Download - Chapter 24

Height and Distance

itroduction 1. Angle

I f a straight line OA rotates about the point ' O ' called r-e vertex from its initial position to the new position O A ' . Then the angle A O A ' , denoted as Z A O A ' , is formed. The aigle may be positive or negative depending up on their rotation. I f the straight line rotates in anticlockwise direction i positive angle is formed and i f it rotates in clockwise direc--on a negative angle is formed. A n angle is measured in :egree (). 1 Quadrants

Let X ' O X and Y O Y ' be two lines perpendicular to :h other. The point ' O ' is called the origin, the line X ' O X is lied X axis and Y O Y ' is called the Y axis. These two lines ide the plane into 4 parts. Each part is called a Q U A D -W T . The part XOY, Y O X ' , X ' O Y ' and Y ' O X are respec-s\y known as 1st, 2nd, 3rd and 4th quadrants. Angle of Elevation

I f an object A ' is above the horizontal line OA we ave to move our eyes in upward direction through an angle

KOA' then the angle A O A ' is called the angle of elevation. I. Angle of Depression

I f an object O is below the horizontal line A ' O ' and *e are standing on the point A ' then we have to move our i>es in downward direction through an angle O ' A ' O . This ingle O ' A ' O is called the angle of depression. 5. Trigonometric Ratio

Let A B C be a right angled triangle. Also let the length : : the sides BC, AC, and A B be a, b and c respectively. Then

A C 1) The ratio

perpendicular b . = = sm6

BC 2) The ratio

A C 3) The ratio -

\nd also remember that

1

sin0

hypotenuse

base a = = cosS

hypotenuse c

perpendicular _ b _

base a

(Hi) c o t e : t a n 0

( iv) t a n 0 = sin 9

cos 6

. COS0 (v) cot9 = 1 ' sin 9 (vi) cos 2 9 + s i n 2 9 = 1

(vii) 1 + tan 2 9 = sec 2 9 (viii) cot 2 9 +1 = cosec 29

6. Values of the trigonometric ratios for some useful angles

4- Ratio/Angle(9)-+ 0 30 45 60 90" sine 0 1

2 1

& S 2

1

cos 6 1. s 2

1 V2

1 2

0

tan 6 0 1 5

1 s 00 -

sec 9 1 2 73

J5 ' ''2'' CO

cosec6 2 42 2 ^ s L i

cote CO 1 1 r 0

Rule 1 Problems Based on Pythagoras Theorem

Phythagoras Theorem => h2 = p2 +b2 (see the figure)

(i) cosec0 = (ii) sec 9 =

Illustrative Example Ex: The father watches his son flying a kite from a dis-

tance o f 80 metres. The kite is at a height o f 150 metres directly above the son. How far is the kite from the father?

Soln: Distance o f the kite from the father = FK

COS0

624 P R A C T I C E B O O K ON Q U I C K E R MATHS

(FKf=(FSf+{SKf

[From the above theorem]

.-. FK = V ( l 5 0 ) 2 + (80 ) 2 = 170 metres.

I * 9 mm

Exercise 1. The father watches his son f ly ing a kite from a distance

o f 3 km. The kite is at a height o f 4 k m directly above the son. How far is the kite f rom the father? a) 5 km b) 1 k m c) 7 k m d) None o f these

2. The father watches his son f ly ing a kite from a distance o f 10 metres. The kite is at a height o f 24 metres directly above the son. H o w far is the kite f rom the father? a) 26 m b ) 2 8 m c ) 2 5 m d) Data inadequate

Answers l . a 2.a

Rule 2 Theorem: A man wishes to find the height of a flagpost which stands on a horizontal plane, at a point on this plane he finds the angle of elevation of the top of the flagpost to

be 9, . On walking x units towards the tower he finds the

corresponding angle of elevation to be 9 2 * Then the height

x tan 9, t an0 , (H) of the flagpost is given by t a n 9 2 - t a n 0 , units and

the value of DB (See the figure given below) is given by

.tan 9,

t an0 , - t a n 0 units.

Illustrative Example Ex: A man wishes to f ind the height o f a flagpost which

stands on a horizontal plane; at a point on this plane he finds the angle o f elevation o f the top o f the flagpost to be 45. On walking 30 metres towards the tower he finds the corresponding angle o f elevation

to be 60. Find the height o f the flagpost. Soln: Detail Method: A B = height o f flagpost = x m

In AABD

tan 60 = AB

BD

BD s ....(i)

tan 4 5 = -AB

BD + DC + 30 = x

V3" = 30

30V3 7 , x = 7lm

0.732 Quicker Method: App ly ing the above theorem, we have

the required height o f the flagpost

30 x tan 45 x tan 60 tan 6 0 - t a n 45

3 0 x ^ 3 x 1 30V3 7 1 m.

V J - l 0.732

Note: 1. The angle o f elevation o f a lamppost changes from

9, to 9 2 when a man walks towards it . I f the height

o f the lamppost is H metres, then the distance trav-

7 / ( t an9 2 - tan 9 ^ elled by man is given by tan 9,. tan 9 2 metres.

2. I f the time for which man walks towards lamppost is given as ' t ' sec then speed o f the man can be calcu-lated by the formula given below.

