ML Aggarwal Solution Class 10 Chapter 20 Heights and Distances Test
Test
Question 1
The angle of elevation of the top of a tower from a point A (on the ground) is 30°. On walking 50 m towards the tower, the angle of elevation is found to be 60°. Calculate
(i) the height of the tower (correct to one decimal place).
(ii) the distance of the tower from A.
Sol :
Let TR be the tower and A is a point on the ground
and angle of elevation of the top of tower = 30°
AB = 50 m
and from B, the angle of elevation is 60°
Let TR = h and AR = x
BR = x – 50
tanθ=TRAR
⇒tan30∘=hx
⇒1√3=hx
⇒x=√3h...(i)
and in right ΔBTR
tan60∘=TRBR=hx−50
⇒√3=hx−50
⇒h=√3(x−50)...(ii)
From (i) and (ii)
h=√3(√3h−50)⇒h=3h−50√3
2h=50√3⇒h=25√3
h=25(1.732)=43.3
Substituting the value of h in (i)
x=√3×25√3=25×3=75
∴Height of tower=43.3 m
and distance of A from the foot of the tower=75 m
Question 2
An aeroplane 3000 m high passes vertically above another aeroplane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the two planes.
Sol :
Let A and B are two aeroplanes
and P is a point on the ground such that
angles of elevations from A and B are 60° and 45° respectively.
AC = 3000 m
Let AB = x
∴ BC = 3000 – x
Let PC = y
⇒y=3000−x
⇒3000√3=3000−x
⇒x=3000−3000√3 [from (i)]
⇒3000−3000×√3√3×√3
3000-1000(1.732)=3000-1732=1268
∴Distance between the two planes=1268 m
Question 3
A 7m long flagstaff is fixed on the top of a tower. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 45° and 36° respectively. Find the height of the tower correct to one place of demical.
Sol :
Let TR be the tower and PT is the flag on it such that PT = 7m
Let TR = h and AR = x
Angles of elevation from P and T are 45° and 36° respectively.
Now in right ∆PAR
⇒1=7+hx
⇒x=7+h...(i)
Again in right ΔTAR
tan36∘=TRAR=hx
⇒0⋅7265=hx
⇒h=x(0⋅7265)...(ii)
From (i) and (ii)
h=(7+h)(0.7265)
⇒h=7×0⋅7265+⋅7265h
⇒h−⋅7265h=7×0⋅7265
⇒(1−⋅7265)h=7×0⋅7265
⇒0⋅2735h=7×0.7265
˙h=7×0.72652735=7×72652735
˙h=7×0.7265.2735=7×72652735
=18.59
∴Height of tower=18.59=18.6 m
Question 4
A boy 1.6 m tall is 20 m away from a tower and observes that the angle of elevation of the top of the tower is 60°. Find the height of the tower.
Sol :
Let AB be the boy and TR be the tower
∴ AB = 1.6 m
Let TR = h
from A, show AE || BR
∴ ER = AB = 1.6 m
TE = h – 1.6
AE = BR = 20 m
tanθ=TEAE
⇒tan60∘=h−1⋅620
⇒√3=h−1⋅620
⇒h−1⋅6=20√3
h=20√3+1⋅6=20(1⋅732)+1⋅6
h=34.640+1.6
=34.64+1.60
h = 36.24
∴ Height of tower = 36.24 m
Question 5
A boy 1.54 m tall can just see the sun over a wall 3.64 m high which is 2.1 m away from him. Find the angle of elevation of the sun.
