Download 10. heights and distances

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10. HEIGHTS AND DISTANCES
LONG ANSWER QUESTIONS
1.
From a point on the level ground the angle of elevation of the top of the hill is α .
After walking through a distance ‘d’ meters towards the foot of the hill the
angles of elevation of the top is found to be β . Find the
height of the hill
B
Solution:
AB be the hill of height h
P be the first point of observation
A
x
Q
d
Q be the second point of observation
The angle of elevation of β
From P and Q are α and β respectively
Given PQ = d let AQ = x
In ∆ le BAQ tan β =
h
⇒ x = h cot β
x
In ∆le BAP tan α =
h
⇒ d + x = h cot α
d+x
d + h cot β = h cot α
d = h {cot α − cot β }
h=
2.
d
cot α − cot β
From the cliff of a mountain 120 meters above the sea level the angles of
depression of two boats on the same side of the mountain are 450 and 300
respectively. Find the distance between the both boats {assume the both the boats
lie along the line from foot of the hill on the same sides of the hill ?
Solution:
Let AB be the hill Ax be the horizontal through the foot of the hill. By parallel to
B
horizontal line passing through the observing.
Y
300
Exp: Let P, Q be the two boats
0
45
Given ∠y BQ = 300
120m
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450
300
X
Q
P
A
P
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From the fig ∠y BP = ∠BPA = 450
BQ = BQA = 300
Let PQ = x
In triangle PQAB
tan 450 =
120
AB
⇒1=
PA
PA
⇒ PA = 120
In triangle BAQ
tan 300 =
x = 120
3.
(
AB
1
120
⇒
=
0
AC
3 x + 120
)
3 −1
The angles of elevation of the top of a mountain from the top and bottom of a
pillar AB of height 5 meters are 300 and 600 respectively. Find the height of the
mountain?
Solution:
Let PQ be the mountain of height ‘h’
AB is the pillar of height 5m
Angles of elevation of Q from A and B are respectively 600 and 300
Let AP = BC = x
Since AB = 5 CQ = h − 5
h-5
h
In triangle QPA tan 600 = → (1)
x
h−5
In triangle QCB tan 30 =
→ (2)
x
30
h
tan 60
3
h
= x ⇒
=
(1) ÷ (2) we have
0
h
−
5
tan 30
 1  h−5


x
 3
2
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h
x
5m
600
0
0
0
A
x
P
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3=
h
15
⇒ 3h − 15 = h ⇒ 2h = 15 ⇒ h =
2
h−5
∴ Height of mountain = 7.5m
4.
From the top of a tree on the bank of a lake an aeroplane in the sky makes an
angle of elevation α and its image in the river makes an angle of depression β .
If the height of the tree from the water surface in a and the height of the
a sin (α + β )
aeroplane in h show that h =
sin ( β − α )
B
h-a
Solution :
h
Let ' β ' the position of aeroplane C b the its image
Q
a
PQ be the tree of height ‘a’ draw QR parallel the AP
Angle of elevation of B from Q is α
h
Angle of depression of C from Q is β
BR = h – a
c
CR= h + Q Let QR= PQ = x
In triangle QRB
tan α =
h−a
→ (1)
x
In triangle QRC
tan β =
h+a
→ (2)
x
(1) ÷ (2) we have
∴
h+a
P
tan α h − a
=
tan β h + a
h+a
tan β
=
tan α
h−a
Using compounds and dividends we have
sin β sin α
+
h+ a −h− a
tan β + tan α
h cos β cos α
=
⇒ =
h +a− h +a
tan β − tan α
a sin β − sin α
cos β cos α
a sin ( β + α )
h sin β cos α + cos β sin α
=
⇒h=
a sin β cos α − cos β sin α
sin ( β − α )
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5.
one end of the ladder is in contact with a wall and the other end is in contact with
level ground making an angle α when the foot of the ladder is moved through a
distance ‘a’ cms away from the wall, the end in contact with the wall slides
through ‘b’ cms along the wall and the angle made by the ladder with the level is
α + β 
now β . Show that a = b tan 

 2 
Solution : -
Let l be the length of the ladder OA = x and OB’ = y where AB is the initial position
of ladder and A’B’ is the final position of ladder
Given AA = a
BB’ = b
B
b
AB = A’B’ = l
B'
In triangle B OA
y
y+b
sin α =
⇒ y + b = l sin α → (1)
l
cos α =
x
⇒ x = l cos α → (2)
2
In triangle B’OA’
sin β =
y
⇒ y = l sin β → (3)
l
cos β =
a+x
⇒ a + x = l cos β → (4)
l
(1) - (3) we have
b = l ( sin α − sin β ) → ( 5)
(4) – (2) we have
a = l ( cos β − cos α ) → ( 6 )
(5) ÷ (6) we have
b l ( sin α − sin β )
=
a l ( cos β − cos α )
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O
x
A a
A'
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b
=
a 



