In the previous section we saw an application of the tangent ratio. In this section, we will see another application.
Many often, to see objects at a higher (elevated) level, we have to raise our heads. Then the line of sight will make an angle with the horizontal. This angle is called the angle of elevation. It is shown in the fig.30.40 below:
• In the above fig., the observer is looking at the top tip point of a conical structure.
• The line of sight is shown in yellow color.
• The horizontal level is shown as a white dashed line.
• The concept of 'line of sight' will be more clear if we consider the view through a telescope. It is shown in the fig.30.41 below:
• In this case, the 'line of sight' can be taken as the axis of the telescope.
• Precision instruments are available for making such observations.
♦ One such instrument is the Theodolite.
♦ It is used for measuring heights and distances using angles.
♦ It's main component is a telescope.
♦ The angles can also be accurately measured.
To see objects at a lower (depressed) level, we have to lower our heads. Then the line of site will make an angle with the horizontal. This angle is called the angle of depression. It is shown in the fig.30.42 below:
• In the above fig., the observer is standing on top of a platform.
• The line of sight is shown in yellow color.
• The horizontal level is shown as a white dashed line.
• He has to look down to see the 'edge of the base' of the conical structure.
♦ When the telescope is looking up, the right side of the bubble level is raised. So the bubble will move to the right side. The angle will be an angle of elevation
♦ When the telescope is looking down, the left side of the bubble level is raised. So the bubble will move to the left side. The angle will be an angle of depression
• A is the position of the eye of the observer.
• B is the position of the object being viewed.
2. Drop a perpendicular from B on to the ground.
♦ For our present discussion the ground is assumed to be horizontal. That is., the ground is assumed to be parallel to the 'horizontal level' shown by the white dashed line
♦ While doing engineering problems, we may have to deal with situations where the ground is not exactly horizontal.
• The perpendicular from B meets the ground at E.
• The perpendicular from B meets the horizontal level at C
• So ∠ACB = ∠DEB = 90o
3. Let the angle of elevation be θ degrees
• From ⊿ABC we get: tan = BC⁄AC
4. In the above step, we have a distance AC. But this distance is same as DE. This DE can be measured on the ground using a tape.
• First measure DF, which is the distance from the observer to the point F, which is the edge of the base of the cone.
• Then add it to FE. This FE is radius of the base of the cone. It is a constant which does not depend on the position of the observer. So it is denoted as k
5. Thus we have: AC = DE = (DF + FE) = (DF +k)
• So from (3), we will get BC.
• To this BC, we must add CE.
• But CE is same as AD, the height of the observer. So we get:
■ Height of the conical structure = BC + AD
• He has to look up to view the top point B of the conical structure. So there is an angle of elevation (θ1).
• He has to look down to view the edge of the base of the conical structure. So there is an angle of depression (θ2).
2. From ⊿AEB we get: tan θ1 = BE⁄AE
• AE = GD = (GC + CD) = (GC + k) Where k is the radius of the base of the cone.
• So we can calculate BE
3. From ⊿AGC we get: tan θ2 = AG⁄CG
• Note that in ⊿ACG, ∠GCA = θ2 (∵ ∠GCA and ∠EAC are alternate interior angles, and hence equal)
4. CG can be easily measured at the field. So from (3) we can calculate AG
■ So we get total height of the conical structure as:
Height = (BE + ED) = (BE + AG)
One more application:
1. From ⊿AGC we get: tan θ2 = AG⁄CG
• Note that in ⊿ACG, ∠GCA = θ2 (∵ ∠GCA and ∠EAC are alternate interior angles, and hence equal)
2. If the height AG (total height of platform and observer) is known, we can calculate CG fro (1).
That is., the distance between the platform and the conical structure can be determined.
This application is very useful if there is any obstruction such as a river or a lake between the platform and the structure.
Solution:
Fig.30.45(a) below shows the rough sketch. AE represents the man. BF represents the tree. AB is the line of sight. AC is the horizontal level. EF is the distance between the man and the foot of the tree.
• Note that, in this problem, the radius of the tree trunk is not given. So we can assume that, the distance 10 m is measured from the position of the man up to the center of the tree trunk. So BF is the center line of the tree trunk.
• The ⊿ABC for calculations are shown in fig.(b). In this triangle we get:
1. tan 40 = BC⁄AC ⟹ BC = AC × tan 40
2. But AC = EF, which is given as 10 m. So we get:
BC = 10 × tan 40 = 10 × 0.8391 = 8.391 m
3. Total height = BC + CF = 8.391+1.7 = 10.091 m
Solved example 30.27
A man 1.8 metres tall stands on top of a light house 25 metres high and sees a ship at see at a depression of 35o. How far is it from the foot of the light house?