Speed o f the man = H t a n 9 2 - tan 9,

t tan 9,. tan 9 2 m/sec

Ex: The angle o f elevation o f a lamppost changes from 30 to 60 when a man walks towards it . I f the height

o f the lamppost is l oV3 metres, f ind the distance

travelled by man. Soln: App ly ing the above theorem, we have

the distance travelled by m a n :

r-f 1 A

10V3 V J - ^ =

V J x - L = 20 metres.

Exercise 1. The angle o f elevation o f a lamppost changes from 30"

to 60 when a man walks 20 m towards it . What is the height o f the lamppost?

MATHS Height and Distance 625

a)8.66m b )10m c) 17.32m d )20m A man is watching from the top o f a tower a boat speed-ing away from the tower. The boat makes an angle o f depression o f 45 with the man's eye when at a distance of 60 metres from the tower. After 5 seconds, the angle of depression becomes 30 . What is the approximate speed o f the boat, assuming it is running in still water?

[SBI Associates P O Exam, 1999] a)32km/hr b)42km/hr c)38km/hr d)36km/hr A man stands at a point P and marks an angle o f 30 with the top o f the tower. He moves some distance towards tower and makes an angle o f 60 with the top o f the tower. What is the distance between the base o f the tower and the point P?

[BSRB Hyderabad P O Exam, 1999] a) 12 units op& units

c) 4 /^3 units d) Data inadequate

The pilot o f a helicopter, at an altitude o f 1200 m finds that the two ships are sailing towards it in the same direction. The angles o f depression o f the ships as ob-served from the helicopter are 60 and 45 respectively. Find the distance between the two ships, a) 407.2 m b)510m c) 507.2 m d) Data inadequate I f the elevation o f the sun changed from 30 to 60 , then the difference between the lengths o f shadows o f a pole 15 m high, made at these two positions is .

9.

a) 7.5 m b ) 1 5 m c) 10V3 15

d ) ^ m

The angles o f elevation o f an aeroplane flying vertically above the ground as observed from two consecutive stones 1 km apart are 45 and 60 . The height o f the aeroplane above the ground in km is .

a) A / 3 + 1

km 3 + V3

b) km

c ) 3 + 7 3 k m d ) V 3 + l k m

A, B, C are three collinear points on the ground such that B lies between A and C and A B = 10 m. I f the angles of elevation o f the top o f a vertical tower at C are respec-tively 30 and 60 as seen from A and B , then the height of the tower is

a) 5V3 m b ) 5 m 10V3

c) r - m 20VJ

d) m

I f the angles o f elevation o f a tower from two points distant a and b (a > b) from its foot and in the same straight line from it are 30 and 60 . Then the height o f the tower is

(a-bp/3 a) yja + b b)

10.

11.

12.

I f from the top o f a tower 50 m high, the angles o f depres-sion o f two objects due north o f the tower are respec-tively 60 and 45 , then the approximate distance be-tween the objects is . a) 11m b ) 2 1 m c ) 3 1 m d ) 4 1 m Two persons standing on the same side o f a tower mea-sure the angles o f elevation o f the top o f the tower as 30 and 45 . I f the height o f the tower is 30 m, the dis-tance between the two persons is approximately a) 52 m b ) 2 6 m c ) 8 2 m d ) 2 2 m I f from the top o f a c l i f f 100 m high, the angles o f depres-sions o f two ships out at sea are 60 and 30 , then the distance between the ships is approximately. a)173m b)346m c)57.6m d) 115.3 m The angles o f depression o f two ships from the top o f the light house are 45 and 30 towards east. I f the ships are 100 m apart, then the height o f the light house is

13.

14.

15.

m 50 50

a )7I7rm b ) V 3 ^ 1 c ) 5 o ( V 3 - l ) m d ) 5 0 ( V 3 + l ) m

The shadow o f a tower standing on a level plane is found to be 60 m longer when the sun's attitude is 30 than when it is 45 . The height o f the tower is . a) 81.96 m b) 51.96 m c) 21.96 m d) None o f these Two observers are stationed due north o f a tower at a distance o f 10 m from each other. The elevation o f the tower observed by them are 30 and 45 respectively. The height o f the tower is . a ) 5 m b)8.66m c)13.66m d ) 1 0 m A boat being rowed away from a c l i f f 150 m high. A t the top o f the c l i f f the angle o f depression o f the boat changes from 60 to 45 in 2 minutes. The speed o f the boat is .

a)2km/hr b)1.9km/hr c)2.4km/hr d)3km/hr

Answers

. c; Hint: Required answer = 20 x tan 60 x tan 30

tan 6 0 - t a n 30

20 m > D

^ = 10V3 =17.32 m 2

626

2. a; Hint: 60 = .xx tan 30

tan 4 5 - t a n 30

x m 60 m

60 x 1

or, X : s.

60x0.732 metres

, 60x0 .732x18 . , .-. reqd speed = = 31.62 32 km/hr

5 x 5

3.d; Hint: Here we use the formula B D (ie C B ) =

xtanG, tan 9 2 - tan 8,

? "C+ ? Here neither the value o f CB nor the values o f x and height o f the tower are given. Hence, required dis-tance cannot be found.