α+β
α −β
l  2 cos
sin
2
2

α + β 
α − β
2 sin 
 sin 
2


 2









b
α + β 
= cot 

a
 2 
α + β 
a = b tan 

 2 
6.
The angles of depression of two mile stones along a road when viewed from an
tan α tan β
areoplane are α , β . Show that the aeroplane is flying at a height of
tan α + tan β
from the road level. Taking the two mile stones to be on opposite side?
Solution: -
Let A be the position of the aeroplane P, Q be two mile stones
Given PQ = 1 mile
Draw XAY parallel to PQ through the observes eye
Given ∠ × AP = ∠APQ = α
∴ ∠XAP = ∠APQ = α
∠YAQ = β
∠YAQ = ∠PQA = β
In triangle PMA
A
x
y
h
tan α =
⇒ PM = h cot α
PM
In triangle tan β =
h
⇒ QM = h cot β
QM
Given PQ = 1
PM + QM = 1 ⇒ h cot α + h cot β = l
h=
1
tan α tan β
=
cot α + cot β tan α + tan β
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h
P
M
Q
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7.
From the point on a bridge across a river the angles of depression of the banks
on opposite sides of the river are 300 and 450 respectively. If the bridge is at a
height of 4 meters from its banks find the width of the river?
Solution : -
Let A, B represent points on the bank on opposite sides of the river so that AB is the
width of the river. Let P be a point on the bridge at a height of 4 meters
i.e., PD = 4 mt
A
In triangle PDA tan 300 =
4
⇒ AD = 4 3
AD
300
h
4
In triangle PDB tan 45 =
⇒
= 1 ⇒ DB = 4
DB
DB
8.
(
450
A
0
AB = AD + DB = 4 3 + 4 = 4
450
300
B
D
)
3 +1
From the top of a light house 300 mts high the angle of dipressino made by two
boats situated in a horizontal line along the foot of the light house are 300 and 450
find the distance between the two boats (i) when they are on the same side (ii)
Q
when they are on either side of the light house?
Y
450
30
Solution : - (i)
300mt
Q
Refer to problem ‘2’
30
450
300
B
And try your self
0
450
P
A
300mt
Solution (ii)
Refer to problem 7
A
P
B
And try your self
9.
From the top of a hill 200 meters high the angle of depression of the top and
bottom of a pillar on the level ground are 300 and 600 respectively. Find the
height of the pillar.?
Solution : -
Let PQ be the full of height 200 mt
AB be the pillar of height h
Q
600 300
Draw QY parallel to horizontal
B
Angle of depression of B from Q is 30
300
h
200
200mt
C
0
h
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600
A
P
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Angle of depression of A from Q is 600
Draw BC parallel to Qy
Since AB = h QC = 200 – h
Let BC = AP = x
In ∆ le QCB tan 300 =
200 − h
x
-------(1)
In ∆ le QPA tan 600 =
100
x
--------(2)
(1) ÷ (2) we have
tan 300 200 − h
1 200 − h
=
⇒ =
0
tan 60
100
3
200
200 = 600 − 3h ⇒ h =
10.
400
3
A pillar of 10 meter height is mounted on a tower from a point on the level
ground, the angles of elevation of the top of the foot of the pillar are 750 and 450
respectively find the height of the tower.?
Solution : -
Let AB be the tower and BC be the pillar of height 10mt
Let P be the point of observation
Angles of elevation from P to B and C are respectively 450 and 750
In triangle BAP tan 450 =
h
-------(1)
x
In triangle CAP tan 750 =
h + 10
x
-------(2)
tan 450
h
=
0
tan 75
h + 10
C
10mt
B
1
h
=
2 + 3 h + 10
(
h
450
)
h + 10 = 2 + 3 h
A
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x
750
P
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(
)
10
=5
3 +1
10 = 1 + 3 h ⇒ h =
11.
(
)
3 −1
From the top of a pillar of a height 80 meters the angles of depression of the top
and foot of the another pillar are 300 and 450 respectively. Find the distance
between the two pillars and the height of the second pillar ?
Solution : -
Let PQ be the pillar of height 80 mts
AB be the second pillar of height h
y
300
450
Draw BX parallel to AP
B
300
QY parallel to AP
0
From Q are 30 and 45
h
450
A
In triangle QXB tan 300 =
80m
x
x
Angles of depression of B and A
0
80-h
x
P
80 − h
-------(1)
x
In triangle QPA
tan 450 =
80
h
----------(2)
(1) ÷ (2) when
tan 300 80 − h
=
tan 450
80
80 3 − h 3 = 80 ⇒ 80
12.
(
)
3 −1 / 3
From a point A on the level ground away from the foot of the tower the angle of
elevation of the top of the tower is 300 . From a point at a height of h meters
vertically above A the angle of elevation of the foot of the tower is 600 . Find the
height of tower in term of h ?
Solution: -
Let PQ be the tower of height ‘x’
Q
A be the first point of observation
600
y
B be the point of a height of ‘h’
h
600
From the point A
P
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B
300
A
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Angle of elevation of Q from A is 300
Angle of depression of P from B is 600
In triangle QPA tan 300 =
x
y
In triangle BAP tan 600 =
h
y
Let AP = 4
x
 y
tan 30
 ⇒1= x⇒x=h
=
0
h
tan 60
3 h
3
y
0
13.
