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HEIGHTS AND DISTANCE
© Department of Analytical Skills
Content
1) Introduction
i. Trigonometry
ii. Trigonometry identities
iii. Values of T ratio
iv. Angle of elevation and depression
2) Problems
i. Two of the sides given
ii. One angle and one side given
iii. Two heights and one angle
iv. Two angles and one height
v. Two angles and two heights
vi. Calculating time and distance
3) Solved problems
4) Practice problems
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1) Introduction
This chapter deals with finding the heights, distances and
angles using trigonometric values
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1.i) Trigonometry:
.
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1.ii) Trigonometry Identities:
sin2 θ + cos2 θ = 1.
1
+ tan2 θ = sec2 θ.
1
+ cot2 θ = cosec2 θ
1.iii) Values of T-ratios:
0°
30°
45°
60°
90°
sin θ
0
1/2
1/√2
√3/2
1
cos θ
1
√3/2
1/√2
1/2
0
tan θ
0
1/√3
1
√3
Not defined
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1.iv) Angle of elevation and depression
x – angle of elevation
y – angle of depression
Note: The base line for angle of elevation and angle of
depression will always be the horizontal line.
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2) Problems
2.i) Two of the sides given
Example: Find the angle of elevation of the sun when the
shadow of a pole of 18 m height is 6√3 m long?
Given:
Perpendicular = 18m
Base
= 6√3 m
Angle
=?
Solution:
tan θ = perpendicular/base
= 18/ 6√3
= 3/√3
= √3
θ
= 60º
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18
θ
6√3
Example 1. A pole of height h = 50 ft has a shadow of
length l = 50.00 ft at a particular instant of time. Find
the angle of elevation (in degrees) of the sun at this
point of time.
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2.ii) One angle and one of the sides given
Example: From a point P on a level ground, the angle of
elevation of the top tower is 30º. If the tower is 100 m high,
then what is the distance of point P from the foot of the tower.
Given:
Perpendicular = 100 m
Angle
= 30º
Base
=?
Solution:
tan θ
tan 30º
1/ √3
Base
= perpendicular/base
= 100/base
= 100/base
= 100 √3
= 173 m
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100
30º
base
Example 2. A man is walking along a straight road. He
notices the top of a tower subtending an angle A = 60o
with the ground at the point where he is standing. If
the height of the tower is h = 25 m, then what is the
distance (in meters) of the man from the tower?
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Example: An observer 1.6 m tall is 203√3 m away from a
tower. The angle of elevation from his eye to the top of the
tower is 30º. Find the height of the tower.
Given:
Base = 203√3 m
Angle = 30º
Height = perpendicular + 1.6 = ?
30º
1.6
Solution:
tan θ
= perpendicular / base
tan 30º
= perpendicular / 203√3
1/ √3
= perpendicular / 203√3
Perpendicular = 203√3 / √3
Perpendicular = 203 m
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203√3
perpendicular
2.iii) Two heights and one angle
Height of the tower = perpendicular + 1.6
= 203 + 1.6
= 204.6m
30º
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2.iv) Two angles and one height
Given:
Perpendicular 1 = 100√3 m
Angle 1
= 60°
Angle 2
= 45°
45º
Perpendicular 2
Example: An aeroplane when 100√3 m high passes vertically
above another aeroplane at an instant when their angles of
elevation at same observing point are 60° and 45° respectively.
Approximately, how many meters higher is the one than the
other?
100√3
Difference b/w the heights =Perpendicular 1 - Perpendicular 2 = ?
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Solution:
Angle 1 = Perpendicular 1/ Base
tan 60° = 100√3 / base
√3
= 100√3 /base
Base
= 100√3 / √3
Base
= 100
Angle 2
= Perpendicular 2/ Base
tan 45°
= Perpendicular 2 / 100
1
= Perpendicular 2 / 100
Perpendicular 2 = 100
Difference b/w the heights =Perpendicular 1 - Perpendicular 2
= 100√3 – 100
= 173 – 100
= 73 m
© Department of Analytical skills
Example 3. Two ships are sailing in the sea on the two
sides of a lighthouse. The angle of elevation of the top
of the lighthouse is observed from the ships are 30º and
45º respectively. If the lighthouse is 100 m high, then
find the distance between the two ships is.