4. c; Hint: A

1200 m

1200 =

x = 1200

xm

xtan 60 x tan 4 5

tan 6 0 - t a n 45

tan 6 0 - t a n 45

tan 60 x tan 45 = 12

- 1

P R A C T I C E B O O K ON Q U I C K E R MATHS

5 c ; Hint:

15 = x x tan 60 x tan 30

tan 6 0 - t a n 30

tan 6 0 - t a n 30 .'. x = 15 x

tan 60 x tan 30

= 15 A/3 x '

6. b; Hint: Height (H) =

= 1 5 V 3 - - l l = ^ = 10V3m.

. I x t a n 4 5 x t a n 60

tan 6 0 - t a n 45

1 k m

l x l x V 3 + \ + X

- \i +\

10 x tan 60 tan 30

km

7. a; Hint: Height tan 6 0 - t a n 30

1 0 m

10xV3x-j=-

- ^ = 5

m

= 1200 - 400V3 = (l 200 - 400 x 1.732) = 507.2 m 8 . b ; H i n t : Height - ( q - A ) t a n o 6 0 x t a n 3 0 tan 6 0 - t a n 30

Height and Distance

a m

9. b; Hint: Required distance (x) - 5 0 tan 6 0 - t a n 45

tan 60 x tan 45

A

10. d; Hint: Required distance = 3 0 x tan 45 x tan 30

A

30 m

:30

1-V3"

11. d; Hint: Required distance (x) = 100|

= 3 0 ( V 3 - l ) * 2 2 m

tan 60 - t a n 30

tan 60 x tan 30

100 m

627

= 100 V3

V 3 x - 4 = . V3 J

200V3 = 115.3 m

12. d; Hint: Required height (H) = 100|

A

tan 45 x tan 30

tan 4 5 - t a n 30

100 m

100!

1

1 1

100

V J - i

100 V3+1 = 5o(V3+l) m

13. a; Hint:

60 m

Required height (H) = 6 0 | tan 45 x tan 30

tan 45 - t a n 30

6 0 x - = x l

! _ _ L " V 3 - 1

60 A / 3 + 1

V J - l V 3 + 1

:30x2.732 = 81.96 m

= 3 o ( V 3 + l )

14. c

15. b; Hint: x tan 6 0 - t a n 45

tan 60 x tan 45 x l 5 0


150 m

= 63.4

distance covered in 2 minutes = 63.4 m

m

.-. speed of the boat : 63.4 60

X

2 1000 1.9km/hr.

km/hr

Rule 3 Theorem: A small boy is standing at some distance from a flagpost. When he sees theflag the angle of elevation formed is 9 . If the height of the flagpost is 'H' units, then the

H

distance of the child from the flagpost is t a n g o units.

Illustrative Example EK A small boy is standing at some distance from a

flagpost. When he sees the flag the angle o f eleva-tion formed is 60. I f the height o f the flagpost is 30 ft, what is the distance o f the child from the flagpost?

AB Soln: Detail Method: = tan 60

BC

3 0 - 1 / 5 o r , - V 3

30 ft

o r , g C = ^ X f x l Q = 1 0 V 3 f t

Quicker Method: Applying the above theorem, we have \ the required distance

J _ 3 0 _ = V 3 x ^ x l O = i o V ?

tan 60 V3 n' Exercise

1- 100A/3 m from the foot o f a c l i f f on level ground, the

angle o f elevation o f the top o f a c l i f f is 30 . Find the height o f this cliff.

4.

a) 100m b ) 5 0 m c) 50V3 m d)300m

A small boy is standing at some distance from a flagpost. When he sees the flag the angle o f elevation formed is 45. I f the height o f the flagpost is 10 ft, what is the distance o f the child from the flagpost?

10 ' a) ft b) 10 ft c ) 1 0 V 3 f t d) None o f these

A small boy is standing at some distance from a flagpost. When he sees the flag the angle o f elevation formed is

30. I f the height o f the flagpost is 24^/3 ft. what is the

distance o f the child from the flagpost?

a)24f t b )48 f t c )72f t d) 24>/3 ft

25-\/3 m from the foot o f a c l i f f on level ground, the

angle o f elevation o f the top o f a c l i f f is 30 . Find the

height o f this cliff.

a) 25 m b ) 7 5 m c) 25A/3 m d) None o f these

45 m from the foot o f a c l i f f on level ground, the angle o f elevation o f the top o f a c l i f f is 60 . Find the height o f this cliff.

45 a ) ^ m b) 45-^3 m c ) 1 3 5 m d) None o f these

Answers

l . a ; H i n t : 100V3 H

m : tan 30

H

100 71m

1 .-. H = 1 0 0 V 3 x t a n 3 0 = 100V3x-_L, = 100 m .

2.b 3.c 4. a 5. b

Rule 4 Theorem: The angles of elevation of top and bottom of a

flag kept on a flagpost from 'x'units distance are 8 , and

9 2 0 respectively. Then (i) the height of the flag is given by

[x(tan9| - t a n 9 2 ) ] units and (ii) the height of the flagpost is

/ / t an9 2 given by

tan 9, - t a n 9 1 units, where h = height of the

flag ie x( tan0, - t a n 9 2 ) .

Height and Distance 629

Note: I f the height o f the flag is not given then, we can calculate the height o f the flagpost directly by the formula given below,

Height o f the flagpost = ( x t a n G 2 ) units.