The angle of elevation of the top of a tower from the foot of the building is twice
the angle of elevation from the top of the building. The height of the building is
50 meters and the height of the tower is 75 meters. Find the angle of elevation of
the top of the tower from the foot of the
B
building?
75m
Let AB be the tower of height 75 mt
PQ be the building of height 50m
Q
x
50m
Draw QX parallel to AP
A
Angles of elevation of B from Q AP
Angle of elevation of B from Q is α
Angle of elevation of B from P is 2α
In triangle BAP tan 2α =
75
--------------(1)
x
In triangle BXQ tan α =
25
-----------------(2)
x
(1) ÷ (2) we have
tan 2α
= 3 ⇒ tan 2α = 3 tan α
tan α
2 tan α
= 3 tan α ⇒ 2 = 3 − 3 tan 2 α
1 − tan α
3 tan 2 α = 1 ⇒ tan α =
1
3
α = 300 ⇒ 2 α = 600
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x
P
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14.
As observed from the top of a light house of height 75m from the sea level the
angles of depressions of two ships are 300 and 450 . It one ship is exactly between
the two ships?
Solution : -
X
H
0
300 45
Refer to Problem 2s
300
And try your self
15.
450
P
Q
L
The angles of elevation of the top of a tower from two points at a distance of 4m
and 9m from the base of the tower and in the same straight line with it are
complementary prove that the height of the tower is
6m
B
Solution: -
AB be the tower f height h
Angle of elevation of B from C and D
Are respectively are α , β
A
4m
9m
Given α + β = 900
α = 900 − β
(
tan α = tan 900 − β
)
tan α = cot β ⇒ tan α =
1
⇒ tan α tan β = 1 ----------------(1)
tan β
In triangle BAC tan α =
h
4
In triangle BAQ tan β =
h
4
tan α tan β = 1 ⇒
h2
=1⇒ h = 6
36
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C
D
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16.
A and B are two points on the line segment joining the feet of two equal pillars.
AB = 30 meters. From A the angles of elevation of top of the pillars are
300 and 600 . From B they are 600 and 300 respectively in the same order find the
height of the pillars and the distance between them?
Solution : -
tan 600 =
S
B
In triangle QPB
h
⇒h=x 3
x
h
h
In ∆le QPA
tan 300 =
P
h
x + 30
x + 30 = h 3
But h = x 3
∴ x + 30 = 3 x ⇒ x = 15
h = 15 3
In triangle SRY
tan 600 =
3=
300
300
600
h
y
15 3
⇒ y = 15
y
Distance between the poles = x + 30 + y = 60m
Height of pillar = 15 3
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x
B
30 A
600
y
R
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16.
Over a tower AB of height h mt there is a flat staff BC. AB and BC are making
equal angles at a point distance d mt from the foot A of the tower. Show that the
(d + h )
h
2
height of the flag staff is
2
d 2 − h2
Solution : -
AB is the tower of height h
BC is the flag staff
C
Let the height of the flat staff be ‘x’
X
P be any point let Ap = d
B
In triangle BAD
h
tan α =
h
d
A
In triangle CAD tan 2α =
h+x
d
2 tan α
h+x
=
2
1−tan α
d
h
2
h+x
d = h + x ⇒ 2h × d
=
2
2
2
h
d
d d −h
d
1− 2
d
2
2hd 2
2hd 2 − hd 2 + h3
−
h
=
x
⇒
x
=
d 2 − h2
d 2 − h2
x=
(
h d 2 + h2
hd 2 + h3
⇒
x
=
d 2 − h2
d 2 − h2
)
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d
P
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17.
ABCD is a monument built vertically on the ground from a point P away from
the moment on the level ground the angles of elevation of AB, AC, AD are
α , β and γ respectively. If AB= a, AC = b , AD = c AP = x and α + β + γ = 1800
then show that ( a + b + c ) x 2 = abc
Solution : -
ABCD is movement
A is on the ground
Given AB = a AC = b
AD = C
P be a point of observation
Given AP = x
α , β , γ are the angles of elevation of B, C, D from
P
In triangle BAP tan α =
a
x
D
C
C
B
In triangle CAP tan β =
In triangle DAP tan γ =
b
x
a
A
c
x
Given that α + β + γ = 1800 ⇒ α + β = 1800 − γ
tan α + tan β
= − tan γ ⇒ tan α + tan β + tan γ = tan α tan β tan γ
1 − tan α tan β
a b c abc
+ + = 3 ⇒ ( a + b + c ) x 2 = abc
x x x
x
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a
P
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18.
An aeroplane is moving one kilometers high from west to east horizontally. From
a point on the ground the angle of elevation of the aeroplane is 600 and after 60
seconds the angles of elevation of the aeroplane is observed as 300 . Find the
speed of the aeroplane in km/hr ?
Solution : -
Let C, E be the positions of aeroplane
P be the point of observation
C
Angle of elevation of C from P is 600
E
1km
600
Angle of elevation of E from P is 300
300
P
Draw CD and EF perpendicular to horizontal line PX
In triangle CDP tan 600 =
⇒ CD =
1
PD
1
3
In triangle EFP tan 300 =
1
PF
PF = 3
Distance travelled by aeropalne in 10 seconds = CE = DF
CE = DF = PD − CD = 3 −
1
2
=
3
3
 2 
2
720
dis tan ce  3 
=
=
× 360 =
= 240 3 km / h
∴ Speed =
time
 10 
3
3


