• ( 1 ) The angle of elevation of the top of a tower as observed from a point in a horizontal line through the foot of the tower is 30°.When the observer moves towards the tower a distance of 100m, he finds the angle of elevation of the top to be 60°. Find the height of the tower and the distance of the first position from the tower.

    • 1) 86.6m, 150m
    • 2) 84.4m, 160m
    • 3) 157.6m, 160m
    • 4) 177.6m, 160m
    • Discussion in forum
      Answer : 1) 86.6m, 150m
      Solution :

      Let AB be the tower.
      AB = h, BD = x, ∠ACB = 30°, ∠ADB = 60°
      In ΔABD,
      tan60° = AB/BD
      √ 3 = h/x
      h = √ 3x.................(eq:1)
      In ΔABD,
      tan30° = AB/BC
      1/√ 3 = h/BD + DC
      BD + DC = h√ 3
      x + 100 = √ 3*√ 3x [from eq:1]
      x + 100 = 3x
      100 = 3x - x
      2x = 100
      x = 100/2
      x= 50
      From eq: 1
      h = √ 3 * 50 = 1.732 * 50 = 86.6 m
      distance from first positon to tower BC = BD + DC = 50 + 100 = 150 m








      discussion


      Answer : 1) 86.6m, 150m

    • ( 2 ) A vertical tower stands on a horizontal plan and is surmounted by a flagstaff of height 30m. At a point on the plan the angle of elevaion of the bottom of the flagstaff is 45° and of the top of the flagstaff is 60°. Determine the height of the tower and the horizontal distance.

    • 1) 41.6m, 41.6m
    • 2) 38.5m, 38.5m
    • 3) 40.98m, 40.98m
    • 4) 52.2m, 52.2m
    • Discussion in forum
      Answer : 3) 40.98m, 40.98m
      Solution :

      Let BC be the height of the tower and DC be the height of the flagstaff.
      DC = 30m.
      In ΔABC,
      tan45° = BC/AB
      1 = BC/AB
      AB = BC ..................(eq:1)
      In ΔABD,
      tan60° = BD/AB
      √ 3 = BD/AB
      AB = (BC+CD)/√ 3
      AB = (BC + 30)/√ 3..................(eq:2)
      BC = (BC + 30)/√ 3 [from eq:1]
      √ 3 BC = BC + 30
      √ 3 BC - BC = 30
      BC(√ 3 - 1)= 30
      BC = 30/0.732 = 40.98 m
      horizontal distance AB = 40.98 m








      discussion


      Answer : 3) 40.98m, 40.98m

    • ( 3 ) A 7m long flagstaff is fixed on the top of a tower on the horizontal plane. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 60° and 45° respectively. Find the height of the tower correct to one place of decimal.

    • 1) 8.65m
    • 2) 9.57m
    • 3) 7.87m
    • 4) 6.75m
    • Discussion in forum
      Answer : 2) 9.57m
      Solution :

      Let BC be the height of the tower and DC be the height of the flagstaff.
      DC = 7m.
      In ΔABC,
      tan45° = BC/AB
      1 = BC/AB
      AB = BC .................(eq:1)
      In ΔABD,
      tan60° = BD/AB
      √ 3 = BD/AB
      AB = (BC+CD)/√ 3
      AB = (BC + 7)/√ 3 ....................(eq:2)
      BC = (BC + 7)/√ 3 [from eq:1]
      √ 3 BC = BC + 7
      √ 3 BC - BC = 7
      BC(√ 3 - 1)= 7
      BC = 7/0.732 = 9.57 m








      discussion


      Answer : 2) 9.57m

    • ( 4 ) The angle of elevation of a jet plan from a point A on tha ground is 60°. After a flight of 15 seconds,the angle of elevationchanges to 30°. If the jet plan is flying at a costant height of 1500√ 3, Find the speed of the plane.

    • 1) 720km/h
    • 2) 580km/h
    • 3) 600km/h
    • 4) 589km/h
    • Discussion in forum
      Answer : 1) 720km/h
      Solution :

      Let D and E be the two position of the aeroplane and speed S m/sec.
      BE = 1500√ 3,
      In ΔABE,
      tan60° = BE/AB
      √ 3 = 1500√ 3/x
      x = 1500√ 3/√ 3 = 1500 .................(eq: 1)
      In ΔACD,
      tan30° = CD/AC
      1/√ 3 =1500√ 3/(x+y)
      x+y = 1500√ 3*√ 3
      1500 + y = 3*1500 [from eq:1]
      y = 4500 - 1500
      y = 3000
      To travel 3000 m aeroplane takes 15 min = speed = distance/Time = (3000/15)*(3600/1000) km/h
      = 720 km/h








      discussion


      Answer : 1) 720km/h

    • ( 5 ) A man standing on the deck of a ship , which is 10m above water level, observes the angle of elevation of top of a hill as 60° and angle of depression of the base of the hill is 30°.Find the distance of the hill from the ship and height of the hill.

    • 1) 17.3m, 40m
    • 2) 18.2m, 30m
    • 3) 15.45m, 50m
    • 4) 17.88m, 50m
    • Discussion in forum
      Answer : 1) 17.3m, 40m
      Solution :

      In ΔBDC
      tan30°=10/x
      => 1/√ 3=10/x
      => x = 10 * √ 3 = 10 * 1.732 = 17.32 m
      In ΔAEC
      tan60°=AE/CF
      => √ 3 = AE/10√ 3 => AE = 30m
      So, Height of the hill = 30 + 10 = 40 m
      Distance of the hill from the ship=17.32 m








      discussion


      Answer : 1) 17.3m, 40m

    • ( 6 ) An aeroplane flying horizontally at height of 2500√ 3 mts above that ground; is observed to be at angle of elevation 60° from the ground. After a flight of 25 seconds the angle of elevation is 30°. Find the speed of the plan in m/sec.