Illustrative Example Ex: The angles o f elevation o f top and bottom o f a flag

kept on a flagpost from 30 metres distance are 45 and 30 respectively. What is the height o f the flag?

Soln: Detail Method: tan 45 50 m

tan 30 = BC

30 r - 3 0 o r , 5 C - - ^

30 Height o f flag AB = 30 = 30 - 1 0 V J

V3 = 30-17 .32= 12.68m

Quicker Method: Apply ing the above theorem, we have the required height = 3 0 ( t a n 4 5 - t a n 3 0 )

= 30 1 - = 12.68 metres.

Exercise 1. An observer standing 72 m away from a building notices

that the angles o f elevation o f the top and the bottom o f a flagstaff on the building are respectively 60 and 45 . The height o f the flagstaff is . a) 124.7 m b) 52.7 m c) 98.3 m d) 73.2 m

2. The angles o f elevation o f top and bottom o f a flag kept on a flagpost from 10 metres distance are 60 and 30 respectively. What is the height o f the flag?

10 20 a) 20fi m b) m c) 10^3 m d) j m

3. The angles o f elevation o f top and bottom o f a flag kept on a flagpost from 45 metres distance are 60 and 45

respectively. What is the height o f the flag?

a ) 2 o ( V 3 - l ) m b ) 4 5 ( V 3 + l ) m

c) 45(A/3 - 1 ) m d) None o f these

Answers 1. b; Hint: Required height = 72(tan 60 - tan 45)

2 .d

72 m

= 72[73-l] = 7 2 x 0 . 7 3 2 = 52.7

3.c

m

Rule 5 Theorem: 'x' units of distance from the foot of a cliff on level ground, the angle of elevation of the top of a cliff b 0 , then the height of the cliff is (x tan 0 ) units.

Illustrative Example Ex: 300 m from the foot o f a c l i f f on level ground, the

angle o f elevation o f the top o f a c l i f f is 30. Find the height o f this cliff .

Soln: Detail Method: Let the height o f the c l i f f A B be xm.

In AABC A

300 m

tan 30 = AB

BC 300

.-. x = ^ = 100V3 =173.20w V3

Quicker Method: Applying the above theorem, we have the required height o f the c l i f f = 300 * tan 30

= 300 x 4 - = 173.20m. V3

Exercise 1. The shadow of a building is 20 m long when the angle o f

elevation o f the sun is 60 . Find the height o f the build-ing.


20 a ) 20A/2 m b) 20A/3 m c) ^ m d) Data inadequate

2. I f a vertical pole 6 m high has a shadow o f length 2 A/3

m, find the angle o f elevation o f the sun. a) 30 b )45 c )60 d )90

3. A ladder leaning against a vertical wall makes an angle o f 45 with the ground. The foot o f the ladder is 3 m from the wall . Find the length o f the ladder.

a)242m b ) 3 A / 2 m c ) 5 m d) 3^3 m

4. The ratio o f the length o f a rod and its shadow is 1 : fi

The angle o f elevation o f the sun is . a) 30 b ) 4 5 c )60 d )90 The angle o f elevation o f a moon when the length o f the shadow o f a pole is equal to its height, is . a) 30 b ) 4 5 c ) 6 0 d ) 9 0 The angle o f elevation o f a tower from a distance 100 m from its foot is 30 . Height o f the tower is .

5.

6.

100 200

d) s m 7.

a) 100A/3 m b) ^ m c) sofi

The altitude o f the sun at any instant is 60 . The height o f the vertical pole that w i l l cast a shadow o f 30 m is

a) 30v3 m b ) 1 5 m 30

Height and Distance 6 3 1

.-. height o f the tree = A B ( A D ) + A C

20 10 30 , n tz = - 7 = + - = r = - 7 = = 10V3 m

V3 V3 V3

Rule 6 Theorem: The horizontal distance between two towers is 'x' units. The angle of depression of the first tower when seen from the top of the second tower is 0 .

(i) If the height of the second tower is ' _y, ' units then the

height of the first tower is given by (y, - JC tan 8) units.

(ii) If the height of thefirst tower is given as ' y2' units then

the height of the second tower is given by ( y 2 + x tan 0 ) .

A

* X

p >

B 2 n d tower 1st tower

Illustrative Example Ex: The horizontal distance between two towers is

50i/3 m. The angle o f depression o f the first tower when seen from the top o f the second tower is 30. I f the height o f the second tower is 160 m, find the height o f the first tower.

Soln: Detail Method: Let A B be the tower 160m high.

50j3~m Let CD be another tower o f height x m Since, A M || PC .-. angle M A C = angle ACP = 30 So, in AAPC

tan 30 = AP I AP

PC S 50fi

.-. AP = 50m

.*. the height o f the other tower = A B - A P = 1 6 0 - 5 0 = H O m .

Quicker Method: Apply ing the above theorem, we

have the required height

= 1 6 0 - 5 0 V 3 x t a n 3 0 = 1 6 0 - 5 0 V 3 x 4 = = l 10 m V3

Exercise 1. A person o f height 2 m wants to get a fruit which is on a

10 __ pole o f h e i g h t m . I f h e stands at a distance o f ^ m

from the foot o f the pole, then the angle at which he should throw the stone, so that it hits the fruit is . a) 15 b ) 3 0 c )45 d )60

2. The distance between two multi-storeyed buildings is 60 m. The angle o f depression o f the top o f the first building as seen from the top o f the second building, which is 150 m high is 30 . The height o f the first build-ing is a) 115.36 m b) 117.85 m c) 125.36 m d) 128.34 m

3. The heights o f two poles are 80 m and 62.5 m. I f the line jo in ing their tops makes an angle o f 45 with the hori-zontal, then the distance between the poles is . a) 17.5 m b) 56.4 m c) 12.33 m d )44m

Answers

L b ; 10 4 .

