1. HEIGHTS AND DISTANCE
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2. Content Introduction Trigonometry Trigonometry identities
Values of T ratio
Angle of elevation and depression
Problems
Two of the sides given
One angle and one side given
Two heights and one angle
Two angles and one height
Two angles and two heights
Calculating time and distance
Solved problems
Practice problems
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3. 1) Introduction
This chapter deals with finding the heights, distances and angles using trigonometric values
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4. 1.i) Trigonometry: .
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5. 1.ii) Trigonometry Identities:
sin2 θ + cos2 θ = 1.
tan2 θ = sec2 θ.
cot2 θ = cosec2 θ
1.iii) Values of T-ratios: 0°
30°
45°
60°
90°
sin θ
1/2
1/√2
√3/2 1
cos θ
tan θ
1/√3 √3
Not defined
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6. 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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7. 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 = 6√3 m
Base = 18 m
Angle = ?
Solution:
tan θ = perpendicular/base
= 6√3 / 18
= √3/3
= 1/√3
θ = 30º
6√3 θ 18
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8. Example 1. A pole of height h = 50 ft has a shadow of length l = 50
Example 1. A pole of height h = 50 ft has a shadow of length l = 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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9. 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 = perpendicular/base
tan 30º = 100/base
1/ √3 = 100/base
Base = 100 √3
= 173 m
100
30º
base
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10. Example 2. A man is walking along a straight road
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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11. 2.iii) Two heights and one angle
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 = ?
Solution:
tan θ = perpendicular / base
tan 30º = perpendicular / 203√3
1/ √ = perpendicular / 203√3
Perpendicular = 200√3 / √3
Perpendicular = 200 m
perpendicular
30º
1.6
203√3
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12. Height of the tower = perpendicular + 1.6 = 200 + 1.6 = 201.6m
30º
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13. 2.iv) Two angles and one height
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?
Given:
Perpendicular 1 = 100√3 m
Angle = 60°
Angle = 45°
Difference b/w the heights =Perpendicular 1 - Perpendicular 2 = ?
100√3
Perpendicular 2
60º
45º
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14. Angle 1 = Perpendicular 1/ Base tan 60° = 100√3 / base
Solution:
Angle 1 = Perpendicular 1/ Base
tan 60° = 100√3 / base
√ = 100√3 /base
Base = 100√3 / √3
Base = 100
Angle = Perpendicular 2/ Base
tan 45° = Perpendicular 2 / 100
= Perpendicular 2 / 100
Perpendicular 2 = 100
Difference b/w the heights =Perpendicular 1 - Perpendicular 2
= 100√3 – 100
= 173 – 100
= 73 m
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15. 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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16. 2.v) Two angles and two heights
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
the top is 30o. What is the height
of the second tower?
Given:
Angle = 90o - 30o = 60o
Angle = 90o - 60o = 30o
Perpendicular = 20m
Height of the second tower = Base 2 – Base 1
30o
Base 1
60o
Base 2 20
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17. Angle 2 = perpendicular/base 2 tan 30o = 20 / base 2
Solution:
Angle 2 = perpendicular/base 2
tan 30o = 20 / base 2
1/√ = 20 / base 2
Base 2 = 20√3
Angle 1 = perpendicular / base 1
tan 60o = 20 / base 1
√ = 20 / base 1
Base 1 = 20 / √3
Height of the second tower = 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
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18. 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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19. 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?
Given:
Angle = 30o
Angle = 60o
Perpendicular = 2000 m
Time = 60 sec
Speed = distance / time = (base 1 – base 2) / time
2000
60o
30o
Base 2
Base 1
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20. Angle 1 = perpendicular / base 1 tan 30o = 2000 / base 1
Solution:
Angle 1 = perpendicular / base 1
tan 30o = 2000 / base 1
1/√ = 2000 / base 1
Base 1 = 2000√3
Angle 2 = perpendicular / base 2
Angle 60o = 2000 / base 2
√ = 2000 / base 2
Base = 2000 / √3
Speed = distance / time
= (base 1 – base 2) / time
= (2000√ / √3) / 60
= 200√3 / 9 m/s
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21. 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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22. 3) Solved examples
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23. Height of the tree = perpendicular + hypotenuse
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.