Solution:
• In fig.30.46 below, AD is the center line of the light house. AB is the line of sight. AC is the horizontal level.
• A perpendicular BC is drawn upwards from B. Thus we get a right triangle ABC.
• Note that in this problem, the radius of the light house is not given. So the man is assumed to be positioned above the center line AD.
• Thus, when we determine AC, we will get the distance of the ship from the center line of the light house.
1. tan 35 = BC⁄AC ⟹ AC = BC⁄tan 35
2. But BC = AD = 25 + 1.8 = 26.8 m
So AC = BC⁄tan 35 = 26.8⁄0.7002 = 38.275 m
3. Distance of the ship = BD = AC = 38.275 m
Solved example 30.28
A boy standing at the edge of a canal sees the top of a tree at an elevation of 70o. The tree is also at the edge of the canal. Stepping 10 m back, he sees it at an elevation of 25o. The boy is 1.5 m tall. How wide is the canal and how tall is the tree?
Solution:
• Consider fig.30.47 below. CF represents the initial position of the boy. B is the top of the tree.
♦ So CB is the initial line of sight
• AE is the final position of the boy.
♦ So AB is the final line of sight.
• Also both the distances AC and EF will be equal to 10 m.
1. Consider ⊿ABD
tan 25 = BD⁄AD ⟹ BD = AD × tan 25 ⟹ BD = (AC+CD) × tan 25 ⟹ BD = (10+CD) × tan 25
2. Consider ⊿CBD
tan 70 = BD⁄CD ⟹ BD = CD × tan 70
3. Equating the values of BD from (1) and (2) we get:
(10+CD) × tan 25 = CD × tan 70
⟹ CD × (tan 70 - tan 25) = 10 × tan 25 ⟹ CD × (2.7475 - 0.4663) = 10 × 0.4663
⟹ CD × 2.2812 = 4.663 ⟹ CD = 2.044
4. So width of the canal = FG = CD = 2.044 m
5. Substituting this value of CD in (2) we get:
BD = 2.044 × tan 70 = 2.044 × 2.7475 = 5.616 m
6. So height of the tree = BD + DG = 5.616+1.5 = 7.116 m
Many often, to see objects at a higher (elevated) level, we have to raise our heads. Then the line of sight will make an angle with the horizontal. This angle is called the angle of elevation. It is shown in the fig.30.40 below:
 |
| Fig.30.40 |
• The line of sight is shown in yellow color.
• The horizontal level is shown as a white dashed line.
• The concept of 'line of sight' will be more clear if we consider the view through a telescope. It is shown in the fig.30.41 below:
 |
| Fig.30.41 |
• Precision instruments are available for making such observations.
♦ One such instrument is the Theodolite.
♦ It is used for measuring heights and distances using angles.
♦ It's main component is a telescope.
♦ The angles can also be accurately measured.
To see objects at a lower (depressed) level, we have to lower our heads. Then the line of site will make an angle with the horizontal. This angle is called the angle of depression. It is shown in the fig.30.42 below:
 |
| Fig.30.42 |
• The line of sight is shown in yellow color.
• The horizontal level is shown as a white dashed line.
• He has to look down to see the 'edge of the base' of the conical structure.
• In a theodolite, a bubble level is attached to know whether the telescope is 'looking up' or 'looking down', or whether it is horizontal. A picture of a bubble level can be seen here.
♦ When the telescope is horizontal, the bubble will be at the center. Then the axis of the telescope will coincide with the 'horizontal level'.♦ When the telescope is looking up, the right side of the bubble level is raised. So the bubble will move to the right side. The angle will be an angle of elevation
♦ When the telescope is looking down, the left side of the bubble level is raised. So the bubble will move to the left side. The angle will be an angle of depression
• Knowing the angles of elevation or depression is the most essential step.
♦ We will want a few additional details which are easily determined at the field.
• Once they are determined, we can easily calculate various heights and distances. Let us see the procedure:
1. Consider fig.30.43 below. The line of site is named as AB. • A is the position of the eye of the observer.
• B is the position of the object being viewed.
 |
| Fig.30.43 |
♦ For our present discussion the ground is assumed to be horizontal. That is., the ground is assumed to be parallel to the 'horizontal level' shown by the white dashed line
♦ While doing engineering problems, we may have to deal with situations where the ground is not exactly horizontal.
• The perpendicular from B meets the ground at E.
• The perpendicular from B meets the horizontal level at C
• So ∠ACB = ∠DEB = 90o
3. Let the angle of elevation be θ degrees
• From ⊿ABC we get: tan = BC⁄AC
4. In the above step, we have a distance AC. But this distance is same as DE. This DE can be measured on the ground using a tape.
• First measure DF, which is the distance from the observer to the point F, which is the edge of the base of the cone.
