Chapter 30.9 - Heights and Distances-Solved examples

In the previous section we saw how heights and distances can be determined using trigonometry. We also saw some solved examples. In this section we will see a few more solved examples.

Solved example 30.29
A 1.75 metre tall man, standing at the foot of a tower, sees the top of a hill 40 metres away at an elevation of 60^o. Climbing to the top of the tower, he sees it at an elevation of 50^o. Calculate the heights of the tower and the hill.
Solution:
1. In the fig.30.48 below:
• AF is the man standing at the foot of the tower
• GH is the man standing on the top of the tower
• GF is the tower and BE is the hill

Fig.30.48

2. In ⊿ADB, tan 60 = ^BD⁄_AD  ⟹ BD = AD ×  tan 60 ⟹ BD = FE × tan 60 = 40 × 1.7321 = 69.28 m
3. In ⊿GCB, tan 50 = ^BC⁄_GC  ⟹ BC = GC ×  tan 50 ⟹ BC = FE × tan 50 = 40 × 1.1917 = 47.67 m
4. Height of the tower = FH = FG - GH
5. FG = BE - BC
• BE = BD + DE = 69.28+1.75 = 71.03 m
• So FG = 71.03 - 47.67 = 23.36 m
6. Substituting this value of FG in (4) we get:
• Height of tower = FH = 23.36-1.75 = 21.61 m
7. Height of hill = BE = BD + DE = 69.28+1.75 = 71.03 m

Solved example 30.30
A 1.5 m tall boy saw the top of a building under construction at an elevation of 30^o. The completed building was 10 m higher and the boy saw it's top at an elevation of 60^o from the same spot. What is the height of the building?
Solution:
1. In the fig.30.49 below:
• AF is the boy
• CE is the building under construction
• BE is the building whose construction is completed

Fig.30.49

2. In ⊿ADB, tan 60 = ^BD⁄_AD  ⟹ AD = ^BD⁄_{tan 60}
3. In ⊿ADC, tan 30 = ^CD⁄_AD  ⟹ AD = ^CD⁄_{tan 30}
4. Equating (2) and (3) we get:
^BD⁄_{tan 60} = ^CD⁄_{tan 30}
5. But we have: CD = BD-10
Substituting this in (4) we get:
^BD⁄_{tan 60} = ^(BD-10)⁄_{tan 30}  ⟹ BD tan 30 = (BD-10) tan 60
⟹ BD tan 30 = BD tan 60 - 10 tan 60 ⟹ BD (tan 60 - tan 30) = 10 tan 60
⟹ BD (1.7320-0.5773) = 10 × 1.7320 ⟹ BD × 1.1547 = 17.32
⟹ BD = 14.999 = 15 m
6. So height of the completed building = BD + DE = 15+1.5 = 16.5 m

Solved example 30.31
When the sun is at an elevation of 40^o, the length of the shadow of a tree is 18 m. What is the height of the tree?
Solution:
Consider fig.30.50(a) below:

Fig.30.50

• A sun ray is falling at point C. 
• The ray makes an angle 40^o with the horizontal. 
    ♦ That is., at that particular time, the sun is at an elevation of 40^o. 
• The sun rays come from a very large distance. So all the sun rays are parallel. 
■ Thus, at any particular time, the angle of elevation of the sun will be the same, whatever be the height of the observer.
• In the fig.a, AB is the ground level. A is a point on the ground. The angle of elevation for the sun will be 40^o at A also. Based on thie above information we can solve this problem.
1. Consider fig.b. EF is the tree. The sun ray passing through the top point E will cast it's shadow at G. So GF is the shadow of the tree
2. We need the ⊿EFG only
tan 40 = ^EF⁄_GF  ⟹ EF = GF ×  tan 40 = 18 × 0.8391 = 15.10 m

Solved example 30.32
A man 1.8 m tall standing at the top of a telephone tower, saw the top of a 10 m high building at a depression of 40^o and the base of the building at a depression of 60^o. What is the height of the tower? How far is it from the building?
Solution:
• In the fig.30.51 below, AE is the man and ED is the telephone tower. BC is the building