    • 1) 200m/sec
    • 2) 300m/sec
    • 3) 500m/sec
    • 4) 600m/sec
    • Discussion in forum
      Answer : 1) 200m/sec
      Solution :

      Let the C and D be the two positions of the aeroplan.
      QB = 2500√ 3 mts
      Let the speed of the aeroplane be S m/sec.
      tan60° = BE/AB
      √ 3 = 2500√ 3 /x
      x = 2500√ 3 /√ 3 = 2500 m
      tan30° = CD/AC
      1/√ 3 = 2500√ 3 /(x + y)
      x + y = 2500√ 3 *√ 3
      2500 + y = 2500*3
      y = 7500 -2500
      y = 5000 m
      to travel 5000 m the aeroplane takes 25 seconds: the speed of the plane = distance/time =5000/25 = 200 m/sec.








      discussion


      Answer : 1) 200m/sec

    • ( 7 ) A boy standing on a horizontal plane finds a bird flying at a distance of 100m from him at an elevation of 30°. A girl standing on the roof of 20 meter high building finds the angle of elevation of the same bird to be 45°. Both the boy and the girl are on opposite sides of the bird. Find the distance of the bird from the girl.

    • 1) 47.6m
    • 2) 40.6m
    • 3) 42.42m
    • 4) 45.45m
    • Discussion in forum
      Answer : 2) 40.6m
      Solution :

      Let CD = X
      In ΔABC
      In sin30°=AC/AB
      => 1/2 = AC/100
      => AC = 50m
      Now, in ΔAFE
      sin45° = AF/AE
      => 1/√ 2 = 30/AE
      => AE = 30√ 2 = 30 * 1.41
      => AE = 42.42m
      Therefore, distance of bird from the girl = 42.42 m.








      discussion


      Answer : 2) 40.6m

    • ( 8 ) Two ships are sailing in the sea on either side of a lighthouse; the angles of depression of two ships as observed from the top of the lighthouse are 60° and 45° respectively. If the distance between the ships is 200(1 + √ 3)/√ 3 meter,find the height of the lighthouse.

    • 1) 500m
    • 2) 200m
    • 3) 300m
    • 4) 400m
    • Discussion in forum
      Answer : 2) 200m
      Solution :

      Let the height of the house AB = h m
      Given distance between the two ships PQ = 200(√ 3 + 1)/√ 3
      Let BQ = x
      PB = (200(√ 3 + 1)/√ 3) - x
      In ΔABQ, θ = 45°
      tan 45° = h/x
      1 = h/x
      x = h ..................(eq:1)
      In ΔABP,
      tan60° = h/[(200(√ 3 + 1)/√ 3) - h]
      => √ 3[(200(√ 3 + 1)/√ 3)- h] = h
      => 200(√ 3 + 1) - √ 3h = h
      => 200(√ 3 + 1) = √ 3h + h
      => 200(√ 3 + 1) = (√ 3 + 1)h
      h = 200(√ 3 + 1)/(√ 3 + 1) = 200m








      discussion


      Answer : 2) 200m

    • ( 9 ) The angle of elevation of a jet fighter from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/hr, find the constant height at which the jet is flying.(Use √ 3=1.732)

    • 1) 3098m
    • 2) 2598m
    • 3) 2000m
    • 4) 5098m
    • Discussion in forum
      Answer : 2) 2598m
      Solution :

      Speed of jet fighter = 720km/h = 720 * 5/18m/sec = 200sec
      Distance covered in 15 sec = 200 * 15 = 3000 m
      => PB = 3000m
      QC = PB = 3000 m
      AC = AQ + QC = y + 3000
      In ΔAQP
      tan60°=PQ/AQ
      => √ 3=h/y
      => h = √ 3y...................(eq: 1)
      In ΔBCA
      tan30° = BC/AC
      => 1/√ 3 = h/(y + 3000)
      =>h= (y+3000)/√ 3........................(eq: 2)
      From equation (eq: 1) & (eq: 2)
      √ 3 y=(y+3000)/√ 3
      => 3y = y+3000
      => 2y = y+3000
      => y = 1500
      On putting y=1500 in (eq: 1),We get
      h = y√ 3 = 15000√ 3 = 1500 * 1.732 = 2598 m








      discussion


      Answer : 2) 2598m

    • ( 10 ) The angle of elevation of an aeroplane from a point A on the ground is 60° . After a flight of 15 seconds horizontally, the angle of elevation changes to 30° . If the aeroplane is flying at a speed of 200m/s , then find the constant height at which the aeroplane is flying.

    • 1) 500√ 3m
    • 2) 600√ 3m
    • 3) 700√ 3m
    • 4) 500√ 5m
    • Discussion in forum
      Answer : 1) 500√ 3m
      Solution :

      Let A be the point of observation.
      Let E and D be positions of the aeroplane initially and after 15 seconds respectively. Let BE and CD denote the constant height at which the aeroplane is fying.
      Given that DAC = 30° and EAB = 60°.
      Let BE = CD = h m
      Let AB = x m
      The distance covered in 15 seconds, ED = 200*15 = 3000m
      Thus, BC = 3000m
      In right angle ΔADC,
      tan 30° = CD/AC
      CD = AC * tan30°
      h = (x + 3000) * 1/√ 3......................(eq:1)
      In ΔAEB,
      tan60° = BE/AB
      √ 3 = h/x .............................(eq:2)
      From eq. 1 & 2, we have
      √ 3*x = (x + 3000) * 1/√ 3
      3x = x + 3000
      x = 1500m
      h = 1500√ 3m [from eq:2]








      discussion


      Answer : 1) 500√ 3m





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