Hint: 2m = y ~ ~ ^ 7 j t a n e

or, t an0x V3 3

i a ' 4 1 or tan0 = x = - = 3 4 fi

10 3

150 m

60 m

Height o f the first building (h) = 150 - 60 tan 30

= 150-20A /3 =115.36 m m

3.a; Hint: 62.5= 80 -Ytan45 1 -


62.5

80 m

\ J C = 80 -62 .5 = 17.5 m

Rule 7 Theorem: Two poles of equal heights stand on either sides of a roadway which is x units wide. At a point of the road-way between the poles, the elevations of the tops of the pole

are 6 , and Q2, then the

(i) heights of the poles: xtanO, t a n 0 2

units and the tan 6, + t a n 0 2 (ii) position of the point P from B (see the figure) =

x tan 0 2 units, and the position of the point Pfrom tan0, + t a n 0 2

/> =

Here, AB = CD = Height of the poles.

Illustrative Example Ex. Two poles o f equal heights stand on either sides o f a

roadway which is 120 m wide. A t a point on the road-way between the poles, the elevations o f the tops o f the pole are 60 and 30. Find the heights o f the poles and the position o f the point.

Soln: Detail Method: Let A B and CD be two poles = x m and P the point on the road. Let BP = y m; then PD = ( 1 2 0 - y ) m

A C

In AABP

t a n 6 0 = = -=>x = yfi ...(>) BP y

In ACDP

tan 30 = CD

=> xJ3 = 1 2 0 - y ....(ii) DP 1 2 0 - y

Combining equations ( i ) and ( i i ) , we get

vVJVJ = 1 2 0 - y => 3 y = 1 2 0 - y => y = 3 0 m

So, from equation ( i ) , x = y VJ = 30\/3 * 52m

Quicker Method: Applying the above theorem, we have

( i ) height o f the po l e :

\20xfix~

^ 3 - = 30V3 mand

120x ( i i ) position o f the point P from B

fi

= 30 m.

Exercise 1. Two poles o f equal heights are standing opposite to

each other on either side o f a road, which is 30 m wide. From a point between them on the road, the angles o f elevation o f the tops are 30 and 60 . The height o f each pole is . a)4.33m b ) 6 . 5 m c ) 1 3 m d ) 1 5 m

2. Two poles o f equal heights stand on either sides o f a roadway which is 20m wide. A t a point on the roadway between the poles, the elevations o f the tops o f the pole are 45 and 30. Find the heights o f the poles.

20

a ) V 3 - 1 m b) 2o(VJ-l)m

c) 10(V3-l) m d) None o f these

3. Two poles o f equal heights stand on either sides o f a roadway which is 50 m wide. A t a point on the roadway between the poles, the elevations o f the tops o f the pole are 60 and 45. Find the heights o f the poles, a) 31.69 m b) 32.96 m c) 31.96 m d) Data inadequate

Answers

1. c; Hint: Required height (h) = 30 x tan 30 x tan 60

tan 60 +tan 30


S + J_ ~ 2 =7.5x1.732

s = 12.975 m x 13 m

2.c 3.a

Rule 8 Consider the following figure,

A

In this figure, 8! and 8 2 are given. A C is given.

To find A B we have fol lowing formula,

AC 1-t a n 8 2

tan 8,

Illustrative Example Lv An aeroplane when 3000 m high passes vertically

above another at an instant when the angles o f eleva-tion at the same observing point are 60 and 45 re-spectively. How many metres lower is one than the other? Detail Method : Let A and B be two aeroplanes, A at a height of3000 m from C and B y m lower than A. Let D be the point o f observation, then angle A D C = 60 and angle BDC = 45 L e t D C = x m In AACD

Soln:

tan 60 = AC 3000

CD

3000 x =

Again, in A B C D

....(i)

BC 3000->> , tan 4 5 = => - - \

CD x .-. ;c = 3 0 0 0 - y . . -( i i ) Combining ( i ) and ( i i ) we get

3000 SOOO-y

y = 3000 1 -73"

3000x0.732

1.732 * 1268m

Quicker M e t h o d : Apply ing the above formula, we have

the required answer : 3000

3000

1 - -tan 45

tan 60

1 a 1268m

Exercise 1. A vertical tower stands on a horizontal plane and is sur-

mounted by a flagstaff o f height 7 m. A t a point on the plane, the angle o f elevation o f the bottom o f the flag-staff is 30 and that o f the top o f the flagstaff is 45 . Find the height o f the tower.

a) 73"-7V3 7 7V3

m b ) V T I m c ) V 3 ^ T m d ) V 3 " " + T m 2. A n aeroplane when 1500 m high passes vertically above

another at an instant when the angles o f elevation at the same observing point are 60 and 30 respectively. How many metres lower is one than the other? a) 1200 m b) 1000 m c)800m d) 1050 m