Fig.30.51

1. In ⊿ACF, tan 60 = ^CF⁄_AF  ⟹ AF = ^CF⁄_{tan 60}
2. In ⊿ABF, tan 40 = ^BF⁄_AF  ⟹ AF = ^BF⁄_{tan 40}
3. Equating (1) and (2) we get:
^CF⁄_{tan 60} = ^BF⁄_{tan 40} 
4. But we have: BF = CF-10
Substituting this in (3) we get:
^CF⁄_{tan 60} = ^(CF-10)⁄_{tan 40}  ⟹ CF tan 40 = (CF-10) tan 60
⟹ CF tan 40 = CF tan 60 - 10 tan 60 ⟹ CF (tan 60 - tan 40) = 10 tan 60
⟹ CF (1.7320-0.8391) = 10 × 1.7320 ⟹ CF × 0.8929 = 17.32
⟹ CF = 19.4 m
5. From (1) we get: AF = ^CF⁄_{tan 60} ⟹ AF = ^19.4⁄_1.7320 = 11.2 m
6. So distance between the tower and the building = CD = AF = 11.2 m 
7. Height of the tower = DE = DA-EA
⟹ DE = CF-EA = 19.4 - 1.8 = 17.6 m

Solved example 30.33
From the top of an electric post, two wires are stretched to either sides and fixed to the ground, 25 m apart. The wires make angles 55^o and 40^o with the ground. What is the height of the post?
Solution:
• In the fig.30.52 below, CD is the post, CA and CB are the wires

Fig.30.52

1. In ⊿ADC, tan 55 = ^CD⁄_AD  ⟹ CD = AD ×  tan 55
2. In ⊿BDC, tan 40= ^CD⁄_BD  ⟹ CD = BD ×  tan 40
3. Equating (1) and (2) we get: AD tan 55 = BD tan 40
4. But BD = (25-AD)
5. Substituting this value of BD in (3) we get:
AD tan 55 = (25-AD) tan 40 ⟹ AD tan 55 = 25 tan 40 - AD tan 40
⟹ AD(tan 55 + tan 40) = 25 × tan 40
⟹ AD(1.4281+0.8391) = 25 × 0.8391 ⟹ AD × 2.2672 = 20.9775
⟹ AD = 9.2526 m
6. Substituting this value of AD in (1) we get: 
CD = 9.2526 × tan 55 = 9.2526 × 1.4281 = 13.214 m    

Solved example 30.34
When the sun is at an elevation of 35^o, the shadow of a tree is 10 m long. What would be the length of the shadow, when the sun is at an elevation of 25^o?
Solution:
• This is a problem similar to the solved example 30.31 that we saw above. See fig.30.53 below:

Fig.30.53

• AC is the sun ray when the elevation of the sun is 35
• CD is the sun ray when the elevation of the sun is 25
• CB is the tree
• We have to calculate BD
1. In ⊿ABC, tan 35 = ^BC⁄_AB  ⟹ BC = AB ×  tan 35 = 10 tan 35 = 10 × 0.7002 = 7.002 m
2. In ⊿DBC, tan 25= ^BC⁄_BD  ⟹ BD = ^BC⁄_{tan 25} = ^7.002⁄_0.4663 = 15.016 m

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This completes our present discussion on Trigonometry. In the next section we will see Coordinates.

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Chapter 30.8 - Heights and Distances using Trigonometry

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:

Fig.30.40

• 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:

Fig.30.41

• 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.

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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

• 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.

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• 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

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• 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

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 = 90^o
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

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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

 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.

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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 40^o. His height is 1.7 metres. What is the height of the tree?
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

• 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 35^o. 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

• 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 70^o. The tree is also at the edge of the canal. Stepping 10 m back, he sees it at an elevation of 25^o. 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

• 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

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In the next section we will see a few more solved examples.

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Trigonometric ratios of 0 and 90 degree Angles

In the previous section we saw solved examples on right triangles with special angles: 30^o, 60^o and 45^o. In this section, we will see two more special angles: 0^o and 90^o.

Consider the circle in fig.1 below.
1. It's center 'O' is at the intersection of two axes: the X axis and the Y axis. These two axes are perpendicular to each other.