Sol :
Let AB be the boy and CD be the wall which is at a distance of 2.1 m
Then AB=1.54 m, CD=3.64 m
BD=2.1 m
Draw AE||BD, then
ED=1.54 m
CE=3.64-1.54=2.1 m
and AE=BD=2.1 m
Now in right ΔCAE
tanθ=CEAE=2⋅12⋅1=2121=1
∴θ=45∘(∵tan45∘=1)
∴ Angle of elevation of the sun =45∘
Question 6
In the adjoining figure, the angle of elevation of the top P of a vertical tower from a point X is 60° ; at a point Y, 40 m vertically above X, the angle of elevation is 45°. Find
(i) the height of the tower PQ
(ii) the distance XQ
(Give your answer to the nearest metre)
⇒y=h√3...(i)
Again in right ΔPYR
tan45∘=PRYR=h−40y⇒1=h−40y
⇒y=h−40...(ii)
From (i) and (ii)
h−40=h√3
⇒√3h−40√3=h
⇒√3h−h=40√3
⇒(1⋅732−1)h
=40(1.732)
⇒⋅732h=69⋅280
h=69⋅280.732=69280732=94.64
∴ Height of the tower =94⋅64 m=95 m
and distance XQ=h−y=95−40
from (ii)
=55 m
Question 7
An aeroplane is flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.
Sol :
A and D are the two positions of the aeroplane ;
AB is the height and P is the point
∴ AB = 1 km,
Let AD = x and PB = y
and angles of elevation from A and D at point P are 60° and 30° respectively.
⇒√3=1y
⇒y=1√3..(i)
Again in right ΔDPC
tan30∘=DCPC=1x+y
⇒1√3=1x+y
⇒x+y=√3..(ii)
From (i) and (ii)
x+1√3=√3
⇒x=√3−1√3=3−1√3=2√3
⇒x=2×√3√3×√3=2(1⋅732)3=3⋅4643 km
Thus this distance covered in 10 seconds
∴Speed of aeroplane (in km/hr)
=3⋅4643×60×6010
=34643×1000×360010
=3464×36300
=3464×12100
=41568100
=415.68 km/hr
Question 8
A man on the deck of a ship is 16 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.
Sol :
Let A is the man on the deck of a ship B and CE is the cliff.
AB = 16 m and angle of elevation from the top of the cliff in 45°
and the angle of depression at the base of the cliff is 30°.
Let CE = h, AD = x, then
CD = h – 16, AD = BE = x
Now in right ∆CAD
⇒1=h−16x
⇒x=h−16...(i)
Again in right ΔADE
tan30∘=DEAD=16x
⇒1√3=16x
⇒x=16√3..(ii)
From (i) and (ii)
h−16=16√3⇒h=16√3+16
⇒h=16(√3+1)=16(1⋅732+1)
=16×2⋅732=43⋅712=43⋅7⋅1 m
and x=h-16=43.71-16=27.71
∴ Distance of cliff =27.71 m
and height of cliff =43.71 m
Question 9
There is a small island in between a river 100 metres wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks and in the line with the tree. If the angles of elevation of the top of the tree from P and Q are 30° and 45° respectively, find the height of the tree.
Sol :
The width of the river (PQ) = 100 m.
B is the island and AB is the tree on it.
Let AB=h and PC=x, then
BQ=100-x
Now in right ΔAPB
tanθ=ABPB
⇒tan30∘=hx
⇒1√3=hx
⇒x=√3h...(i)
Again in right ΔABQ
tan45∘=ABBQ=h100−x
⇒1=h100−x
⇒h=100−x..(i)
From (i) and (ii)
h=100−√3h⇒h+√3h=100
⇒(1+1⋅732)h=100⇒h=1002⋅732
∴h=100×10002732=1000002732=36.6
∴ Height of the tree =36⋅6 m
Question 10
A man standing on the deck of the ship which is 20 m above the sea-level, observes the angle of elevation of a bird as 30° and the angle of depression of its reflection in the sea as 60°. Find the height of the bird
Sol :
Let P is the man standing on the deck of a ship
which is 20 m above the sea level and B is the bird.
Now angle of elevation of the bird from P = 30°
and angle of depression from P to the shadow of the bird in the sea
tan30∘=BCPC⇒1√3=hx⇒x=√3h m...(i)
Similarly in right ΔPCR
tan60∘=CRPC
⇒√3=h+40x
h+40√3h=√3 [from (i)]
h+40=√3×√3h=3h
3h-h=40
⇒2h=40
⇒h=402=20
From sea level height of the brid
=20m+h
=20+20=40 m
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