chapter 24 chapter 24 spirochetes spirochetes chapter 24 chapter 24 spirochetes spirochetes.
Chapter 24
Transcript of Chapter 24
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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
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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?
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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
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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
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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
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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
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628 P R A C T I C E B O O K ON Q U I C K E R MATHS
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 ) .
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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.
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630 P R A C T I C E B O O K ON Q U I C K E R MATHS
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
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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 -
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632 P R A C T I C E B O O K ON Q U I C K E R MATHS
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
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Height and Distance 633
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)
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634 P R A C T I C E B O O K ON Q U I C K E R MATHS
.-. 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
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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
-
Height and Distance 637
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
-
638 PRACTICE BOOK ON QUICKER MATHS
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
-
640 PRACTICE BOOK ON QUICKER MATHS
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