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









Now in right ΔATR,

tanθ=TRAR

tan30=hx

13=hx

x=3h...(i)

and in right ΔBTR

tan60=TRBR=hx50

3=hx50

h=3(x50)...(ii)

From (i) and (ii)

h=3(3h50)h=3h503

2h=503h=253

h=25(1.732)=43.3

Substituting the value of h in (i)

x=3×253=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












and B are 60° and 45° respectively 
AC=3000 m
Let AB=x
∴BC=3000-x
Let PC=y

Now in right ΔAPC
tanθ=ACPC

tan60=3000y

3=3000y

y=30003...(i)

Again in right ΔBPC
tan45=BCPC1=3000xy

y=3000x

30003=3000x

x=300030003 [from (i)]

30003000×33×3

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










tanθ=PRAR

tan45=7+hx

1=7+hx

x=7+h...(i)

Again in right ΔTAR

tan36=TRAR=hx

07265=hx

h=x(07265)...(ii)

From (i) and (ii)

h=(7+h)(0.7265)

h=7×07265+7265h

h7265h=7×07265

(17265)h=7×07265

02735h=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











Now in right ΔTAE

tanθ=TEAE

tan60=h1620

3=h1620

h16=203

h=203+16=20(1732)+16

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=3.6 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=2121=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)

Sol :












Let PQ be the tower and let PQ = h
and XQ = YR = y
XY = 40 m
∴ PR = h – 40












Now in right ΔPXQ

tanθ=PQXQ

tan60=hy3=hy

y=h3...(i)

Again in right ΔPYR

tan45=PRYR=h40y1=h40y

y=h40...(ii)

From (i) and (ii)

h40=h3

3h403=h

3hh=403

(17321)h

=40(1.732)

732h=69280

h=69280.732=69280732=94.64

Height of the tower =9464 m=95 m

and distance XQ=hy=9540

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.










Draw DC PB produced.
DC=AB=1 km

Now in right ΔAPB,

tanθ=ABPBtan60=1y

3=1y

y=13..(i)

Again in right ΔDPC

tan30=DCPC=1x+y

13=1x+y

x+y=3..(ii)

From (i) and (ii)

x+13=3

x=313=313=23

x=2×33×3=2(1732)3=34643 km

Thus this distance covered in 10 seconds

∴Speed of aeroplane (in km/hr)

=34643×60×6010

=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













tanθ=CDAD

tan45=h16x

1=h16x

x=h16...(i)

Again in right ΔADE

tan30=DEAD=16x

13=16x

x=163..(ii)

From (i) and (ii)

h16=163h=163+16

h=16(3+1)=16(1732+1)

=16×2732=43712=4371 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.










Angles of elevation from A to P and Q are 30° and 45° respectively

Let AB=h and PC=x, then

BQ=100-x

Now in right ΔAPB

tanθ=ABPB

tan30=hx

13=hx

x=3h...(i)

Again in right ΔABQ

tan45=ABBQ=h100x

1=h100x

h=100x..(i)

From (i) and (ii)

h=1003hh+3h=100

(1+1732)h=100h=1002732

h=100×10002732=1000002732=36.6

Height of the tree =366 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













Let BC=h, PQ=20=CA, AR=(h+20) m
and CR=h+20+20=h+40 m
Let PC=QA=x

Now in right ΔPCB

tan30=BCPC13=hxx=3h m...(i)

Similarly in right ΔPCR

tan60=CRPC

3=h+40x

h+403h=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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