Fig.1

2. Consider the radius OB₁.
• It makes an angle θ₁ with the horizontal axis.
3. From B₁, a perpendicular B₁A₁ is dropped onto the horizontal axis.
• So we get a right triangle:  ⊿OA₁B₁
4. Now, if we increase the value of θ, with O as the pivot point, the radius OB₁ will rotate about O.
• Let the new value of θ be θ₂.
• B₁ will move to B₂ and A₁ will move to A₂.
• Here also we get a right triangle: ⊿OA₂B₂
5. Let us take trigonometric ratios in both the triangles:
• In ⊿OA₁B₁, sin θ₁ = ^A₁B₁⁄_OB1
• In ⊿OA₂B₂, sin θ₂ = ^A₂B₂⁄_OB2
6. But denominators are the same because OB₁ = OB₂ (since radii of the same circle)
• Also A₁B₁ < A₂B₂ (since A₂B₂ is nearer to the Y axis)
7. Since denominators are the same, from (5), we can easily see that sin θ₁ is less than sin θ₂
■ In general we can say that, when the angle increases, the sine value increases

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Now we will see the extreme values.
• We have seen that when the angle increases, the sine value increases. Then what is the maximum value?
To find the answer, consider the fig.2 below:

Fig.2

1. The angle θ is now increased to θ₃. This θ₃ is very nearly equal to 90^o.
2. Consider the right triangle ⊿OA₃B₃
• We can see that the opposite side A₃B₃ is very nearly equal to the hypotenuse OB₃
• When θ₃ is exactly equal to 90, A₃B₃ will become exactly equal to OB₃
• So when θ₃ is very close to 90, we can consider A₃B₃ to be very close to OB₃.
    ♦ In such a condition, the difference in lengths between A₃B₃ and OB₃ will be very small like 0.000001 cm
    ♦ Even when the difference in lengths is such a low value, the ⊿OA₃B₃ will exist
   ♦ But if the difference become zero, the ⊿OA₃B₃ will disappear because, A₃B₃ will coincide with OB₃.
•For practical purposes, when the difference between A₃B₃ and OB₃ is very small, we can take:
θ₃ = 90^o and A₃B₃ = OB₃.
3. we will get:
sin θ₃ = sin 90 = ^A₃B₃⁄_OB3 = 1 (since numerator = denominator)
■ That is., we can take sin 90 = 1

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Now we will see the other extreme:

• We have seen that when the angle increases, the sine value increases. It increases upto a maximum of 1. 
• That means, when angle decreases, the sine value decreases. 
• Then what is the minimum value?
To find the answer, consider the fig.3 below:

Fig.3

1. θ is now decreased to θ₀. This θ₀ is very nearly equal to 0^o
2. Consider the right triangle: ⊿OA₀B₀
• We can see that the opposite side A₀B₀ is now very small
• When θ₀ is exactly equal to 0, the length of the opposite side A₀B₀ will become exactly equal to 0 cm
• So when θ₀ is very close to 0, we can consider A₀B₀ to be very close to 0 cm
    ♦ In such a condition, the length of A₀B₀ will be very small like 0.000001 cm
    ♦ Even when the length is such a low value, the ⊿OA₀B₀ will exist
   ♦ But if the length become zero, the ⊿OA₀B₀ will disappear because, A₀B₀ will coincide with OB₀.
•For practical purposes, when the length of A₀B₀ is very small, we can take:
θ₀ = 0^o and A₀B₀ = 0 cm
3. we will get:
sin θ₀ = sin 0 = ^A₀B₀⁄_OB0 = 0 (since numerator = 0)
■ That is., we can take sin 0 = 0

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So we got three information:

• When the angle increases, sine value increases
• The minimum value is: sin 0 = 0
• The maximum value is: sin 90 = 1

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Next we will consider cosine. The same three figs. can be used for this purpose also

1. Consider fig.1 above. For the angle θ, the opposite sides are A₁B₁, A₂B₂ … so on
• As θ increases, these opposite sides increases
2. But the adjacent sides are OA₁, OA₂ . . . so on
• As θ increases, these adjacent sides decreases.
3. Let us see the effect of this decrease when we take the cosine ratios:
• In ⊿OA₁B₁, cos θ₁ = ^OA₁⁄_OB1
• In ⊿OA₂B₂, cos θ₂ = ^OA₂⁄_OB2
4. But denominators are the same because OB₁ = OB₂ (since radii of the same circle)
• Also OA₂ < OA₁ (since A₂ is nearer to the origin than A₁)
5. Since denominators are the same, from (3), we can easily see that cos θ₂ is less than cos θ₁
■ In general, we can say that, when the angle increases, the cosine value decreases

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Now we will see the extreme values for cosine.