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Base 1
Example: Two towers face each other separated by a
distance d = 20 m. As seen from the top of the first tower,
the angle of depression of the second
tower's base is 60o and that of
30o
the top is 30o. What is the height
60o
of the second tower?
Given:
Angle 1
= 90o - 30o = 60o
Angle 2
= 90o - 60o = 30o
Perpendicular = 20m
Height of the second tower = Base 2 – Base 1
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20
Base 2
2.v) Two angles and two heights
Solution:
Angle 2
tan 30o
1/√3
Base 2
= perpendicular/base 2
= 20 / base 2
= 20 / base 2
= 20√3
Angle 1 = perpendicular / base 1
tan 60o = 20 / base 1
√3
= 20 / base 1
Base 1 = 20 / √3
Height of the second tower =
=
=
=
=
=
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Base 2 – Base 1
20√3 – 20 / √3
20√3 - 20√3 / 3
20√3 ( 1 – 1/3)
20√3 (2/3)
40√3 / 3
Example 4. To a man standing outside his house, the
angles of elevation of the top and bottom of a window
are 60° and 45° respectively. If the height of the man is
180 cm and he is 5 m away from the wall, what is the
length of the window?
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2.vi) Calculating time and speed
Example: You are stationed at a radar base and you observe
an unidentified plane at an altitude h = 2000 m flying
towards your radar base at an angle of elevation = 30o. After
exactly one minute, your radar sweep reveals that the plane
is now at an angle of elevation = 60o maintaining the same
altitude. What is the speed (in m/s) of the plane?
60o
30o
Base 2
2000
Given:
Angle 1
= 30o
Angle 2
= 60o
Perpendicular = 2000 m
Base 1
Time
= 60 sec
Speed
= distance / time = (base 1 – base 2) / time
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Solution:
Angle 1
tan 30o
1/√3
Base 1
= perpendicular / base 1
= 2000 / base 1
= 2000 / base 1
= 2000√3
Angle 2 = perpendicular / base 2
Angle 60o = 2000 / base 2
√3
= 2000 / base 2
Base 2
= 2000 / √3
Speed =
=
=
=
distance / time
(base 1 – base 2) / time
(2000√3 - 2000 / √3) / 60
200√3 / 9 m/s
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Example 5. A balloon leaves the earth at a point A and
rises vertically at uniform speed. At the end of 2
minutes, John finds the angular elevation of the
balloon as 60°. If the point at which John is standing is
150 m away from point A, then what is the speed of the
balloon?
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3) Solved examples
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A. 30 m
B. 60 m
C. 75 m
perpendicular
Q 1. A vertical tree is broken by the wind. The top of the tree
touches the ground and makes an angle 30º with it. If the top of
the tree touches the ground 30√3 m away from its foot, then find
the actual height of the tree.
D. 90 m
Given:
30º
30√3
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Solution:
tan θ
= perpendicular / base
tan 30º
= perpendicular / 30√3
1 / √3
= perpendicular / 30√3
Perpendicular = 30
Cos 30º
= hypotenuse / base
√3 / 2
= hypotenuse / 30√3
Hypotenuse = 30(3)/2
Hypotenuse = 45
Height of the tree = perpendicular + hypotenuse
= 30 + 45
= 75 m
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30º
A. 22 m
B. 44 m
Given:
Angle 1 = 30°
Angle 2 = 90° - 60° = 30°
C. 33 m
D. 16.5 m
11
Height of the tree = Height of the wall + Base 2 = ?
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30º
Base 1
Base 2
Q 2. The angles of depression and elevation of the top of a wall 11 m
high from top and bottom of a tree are 60° and 30° respectively. What
is the height of the tree ?
60º
Solution:
tan θ1 = perpendicular / base 1
tan 30º = 11 / base 1
1 / √3 = 11 / base 1
Base 1 = 11 √3
tan θ2 = perpendicular / base 2
tan 30º = 11√3 / base 2
1 / √3 = 11√3 / base 2
Base 2 = 33
Height of the tree = Height of the wall + Base 2
= 11 + 33
= 44 m
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60o
24
perpendicular
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Base 2
45o
Base 1
Q 3. A ship of height h = 24 m is sighted from a lighthouse. From
the top of the lighthouse, the angle of depression to the top of the
mast and the base of the ship equal 30o and 45o respectively. How
far is the ship from the lighthouse (in meters)?