3. A n aeroplane when 1000 m high passes vertically above another at an instant when the angles o f elevation at the same observing point are 45 and 30 respectively. How many metres lower is one than the other? a) 442.6 m b) 424.6 m c) 482.6 m d) 444.6 m

Answers 1. a; H i n t : Applying the given rule, we have

the whole height (ie tower + flagstaff)


.-. height o f the tower (h) = f 7 - 7 = 4 ^ - 7

-\

i

VJ - i m 2.b 3. a

See the following figure

A

Rule 9

In this figure, AC = x

tan0 = - then, b

since = =>AB = ACx~ = 102x = 90m AC 17 17

Quicker M e t h o d : Applying the above formula we have

the required answer = = _x 102 = 90 metres. V l 5 2 + 8 2

Exercise 1. The length o f a string between a kite and a point on the

ground is 85 m. I f the string makes an angle 8 with the

level ground such that tan 8 = , how high is the kite, 8

when there is no slack in the string? a) 78.05 m b ) 7 5 m c) 316 m d) Data inadequate

2. The length o f a string between a kite and a point on the ground is 25 m. I f the string makes an angle a with the

4 level ground such that a = , how high is the kite?

a) 20 m b ) 1 5 m c ) 2 4 m d ) 1 6 m

( 0 AB-

(ii) BC =

4a2+b2

J a2 +b2

and

Illustrative Example Ex: The length o f a string between a kite and a point on

the ground is 102 m. I f the string makes an angle a

_ 15 with the level ground such that

Height and Distance

M A

635

Illustrative Example E K The angles o f depression o f two ships from the top o f

a lighthouse are 45 and 30. I f the ships are 120 m apart, find the height o f the lighthouse. Detail Method: Let A B , the height o f the lighthouse = xm

Soln:

M 3 0 y V 4 5

X

4 5 ( \

N

y B 120 120 m

Since M N || PQ .-. angle M A P = angle APB == 30 and angle

N A Q = angle A Q B = 45 Let the length between P and B be y m. So, the length between B and Q is (120 - y) m. In A A B P

tan 30 = AB

BP

U y = xfi ....(i)

Again, in AABQ

BQ

1 2 0 - y

J _ x

y

tan 45 = 120->>

=> x = izu-y ( i i )

Combining equations ( i ) and ( i i ) , we get

x = \20-xfi

or, x(l + V3")=120

120 x =

1 + V3 44m

Quicker Me thod : Applying the above theorem, we have

the required answer : 120 x tan 30

tan 45+ tan 30

' 20 *, * 44 metres.

1 + V3

Exercise

1. From the top o f a c l i f f ] 00\/3 m high the angles o f de-

pression o f two boats which are due south o f observer

are 60 and 30. Find the distance between the two boats.

a) 400 m b)250m c) 200^3 d) 400^3 m

A landmark on a river bank is observed from two points

A and B on the opposite bank o f the river. The lines o f

sight make equal angles (45) with the bank o f the river.

I f A B = 1 km, then the width o f the river is .

2.

a) 2 km 1

b) - km 2

, 3V2 c) km d) km

; 2

The angles o f elevation o f the top o f a tower 40 m high from two points on the level ground on its opposite sides are 45 and 60 . The distance between the two points in nearest metres is . a)60m b ) 6 1 m c ) 6 2 m d )63m Two boats approach a light house in mid-sea from oppo-site directions. The angles o f elevation o f the top o f the light house from the two boats are 30 and 45 respec-tively. I f the distance between the two boats is 100 m, the height o f the light to house is a) 36.6 m b) 73.2m c) 136.6m d)68.3m

Answers

l a ; Hin t 100V3 x tan 60 x tan 30

,-. .x = 100V3| f tan 60 + tan 30

' [ t a n 60 x tan 30

= 100A/3 fi_ = 400 m

636 PRACTICE BOOK ON QUICKER MATHS

b; Hint:

P f ' f d m a r k ) R ive r Bank

R ive r Bank

1 km Required width (PO)

1 x tan 45 x tan 45 l x l x l 1 gas lcTT)

i 4 5 4 tan 45 1 + 1 2 3. d; Hint: Required dis.zuce (x)

"tan 45 +tan 60

tan 45 x tan 60 B

*40

40 n \ / V 5 6 0 \

x

' l + V T

4. a; Hint: Required height ( h ) !

B

40 * 63 m

100 x tan 45 x tan 30 tan 45 + tan 30

= 5o(V3-l) = 50x0.732 = 36.6 m

Rule 11 Thoerem: From the top and bottom of a building of height h units, the angles of elevation of the top of a tower are a

and P respectively, then the (i) height of the tower is given

/?tanp by

tan P - tan a units, (ii) distance between the building

and the tower is given by tan P - tan a y

/itana

units and (Hi) RM

units. (Seetheflgure)-^_xma

Note: I f height o f the tower is given as ' H ' units, then the distance between the building and the tower is giver.

H by _ ~ r7 units.

3 tanp

R

I I

Illustrative Example Ex: From the top and bottom o f a building o f height 11'.

metres, the angles o f elevation o f the top o f a towe-are 30 and 45 respectively. Find the height o f the tower.

