#### Data Interpretation

• ( 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
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

• ( 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
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

• ( 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
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

• ( 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
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

• ( 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
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

• ( 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
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

• ( 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
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

• ( 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
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

• ( 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
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

• ( 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
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
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