Given:
Angle 1 = 90o - 30o = 60o
Angle 2 = 90o - 45o = 45o
Perpendicular = ?
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Solution:
tan θ 2
= perpendicular / base 2
tan 45o
= perpendicular / base 2
1
= perpendicular / base 2
perpendicular = base 2
tan θ 1 = perpendicular / base 1
tan 60o = perpendicular / base 2 – 24
√3
= perpendicular / base 2 – 24
w.k.t perpendicular = base 2
√3
= perpendicular / perpendicular – 24
√3 perpendicular – 24√3
= perpendicular
√3 perpendicular – perpendicular = 24√3
perpendicular (√3 – 1)
= 24√3
perpendicular
= 24√3 / (√3 – 1) m
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Q 4. A simple pendulum of length 40 cm subtends 60o at the vertex
in one full oscillation. What will be the shortest distance between
the initial position and the final position of the bob? (between the
extreme ends)
40
60º 60º
40
perpendicular 1 perpendicular 2
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Given:
Angle 1
= angle 2
= 60o
Hypotenuse 1 = hypotenuse 2
= 40
Perpendicular 1 = perpendicular 2
Distance b/w the two ends = perpendicular 1 + perpendicular 2 = ?
Solution:
sin θ 1
= Perpendicular 1 / hypotenuse 1
sin 60o
= perpendicular 1 / 40
√3 / 2
= perpendicular 1 / 40
perpendicular 1 = 20√3
Distance b/w the two ends = perpendicular 1 + perpendicular 2
= 2 (perpendicular 1)
= 2 (20√3)
= 40√3
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© Department of Analytical skills
θ
100
90o – θ
100
Perpendicular 2
Perpendicular 1
Q 5. Two vertical poles are 200 m apart and the height of one is
double that of the other. From the middle point of the line joining
their feet, an observer finds the angular elevations of their tops to
be complementary. Find the height of the tallest pole.
Given:
Base 1
= Base 2 = 100 m
Angle 1
=θ
Angle 2
= 90o – θ
Perpendicular 2 = 2 (perpendicular 1) ------------ 1
Height of the tallest pole = perpendicular 2
Solution:
tan θ 1 = perpendicular 1 / base 1
tan θ = perpendicular 1 / 100
---------------- 2
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tan θ 2
= perpendicular 2 / base 2
tan (90o – θ) = perpendicular 2 / 100
cot θ
= perpendicular 2 / 100
1 / tan θ
= perpendicular 2 / 100
Substitute 1 and 2 in the previous equation
100 / perpendicular 1 = 2 perpendicular 1 / 100
perpendicular 1 ^2 = 100^2 / 2
perpendicular 1
= 100 / √2
= 100√2 / 2
= 50√2
Perpendicular 2 = 2 (50√2)
= 100√2
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4) Practice problems
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1. The angle of elevation of the sun, when the length
of the shadow of a tree 3 times the height of the tree,
is:
A.
B.
C.
D.
30º
45º
60º
90º
© Department of Analytical Skills
2. From a tower of 80 m high, the angle of depression
of a bus is 30°. How far is the bus from the tower?
A.
B.
C.
D.
80 m
80√3 m
80√3 / 3 m
240 m
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3. Two men on the same side of a tall building notice
the angle of elevation to the top of the building to be
30o and 60o respectively. If the height of the building is
known to be h =80 m, find the distance (in meters)
between the two men.
A.
B.
C.
D.
80√3 m
80√3 / 3 m
80 (√3 – 1) m
80√3 – 80 / √3 m
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4. The top of a 15 metre high tower makes an angle of
elevation of 60° with the bottom of an electronic pole
and angle of elevation of 30° with the top of the pole.
What is the height of the electric pole?
A.
B.
C.
D.
5 metres
8 metres
10 metres
12 metres
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