• We have seen that when the angle increases, the cosine value decreases. Then what is the minimum value?
To find the answer, consider the fig 2 above
1. The angle θ is now increased to θ₃. This θ₃ is very nearly equal to 90^o.
2. Consider the right triangle ⊿OA₃B₃
• We can see that the adjacent side OA₃ is now very small
• When θ₃ is exactly equal to 90, the length of the adjacent side OA₃ will become exactly equal to 0 cm
• So when θ₃ is very close to 90, we can consider OA₃ to be very close to 0 cm
    ♦ In such a condition, the length of OA₃ will be very small like 0.000001 cm
    ♦ Even when the length is such a low value, the ⊿OA₃B₃ will exist
   ♦ But if the length become zero, the ⊿OA₃B₃ will disappear because, A₃B₃ will coincide with OB₃.
•For practical purposes, when the length of OA₃ is very small, we can take:
θ₃ = 90^o and OA₃ = 0 cm
3. we will get:
cos θ₃ = cos 90 = ^OA₃⁄_OB3 = 0 (since numerator = 0)
■ That is., we can take cos 90 = 0

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Now we will see the other extreme for cosine:

• We have seen that when the angle increases, the cosine value decreases. It decreases upto a minimum of zero. 
• That means, when angle decreases, the cosine value increases. 
• Then what is the maximum value?
To find the answer, consider the fig.3 above.
1. θ is now decreased to θ₀. This θ₀ is very nearly equal to 0^o 
2. Consider the right triangle: ⊿OA₀B₀
• We can see that the adjacent side OA₀ is very nearly equal to the hypotenuse OB₀
• When θ₃ is exactly equal to 0, OA₀ will become exactly equal to OB₀
• So when θ₃ is very close to 0, we can consider OA₀ to be very close to OB₀.
    ♦ In such a condition, the difference in lengths between OA₀ and OB₀ will be very small like 0.000001 cm
    ♦ Even when the difference in lengths is such a low value, the ⊿OA₀B₀ will exist
   ♦ But if the difference become zero, the ⊿OA₀B₀ will disappear because, OA₀ will coincide with OB₀.
• For practical purposes, when the difference between OA₀ and OB₀ is very small, we can take:
θ₃ = 0^o and OA₀ = OB₀.
3. we will get:
cos θ₃ = cos 0 = ^OA₀⁄_OB0 = 1 (since numerator = denominator)
■ That is., we can take cos 0 = 1

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So we got three information:

• When the angle increases, cosine value decreases
• The minimum value is: cos 90 = 0
• The maximum value is: cos 0 = 1

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Next we will consider tangent ratio. This does not need lengthy calculations because we know that:
tan θ = ^{sin θ}⁄_{cos θ}
So we get:
• tan 0 = ^{sin 0}⁄_{cos 0} = ⁰⁄₁ = 0
• tan 90 = ^{sin 90}⁄_{cos 90} = ¹⁄₀ . So tan 90 is not defined because it is a division by zero.

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So we have completed this present discussion on the two special angles 0^o and 90^o.

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Chapter 30.7 - Problems related to Trigonometric ratio Tan

In the previous section we saw trigonometry related to circles. In the discussion so far in this chapter, we have seen problems related to sine and cosine. We have seen the basic details of tangent also. Details here. In this section, we will see  some problems related to tangent.

An example:
• Some steps are built on the side of a hill. See fig.30.33(a) below:

Fig.30.33

• The steps lead up to the top of the hill. But all the steps are not shown in the fig. We need only the first few steps for our present problem.
• A cone (coloured in yellow) is placed on the third step. We want to know the level at which the base of the cone is situated. In other words, what is the value of 'h'?
• For solving this problem, we are given two information:
    ♦ The tread of each step is 20 cm
    ♦ The angle which the side of the hill makes with the horizontal ground surface is 35^o
Solution:
1. Drop a perpendicular  CB from the edge of the third step onto the ground surface. This is shown by the red dashed line in fig.(b)
• Now we have a right triangle: ⊿ABC. 
2. Base AB will be equal to 3 times the tread. So we get AB = 3 × 20 = 60 cm
3. In this right triangle,
tan 35 = ^{opposite side}⁄_{adjacent side}  = ^BC⁄_AB  = ^h⁄₆₀  ⟹ h = 60 × tan 35 ⟹ h = 60 × 0.7002 = 42.01 cm

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Now we will see a few solved examples like this:

Solved example 30.21

An iron rod of unknown length, leans against a wall. The bottom end of the iron rod is 2 m away from the wall. Also it makes an angle of 40^o with the floor. See fig.30.34(a) below:

Fig.30.34

How high is the top end of the iron rod from the ground?
Solution:
1. The wall, iron rod and the ground surface together will give a right triangle as shown in fig.(b)
We have tan 40 = ^{opposite side}⁄_{adjacent side}  = ^h⁄₂  ⟹ h = 2 × tan 40 ⟹ h = 2 × 0.8391 = 1.68 m

Solved example 30.22 
Three rectangles are cut along the diagonals as shown in the figs.30.35(a, b and c) below:

Fig.30.35

One large rectangle and two smaller rectangles. The two smaller rectangles are identical. The 6 pieces so obtained are rearranged to form a regular pentagon. See. fig.(d). If the sides of the pentagon are to be 30 cm, what should be the length and breadth of the rectangles  
Solution:
1. First we want the interior angles of the pentagon. Since it is a regular pentagon, all sides will be equal. All angles will also be equal.
• We have: Sum of interior angles of any regular polygon = (n-2)×180. 
Where n is the number of sides.
• For a regular pentagon, n = 5. So the sum of all the interior angles = (5-2)×180 = 3×180 = 540^o
    ♦ So angle at each vertex = ⁵⁴⁰⁄₅ = 108^o
2. Now consider the yellow and magenta triangles in fig.30.36 below:

Fig.30.36

3. They are identical. So the edge AB will bisect the interior angle (at vertex A) of the pentagon. So we get: ∠CAB = ∠DAB = ¹⁰⁸⁄₂ = 54^o
4. Consider ⊿ABC. 
sin 54 = ^{opposite side}⁄_hypotenuse  = ^BC⁄_AC = ^BC⁄₃₀  ⟹ BC = 30 × sin 54 
⟹ BC = 30 × 0.8090 = 24.27 cm
■ So we get: Length of the small rectangles = 24.27 cm
5. Again consider ⊿ABC. 
cos 54 = ^{adjacent side}⁄_hypotenuse  = ^AB⁄_AC = ^AB⁄₃₀  ⟹ AB = 30 × cos 54 
⟹ AB = 30 × 0.5878 = 17.63 cm
■ So we get: Width of the small rectangles = 17.63 cm
6. Now consider the vertex C of the pentagon. The yellow and green triangles are identical. So each of them will contribute 36^o at the vertex C
7. The blue and red triangles are identical. So they will contribute equally at vertex C
So we can write:
2×36 + 2 × contribution from blue triangle = 108 
⟹ 72 + 2 contribution from blue triangle = 108 
⟹ 2 × contribution from blue triangle = 36 
⟹ contribution from blue triangle = ³⁶⁄₂  = 18^o
• So contribution from red triangle is also 18^o
8. Now consider triangle DEC. It is right angled at E. So we get:
∠CDE + 90 + 18 = 180 ⟹ ∠CDE + 108 = 180 ⟹ ∠CDE = 72^o
9. tan 72 = ^CE⁄_DE
• But DE = half of the side of the pentagon = 15 cm
• So we get: tan 72 = ^CE⁄₁₅ ⟹ CE = 15 × tan 72 ⟹ CE = 15 × 3.078 = 46.17 cm
■ So length of the large rectangle is 46.17 cm and it's width is 15 cm

Solved example 30.23
In the fig.30.37(a) below, the vertical lines are equally spaced.