 360 
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D
F
x
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19.
A pillar is leaning towards east and α , β are the angles of elevation of the top of
the pillar from points due west of the pillar at distance a and b respectively.
Show that the angle between the pillar and the horizontal is


b−a
tan −1 

 b cot α − a cot β 
Solution : B
Let AB be the pole inclined towards east
P, Q are the points of observations
h
At a distance of a and b from A respectively
Angles of elevation from P, Q to B are α , β
Draw BM perpendicular to AQ let BM = h AM = x
In triangle AMB tan θ =
h
⇒ x = h cot θ → (1)
x
In triangle PMB tan α =
h
⇒ a + x = h cot α → (2)
a+x
In triangle QMB tan β =
h
⇒ b + x = h cot β → (3)
b+x
(2) ------(1) we have a + x − x = h {cot α − cot θ } → (4)
(3) -------(1) we have b + x − x = h {cot β − cot θ } → (5)
(4) ÷ (5) we have
a cot α − cot θ
=
b cot β − cot θ
a cot β −a cot θ = b cot α − b cot θ
( b − a ) cot θ = b cot α − a cot β
cot θ =
b cot α − a cot β
b−a
⇒ tan θ =
b−a
b cot α − a cot β

b−a

 b cot α −a cot β 

θ = tan −1 
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Q
P
P
a A
x
M
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20.
From a point B on the level ground away from the foot of the hill AD, the top of
the hill makes an angles of elevation α from the point B, the point C is reached
by moving a distance d along a slant/slope which makes an angle ϑ with the
horizontal. If B is the angle of elevation of the top of the hill from C, find the
height of the hill ?
Solution : -
AD be the hill b is the point of observation
A
BC is the inclined plane
Draw CE parallel to BD
Angle of elevation of A from B is α
E
Angle of elevation of A from C is β
Angle of inclination of BC with BD is ϑ
In ∆ le ADB
∠BAD 900 − α
In ∆le CEA
∠CAE = 900 − β
In ∆le BCA
∠ABC = α − ϑ
∠BAC = ( 90 − α ) − ( 90 − β ) = β − α
∠BAC = 1800 − {∠BAC + ∠ABC}
= 1800 − { α − ϑ + β − α }
= 1800 − ( β − ϑ )
Using since rule
BC
AB
=
0
sin ( β − α ) sin180 − ( β − ϑ )
d
sin ( β − α )
=
d sin ( β − ϑ )
AB
⇒ AB =
sin ( β − ϑ )
sin ( β − α )
In triangle BDA
sin α =
AD
⇒ AD = AB sin α
AB
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C
B
D
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∴ AD =
21.
d sin α sin ( β − ϑ )
sin ( β − α )
From a point on the slope of a hill, the angle of elevation of the top of the hill is
450 . After walking 200 mt from that point on a slope of 150 angle with
horizontal the same point on the top of the hill makes an angle of elevation 600
show that the height of the hill is 100
(
6+ 6
)
A
Solution : 600
Do the problem again by taking α = 450 , ϑ = 150 , β = 600
200mt
E
C
and d = 200 m in the above problem
0
450
15
B
