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( 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

Show Answer Report Discussion in forumAnswer : 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

Show Answer Report Discussion in forumAnswer : 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

Show Answer Report Discussion in forumAnswer : 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

Show Answer Report Discussion in forumAnswer : 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

Show Answer Report Discussion in forumAnswer : 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

Show Answer Report Discussion in forumAnswer : 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

Show Answer Report Discussion in forumAnswer : 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

Show Answer Report Discussion in forumAnswer : 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

Show Answer Report Discussion in forumAnswer : 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

Show Answer Report Discussion in forumAnswer : 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