Soln: Applying the above theorem, we have

120 x tan 45 the height o f the tower =

tan 4 5 - t a n 30

120

1 -fi

120x1.732

.732 * 284 metres

Exercise 1. A tower is 30 m high. A n observer from the top of tr*

tower makes an angle o f depression o f 60 at the base I a building and angle o f depression o f 45 at the top o f the building, what is the height o f the building? A > : find the distance between building and tower.

10 a) 12.6m, ^ j m b) 12.6m, 17.3m

c) 12 m, io-y/3 m d) Data inadequate

2. The top o f a 15 metre-high tower makes an angle of el-evation o f 60 with the bottom o f an electric pole anc x angle o f elevation o f 30 with the top o f the pole. W :m is the height o f the electric pole?

[SBI Bank P O Exam, 1 fA a ) 5 m b ) 8 m c ) 1 0 m d ) 1 2 m

3. From the top o f a building 30 m high, the top and bon: o f a tower are observed to have angles o f depress m 30 and 4 5 respectively. The height o f the tower


a) 15(l + V3") m

1

b) 3 o ( V 3 - l ) m

d) 30! + - ^ J i n d ) 3 0 ^ 1 - - ) = - 1;

From the foot o f a tower the angle o f elevation o f the top of a column is 60 and from the top o f the tower which is 25 m high, the angle of elevation is 30. The height o f the column is .

a) 37.5 m b) 42.5 m c) 43.3 m d) 14.4 m

A n s w e r s : b; Hint: Here, a = 45 and P = 60

45,

60

h = ?

H = 3 0 m

Now applying the given rule,

3 Q _ Man 60

tan 6 0 - t a n 45

tan 6 0 - t a n 45 .. h = x30 = 30 12.6 m

tan 60 h = height o f the building. Distance between the building and the tower

30 = [See Note o f the given rule]

tan 60

= ^ = 10V3 =17.3 m

Note: The distance between building and tower can also be

found by using the given rule ( i i ) , .-. required distance

30

c; Hint: 1 5 =

30

- i "73

/i(tan 60)

tan 60 - t a n 30 A

10^3 = 17.3

15 m

:.h = _ 1 5 x ( t a n 6 0 - t a n 3 0 )

tan 60

15x V3

3. d; H i n t : Here, we have to find h = ? 30

30 m

H is given = 30

a = 30 and P = 45 Now, applying the given rule, we have

/ ? x t a n 4 5 30 =

tan 4 5 - t a n 30

.-. h = 30] tan 4 5 - t a n 30

30

y tan 45 4. a; Hint: Required height o f the column (H)

25 x tan 60 " tan 60 - tan 30

B

m

25 m 60

25V3

A

25x3

3 - 1

H

75 = 37.5 m

Miscellaneous 1. A balloon is connected to a meteorological station by a

cable o f length 200 m, inclined at 60 to the horizontal. Find the height o f balloon from the ground. Assuming that there is no slack in the cable.

200 100 a) 200V3 m b ) ^ = m c) \00j3 d) m

2. From the top o f a c l i f f 25 m high the angle o f elevation o f a tower is found to be equal to the angle o f depression o f the foot o f the tower. Find the height o f the tower. a) 50 m b ) 7 5 m c )60m d) Data inadequate

3. In a rectangle, i f the angle between a diagonal and a side is 30 and the length o f diagonal is 6 cm, the area o f the

rectangle is .

a) 9 cm 2 b) 9^3 cm 2 c )27cm 2 d )36cm 2

4. The length o f a string between a kite and a point on the


ground is 90 m. The string makes an angle of 60 with the level ground. I f there is no slack in the string, the height o f the kite is .

a) 90VJ m b) 45^3 m c)180m d ) 4 5 m

5. From the top o f a pillar o f height 20 m, the angles o f elevation and depression o f the top and bottom of an-other pillar are 30 and 45 respectively. The height o f the second pillar (in metres) is

a ) 2 0 ( f - ' > b ) i o c)ioV3" ot/M V3 V3

6. The angles o f elevation o f the top o f a tower from two points distant 30 m and 40 m on either side from the base and in the same straight line with it are complementary. The height o f the tower is . a) 34.64 m b) 69.28 m c) 23.09 m d) 11.54 m Two posts are k metres apart and the height o f one is double that o f the other. I f from the middle point o f the line joining their feet, an observer finds the angular el-evations o f their tops to be complementary, then the height (in metres) o f the shorter post is .

7.

k

a ) 2V2" b) c) kfi k

d) 8. The banks o f a river are parallel. A swimmer starts from a

point on one o f the banks and swims in a straight line inclined to the bank at 45 and reaches the opposite bank at a point 20 m from the point opposite to the start-ing point. The breadth o f the river is . a) 20 m b) 28.28 m c) 14.14 m d ) 4 0 m

9. The angle o f elevation o f an aeroplane from a point on the ground is 45 . After 15 second's flight, the elevation changes to 30 . I f the aeroplane is flying at a height o f 3000 m, the speed o f the plane in km per hour is . a)304.32 b) 152.16 c)527 d)263.5

10. A man on a c l i f f observes a fishing trawler at an angle o f depression o f 30 which is approaching the shore to the point immediately beneath the observer with a uniform speed. 6 minutes later, the angle o f depression o f the trawler is found to be 60 . The time taken by the trawler to reach the shore is .

a) 3 V3 m i n b) ^3 nun c) 1.5 min d) 3 min 11. A flagstaff o f height (1/5) o f the height o f a tower is

mounted on the top o f the tower. I f the angle o f eleva-tion of the top o f the flagstaff as seen from the ground is 45 and the angle o f elevation o f the top o f the tower as seen from the same place is 6 , then the value o f tan 6 is

3 ) I b) 5V3

c) d) 6 ''6 ~' 5 12. I f the angle o f elevation o f a cloud from a point 200 m

above a lake is 30 and the angle o f depression o f its reflection in the lake is 60 , then the height o f the clouc above the lake, is

a) 200 m b)500m c ) 3 0 m d) None of these

Answers 1. c; Hint: Let B be the balloon and A B be the vertica:

height.

