{\sqrt{3}}{2} = \frac{AD}{BD} \Rightarrow BD = 4\sqrt{3} m.

    Thus, height of the tree AC=AB+BC=AB+BD=203AC = AB + BC = AB + BD = \frac{20}{\sqrt{3}} m.

  50. The diagram is given below:

    50th problem

    Let ABAB be the building with height 1010 m. Let BCBC be the flag with height hh m. Also, let distance between PP and foot of the building as AP=xAP = x m. The angle of elevation of top of the building is 3030^\circ and that of the flag is 4545^\circ.

    In ABP,tan30=13=ABAP=10xx=103\triangle ABP, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AP} = \frac{10}{x} \Rightarrow x = 10\sqrt{3} m.

    In ACP,tan45=1=ACAP=10+hxh=10(31)\triangle ACP, \tan45^\circ = 1 = \frac{AC}{AP} = \frac{10 + h}{x} \Rightarrow h = 10(\sqrt{3} - 1) m.