Fig.30.37

Prove that their heights are in arithmetic sequence
Solution:
1. In fig.(b), the vertical lines are marked as AE, BF, CG etc.,
2. These vertical lines are drawn at an equal spacing of 's' cm
• Let the initial distance OA be 'a' cm
3. Consider the first triangle OAE. Since AE is vertical, triangle OAE is right angled at A
• So we get: tan 40 = ^AE⁄_a ⟹ AE = a tan 40
4. Consider the second triangle OBF. Since BF is vertical, triangle OBF is right angled at B
• So we get: tan 40 = ^BF⁄_(a+s) ⟹ BF = (a+s) tan 40
5. Consider the third triangle OCG. Since CG is vertical, triangle OCG is right angled at C
• So we get: tan 40 = ^CG⁄_(a+s+s) ⟹ BF = (a+s+s) tan 40 ⟹ BF = (a+2s) tan 40
6. Consider the fourth triangle ODH. Since DH is vertical, triangle ODH is right angled at D
• So we get: tan 40 = ^DH⁄_(a+s+s+s) ⟹ DH = (a+s+s+s) tan 40 ⟹ DH = (a+3s) tan 40
7. Continuing like this, if there are more vertical lines, we will get their heights as:
(a+4s) tan 40, (a+5s) tan 40 etc.,
8. Let us write the heights as a sequence:
a tan 40, (a+s) tan 40, (a+2s) tan 40, (a+3s) tan 40 . . . ,
9. In the above sequence, only 4 terms are written. We can write more terms if we want, just by increasing the coefficient of 's'.
• Now we have to prove that this sequence is an arithmetic sequence
10. Fourth term - third term = (a+3s) tan 40 - (a+2s) tan 40
= tan 40[(a+3s)-(a+2s)]
= tan 40[a+3s-a-2s]
= tan 40[s]
= s tan 40
11. Third term - second term = (a+2s) tan 40 - (a+s) tan 40
= tan 40[(a+2s)-(a+s)]
= tan 40[a+2s-a-s]
= tan 40[s]
= s tan 40
12. Second term - first term = (a+s) tan 40 - (a) tan 40
= tan 40[(a+s)-(a)]
= tan 40[a+s-a]
= tan 40[s]
= s tan 40
13. So we find that the difference the difference between any term and it's preceding term is a constant.
• So it is an arithmetic sequence
• The common difference is s tan40.

Solved example 30.24
One side of a triangle is 6 cm and the angles at it's ends are 40 and 65. Calculate it's area
Solution:
1. The given data is shown in fig.30.38(a) below:

Fig.30.38

2. Let us name the triangle as ABC. Drop a perpendicular CD from the vertex C on to the base AB
• Then area of ΔABC = ¹⁄₂ × AB × CD
• So we have to find CD
3. In ΔADC we have:
tan 40 = ^CD⁄_AD ⟹ AD = ^CD⁄_{tan 40}
4. In ΔBDC we have:
tan 65 = ^CD⁄_BD ⟹ BD = ^CD⁄_{tan 65}
5. But AB = AD + BD ⟹ 6 = [^CD⁄_{tan 40} + ^CD⁄_{tan 65}] ⟹ 6 = [^CD⁄_{tan 40} + ^CD⁄_{tan 65}]
⟹ 6 = [^CD⁄_0.8391 + ^CD⁄_2.1446]
[^(CD×2.1446)⁄_{(0.8391× 2.1446)} + ^(CD×0.8391)⁄_{(0.8391×2.1446)}] = 6
⟹  [^(CD×2.1446)⁄_{(0.8391× 2.1446)} + ^(CD×0.8391)⁄_{(0.8391×2.1446)}] = 6
⟹ [(CD×2.1446) + (CD×0.8391)] = 6×0.8391×2.1446
⟹ (2.9837 × CD) = 10.7972
⟹ CD = 3.62 cm
6. So area = ¹⁄₂ × AB × CD = ¹⁄₂ × 6 × 3.62 = 10.86 cm²

Solved example 30.25
In the fig.30.39(a) below, prove that h = √(xy)

Fig.30.39

Solution:
1. The given triangle is named as ABC in fig.(b). Drop a perpendicular CD from the vertex C onto the base AB
2. ADC is a right triangle.
We have: ∠CAD + 90 + ∠ACD = 180 ⟹ 30 + 90 + ∠ACD = 180^o ⟹ ∠ACD = 60^o
3. So it is a 30, 60, 90 triangle
• CD is the shortest side (opposite to the smallest angle 30)
• AD is the medium side (opposite to the medium angle 60)
• AC is the largest side (opposite to the largest angle 90)
4. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• AD = √3 × CD ⟹ x = √3 × h = h√3 cm
5. BDC is a right triangle.
We have: ∠CBD + 90 + ∠BCD = 180 ⟹ 60 + 90 + ∠BCD = 180^o ⟹ ∠BCD = 30^o
6. So it is a 30, 60, 90 triangle
• BD is the shortest side (opposite to the smallest angle 30)
• CD is the medium side (opposite to the medium angle 60)
• BC is the largest side (opposite to the largest angle 90)
7. In a 30, 60, 90 triangle, medium side is √3 times the shortest side. So we get:
• CD = √3 × BD ⟹ h = √3 × y = y√3 ⟹ y = ^h⁄_(√3)
8. From (4) we have: x = h√3
From (7) we have: y = ^h⁄_(√3)
9. So xy = h√3 × [^h⁄_(√3)] = h² ⟹ h = √(xy)

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In the next section we will see another application of tangent.

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