A. 30 m B. 60 m C. 75 m D. 90 m
Given:
Angle = 30º
Base = 30√3 m
Height of the tree = perpendicular + hypotenuse
hypotenuse
perpendicular
30º
30√3
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24. tan θ = perpendicular / base tan 30º = perpendicular / 30√3
Solution:
tan θ = perpendicular / base
tan 30º = perpendicular / 30√3
1 / √ = perpendicular / 30√3
Perpendicular = 30
Cos 30º = hypotenuse / base
√3 / = hypotenuse / 30√3
Hypotenuse = 30(3)/2
Hypotenuse = 45
Height of the tree = perpendicular + hypotenuse =
= 75 m
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25. Height of the tree = Height of the wall + Base 2 = ? Solution:
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 ?
A. 22 m B. 44 m C. 33 m D m
Given:
Angle 1 = 30°
Angle 2 = 90° - 60° = 30°
Height of the tree = Height of the wall + Base 2 = ?
Solution:
tan θ1 = perpendicular / base 1
tan 30º = 11 / base 1
1 / √3 = 11 / base 1
Base 1 = 11 √3
60º
30º
Base 2 11
30º
Base 1
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26. tan θ2 = perpendicular / base 2 tan 30º = 11√3 / base 2
Height of the tree = Height of the wall + Base 2 =
= 44 m
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27. tan θ 2 = perpendicular / base 2 tan 45o = perpendicular / base 2
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 = ?
Solution:
tan θ = perpendicular / base 2
tan 45o = perpendicular / base 2
= perpendicular / base 2
perpendicular = base 2
45o
Base 1
60o
Base 2 24
perpendicular
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28. tan θ 1 = perpendicular / base 1 tan 60o = perpendicular / base 2 – 24
w.k.t perpendicular = base 2
√ = perpendicular / perpendicular – 24
√3 perpendicular – 24√ = perpendicular
√3 perpendicular – perpendicular = 24√3
perpendicular (√3 – 1) = 24√3
perpendicular = 24√3 / (√3 – 1) m
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29. Hypotenuse 1 = hypotenuse 2 = 40 Perpendicular 1 = perpendicular 2
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)
Given:
Angle = angle = 60o
Hypotenuse = hypotenuse = 40
Perpendicular 1 = perpendicular 2
Distance b/w the two ends = perpendicular 1 + perpendicular 2 = ?
Solution:
sin θ = Perpendicular 1 / hypotenuse 1
sin 60o = perpendicular 1 / 40
√3 / = perpendicular 1 / 40
perpendicular 1 = 20√3 40
60º
60º 40
perpendicular 1
perpendicular 2
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30. Distance b/w the two ends = perpendicular 1 + perpendicular 2
= 2 (20√3)
= 40√3
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31. Perpendicular 2 = 2 (perpendicular 1) ------------ 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 = Base 2 = 100 m
Angle = θ
Angle = 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 /  2
Perpendicular 2
Perpendicular 1 θ
90o – θ
100
100
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32. 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 = 100 / √2
= 100√2 / 2
= 50√2
Perpendicular 2 = 2 (50√2)
= 100√2
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33. 4) Practice problems
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34. 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:
30º
45º
60º
90º
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35. 2. From a tower of 80 m high, the angle of depression of a bus is 30°
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?
80 m
80√3 m
80√3 / 3 m
240 m
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36. 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.
80√3 m
80√3 / 3 m
80 (√3 – 1) m
80√3 – 80 / √3 m
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37. 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?
5 metres
8 metres
10 metres
12 metres
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