22.
D
A church tower AB standing on a level plane is surrounded by a spire BC of the
1
same height as the tower D is a point in AB such that AD = AD. At a point on
3
the plane 100 mt from the foot of the tower the angles subtended by AD and BC
are equal. Find the height of tower ?
Solution :-
In triangle OQP
C
h
tan θ =
300
h
B
D
In triangle BAP
tan (θ + Q ) =
h
100
P
In triangle CAP
tan ( 2θ + Q ) =
2h
h
=
100 50
tan ( 2θ + Q ) =
h
h
⇒ tan {θ + θ + Q} =
50
50
tan θ + tan (θ + Q )
1 − tan θ tan (θ + Q )
=
h
50
h
h
+
300 100 = h ⇒ 400 h × 30000 = h
h2
50
30, 000 30000 − h 2 50
1−
30, 000
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100mt
A
h
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20, 000 = 30, 000 − h 2 ⇒ h 2 = 10, 000 ⇒ h = 100mt
23.
A man observes a tower AB of height h from a point P on the ground the moves a
distance ‘d’ towards the foot of the tower and finds that the angle of elevation is
3d
in the same direction and the angle of
doubled. He further moves a distance
4
elevation is three times that at P. Prove that 36h 2 = 35d 2
Solution : In triangle PQB
∠PQB = 180 − 2α
B
∠PBQ = 180 − {α + 180 − 2α }
=α
d
180-R
180-k
∴∠BPQ = ∠PBQ = α
∴ BQ = BQ = d {sides opposite to equal angles are equal}
In triangle BQR ∠QRB = 180 − 3α
∠QBR = 1800 − {2α + 180 − 3α } = α
Using since rule in this triangle BQR
BQ
QR
d
sin 3α
=
⇒
=
3d
sin (180 − 3α ) sin α
sin α
4
4
4
= 3 − 4 sin 2 α ⇒ 4sin 2 α = 3 −
3
3
sin 2 α =
5
12
cos 2 α =
7
12
In ∆ le QAB
sin 2α =
AB
h
⇒ sin 2α =
BQ
d
h = 2d sin α cos α ⇒ h 2 = 4d 2sin 2 α cos 2 α
h2 = 4 d 2 ×
5
7
× ⇒ 36h 2 = 35d 2
12 3 12
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P
d
Q
R
A
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24.
An observer finds that the angle of elevation of the tower is θ , an advancing ‘a’
meters towards the tower, the elevation is 450 and an advancing b meter near
the elevation is 900 − θ . Find the height of the
B
tower.?
Solution : h
AB be the tower of height h
900 - Q
450
P the first point of observation
P
a
Q
b
From which the angle of elevation is 'θ '
R
x
A
Q be the second point such that PQ = a and angle of elevation from Q is 450
R be the third pant such that QR = b and angle of elevation is 900 − θ let AR = x
In triangle BAR tan ( 900 − θ ) =
In triangle QAB tan 450 =
In triangle PAB tan θ =
h
h
⇒ cot θ = ⇒ x = h tan θ -------(1)
2
x
h
⇒ h = b + x ---------(2)
b+x
h
⇒ a + b + x = h cot θ -------(3)
a+b+ x
(2) – (1) we have
(3) – (2) we have
b = h (1 − tan θ ) ------(4)
a = h {cot θ − 1} -------(5)
( 4)
(5)
we have
b h (1− tan θ )
=
a h ( cot θ − 1)
h {1 − tan θ }
b
b
⇒ = tan θ
=
a
a
 1

− 1
h
 tan θ

ab
 b
Sub tan θ is (3) we have b = h 1 −  ⇒ h =
a−b
 a
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