Let C be the meteorological station and CB be the cable.

Then, BC = 200 m and Z A C B = 60

A B V3 Then = sin 60 = -

BC 2

A B 73 r ' 2 O T = T "I AB=100V3 m

2. a; Hint: Let A B be the c l i f f and CD be the tower. Then A B = 2 5 m p

From B draw

B E 1 C D , B r ^ 2 -

Let Z E B D = Z A C B = a

DE A B Now, = tana and = tana

DE A B = - So,DE = A B [ v B E = A C ]

.'. C D = CE + DE = A B + A B = 2 A B = 50 m

3.b; H i n t : L e t A B C D be the rectangle in whic'r

ZBAC = 30 and A C = 6 cm

AB AB r

BC 1 = s i n 3 0 = - :

AC 2

BC BC = 3 cm

eight and Distance 639

.-. Area o f the rectangle = ABxBC = 9-^3 c m 2

b; Hint: Let K be the position o f the kite and HK be the string so that

K

HK = 9 0 m & ZAHK = 60

= s i n 6 0 = ^ ^ m - r HK 2 90 2 = > A K ~ 4 5 a / 3

* Height o f the kite = 45^3

Hint: Let AB and CD be two pillars in which AB = 20 m. Let BE1DC Then,

ZDBE = 30 and ZEBC = ZACB = 45

Let DE = x Clearly, EC - AB = 20 m

= c o t 4 5 = l AB B

AC = \=> AC = 20 m

20

.-. BE = AC = 20m

DE _ 1 1 20 N o w , ^ = t a n 3 0 ~ ^ = - ^ = - 7 y m

20 2o(V3+l) .-. HeightofpillarCD = 2 0 + x = 2 0 + rj^=L

Hint: Let AB be the tower and C, D be the points o f observation.

Then, AC = 30 m & AD = 40 m

Let ZACB = 0 Then, ZADB = 90 - 0

xi AB AB Now, tan 0 = = AC 30

t a n ( 9 O - 0 ) = AB AB

AD 40

or cot 0 = AB

40

.'. tan 9 x cot 0 = AB1

1200

7. a;

8.c;

or, AB = Vl200 = 2 o V 3 =(20xl .732)m = 34.64m

Hint: Let and CD be the two posts such that AB =

2 CD. Let M be the midpoint o f CA. Let ZCMD = 9

and ZAMB = 9 0 - 6 .

Clearly, C M = M 4 = ^-

Let CD = A Then, AB = 2A

Now, = tan(90 - 0) = cot 0 AM

D

= > c = COt 0

4/; = > C O t 0 = . . . . ( / )

A C

CD fo = tan 0 => tan 0 =

CM

Ah 2h Mult ip lying (/') and (/'/'), we get ~rx~r

K K

:.h = => h = 2V2

metres

Hint: Let A be the starting point and B, the end point o f the swimmer. Then,

A

C B

AB = 20 m and ZBAC = 45

BC . . . . 1 BC 1 Now " = s i n 4 5 = - 7 = => = ~F= IN0W' AB V2 20 V2

^ c = 2 0 x V 2 = ] 4 ] 4 m 2

9. c; Hint: Let A and B be the two positions o f the plane and let O be the point o f observation and OD be the horizontal. Draw ACLOD & BDIOD.

300Qm


Then, ZDOB = 3 0 ,

ZDOA = 45 &AC = BD = 3000 m.

Let AB = h.

. : = cot 30 = S =>OD = (3000 x S) DB

m

= cot 45 = 1 => OC = 3000 m AC

Distance covered in 15 sec = AB = CD = OD-OC

= (3000V3 - 3000) M = 2196 m

.-. Speed o f the plane

'2196 1 ) Q x 60 x 60 I k m / h r = 527 km/hr

10. d; Hint: Let AB be the c l i f f and C and D be the two positions o f the fishing trawler.

Then, ZACB = 30 and ZADB = 60

Let AB = h

h

And, = col30 = fi => AC = fih AB

AD Now, r- = c o t 6 0 =

f CD = AC- AD = fih-

2h JL

Let u m/min be the uniform speed o f the trawler.

Distance covered in 6 min = 611 metres.

2h :. CD = 6u : 6u => h = 3A/3!(

, n A 3V3 w . Now, A D = ~ ^ = = 3 w

Time taken by trawler to reach A

Distance AD 3u 3 m i n .

speed u

11. c; Hint: Let AB be the tower and BC the flagstaff.

Then, BC = h Let O be the observer. 5

Then, ZAOC = 45 and Z/4