• Then add it to FE. This FE is radius of the base of the cone. It is a constant which does not depend on the position of the observer. So it is denoted as k
5. Thus we have: AC = DE = (DF + FE) = (DF +k)
• So from (3), we will get BC.
• To this BC, we must add CE.
• But CE is same as AD, the height of the observer. So we get:
■ Height of the conical structure = BC + AD
Another application:
1. In fig.30.44 below, the observer is standing on top of a platform. • He has to look up to view the top point B of the conical structure. So there is an angle of elevation (θ1).
• He has to look down to view the edge of the base of the conical structure. So there is an angle of depression (θ2).
 |
| Fig.30.44 |
• AE = GD = (GC + CD) = (GC + k) Where k is the radius of the base of the cone.
• So we can calculate BE
3. From ⊿AGC we get: tan θ2 = AG⁄CG
• Note that in ⊿ACG, ∠GCA = θ2 (∵ ∠GCA and ∠EAC are alternate interior angles, and hence equal)
4. CG can be easily measured at the field. So from (3) we can calculate AG
■ So we get total height of the conical structure as:
Height = (BE + ED) = (BE + AG)
One more application:
1. From ⊿AGC we get: tan θ2 = AG⁄CG
• Note that in ⊿ACG, ∠GCA = θ2 (∵ ∠GCA and ∠EAC are alternate interior angles, and hence equal)
2. If the height AG (total height of platform and observer) is known, we can calculate CG fro (1).
That is., the distance between the platform and the conical structure can be determined.
This application is very useful if there is any obstruction such as a river or a lake between the platform and the structure.
Now we will see some practical problems:
Solved example 30.26
A man standing 10 metres away from the foot of a tree sees it's top at an elevation of 40o. His height is 1.7 metres. What is the height of the tree?Solved example 30.26
Solution:
Fig.30.45(a) below shows the rough sketch. AE represents the man. BF represents the tree. AB is the line of sight. AC is the horizontal level. EF is the distance between the man and the foot of the tree.
 |
| Fig.30.45 |
• The ⊿ABC for calculations are shown in fig.(b). In this triangle we get:
1. tan 40 = BC⁄AC ⟹ BC = AC × tan 40
2. But AC = EF, which is given as 10 m. So we get:
BC = 10 × tan 40 = 10 × 0.8391 = 8.391 m
3. Total height = BC + CF = 8.391+1.7 = 10.091 m
Solved example 30.27
A man 1.8 metres tall stands on top of a light house 25 metres high and sees a ship at see at a depression of 35o. How far is it from the foot of the light house?
Solution:
• In fig.30.46 below, AD is the center line of the light house. AB is the line of sight. AC is the horizontal level.
 |
| Fig.30.46 |
• Note that in this problem, the radius of the light house is not given. So the man is assumed to be positioned above the center line AD.
• Thus, when we determine AC, we will get the distance of the ship from the center line of the light house.
1. tan 35 = BC⁄AC ⟹ AC = BC⁄tan 35
2. But BC = AD = 25 + 1.8 = 26.8 m
So AC = BC⁄tan 35 = 26.8⁄0.7002 = 38.275 m
3. Distance of the ship = BD = AC = 38.275 m
Solved example 30.28
A boy standing at the edge of a canal sees the top of a tree at an elevation of 70o. The tree is also at the edge of the canal. Stepping 10 m back, he sees it at an elevation of 25o. The boy is 1.5 m tall. How wide is the canal and how tall is the tree?
Solution:
• Consider fig.30.47 below. CF represents the initial position of the boy. B is the top of the tree.
♦ So CB is the initial line of sight
 |
| Fig.30.47 |
♦ So AB is the final line of sight.
• Also both the distances AC and EF will be equal to 10 m.
1. Consider ⊿ABD
tan 25 = BD⁄AD ⟹ BD = AD × tan 25 ⟹ BD = (AC+CD) × tan 25 ⟹ BD = (10+CD) × tan 25
2. Consider ⊿CBD
tan 70 = BD⁄CD ⟹ BD = CD × tan 70
3. Equating the values of BD from (1) and (2) we get:
(10+CD) × tan 25 = CD × tan 70
⟹ CD × (tan 70 - tan 25) = 10 × tan 25 ⟹ CD × (2.7475 - 0.4663) = 10 × 0.4663
⟹ CD × 2.2812 = 4.663 ⟹ CD = 2.044
4. So width of the canal = FG = CD = 2.044 m
5. Substituting this value of CD in (2) we get:
BD = 2.044 × tan 70 = 2.044 × 2.7475 = 5.616 m
6. So height of the tree = BD + DG = 5.616+1.5 = 7.116 m
In the next section we will see a few more solved examples.
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