51 
 

 American Scientific Research Journal for Engineering, Technology,  and Sciences  (ASRJETS) 
ISSN (Print) 2313-4410, ISSN (Online) 2313-4402 

© Global Society of Scientific Research and Researchers  
http://asrjetsjournal.org/  

 

A Method of Finding the Distance between Two Places on 

the Surface of the Earth 

F.T. Zohoraa*, M. Sharmin Akterb 

a,bDepartment of General Educational Development, Daffodil International University, Dhaka 1207, 

Bangladesh 
aEmail: fatema.ged@diu.edu.bd 

bEmail: sharmin.ged@diu.edu.bd 

 

 

 

Abstract 

Differential Geometry is the language of modern physics and provides us with a new technique of solving the 

problems related to the calculation of the plane and surface with parameters. There is a lot of application of 

Differential Geometry in calculating some of the geographical problems. Finding the distance between my 

house to my friend’s house in the next block is very easy but the approach to calculate the distance between two 

places holds true only to a certain extent. In this paper we have tried to build a method for finding the distance 

between two places on the surface of the earth using differential Geometry. 

Keywords: Great circle; Sphere; Latitude and Longitude. 

1. Introduction 

It is imagined that the earth is a perfect sphere with an axis around which it spins. The end of the axis are the 

North and South Pole. Every  places  of  the  earth  has its latitude  and  longitude  which  are  considered  as the  

arcs of the circles. The shortest distance between two places that covered by great circle as geodesics on a 

sphere are the arcs of great circle [5]. With  the  help of spherical trigonometry and by  using latitude  and  

longitude we can  calculate  the  distance  and  direction  from  one  place to  another  on  the  surface  of  the  

earth  with  respect  to the geographical north pole or south pole. 

------------------------------------------------------------------------ 

* Corresponding author.  

http://asrjetsjournal.org/


American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2018) Volume 49, No  1, pp 51-60 

 

52 
 

1.1. Sphere 

A sphere is a solid figure such that every point of its surface is equally distant from a fixed   point with in it, 

which is called center of the sphere. The straight line drawn through the center and terminated both ways by the 

sphere is called a diameter and any straight line joining the center of sphere to any point on the surface is called 

a radius of the sphere. 

1.2. Theorem  

1.3. If the arc S of a circle with radius r subtends an angles 𝜽𝜽 at the center then 𝑺𝑺 = 𝒓𝒓𝜽𝜽 where 𝜽𝜽 is measured 

in radians 

 

Figure 1 

2. Spherical Triangle 

A spherical triangle [3] is  the  proportion  of  a  sphere  bounded  by  three  arcs  of great circles. The  arcs  are  

its  sides  and spherical angles  between  the  arcs  are  its  three angles.  

2.1. Cosine Formula 

 

Figure 2 

To find the value of cosine of an angle of a spherical  triangle in  terms of cosines and sines of the sides [3].  

 ABC is spherical triangle, O the center of sphere. Draw two tangents at  A to the  arc  AB  and  AC  which  

intersect  the  lines  OB  and  OC  extended  to  D  and E respectively.  



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2018) Volume 49, No  1, pp 51-60 

 

53 
 

Join DE. Then  ∠A = ∠DAE.  

Let  ∠𝐷𝐷𝐷𝐷𝐷𝐷 = 𝑎𝑎. Now, in the triangle DOE 

𝐷𝐷𝐷𝐷2 = 𝐷𝐷𝐷𝐷2 + 𝐷𝐷𝐷𝐷2 − 2𝐷𝐷𝐷𝐷.𝐷𝐷𝐷𝐷. cos𝐷𝐷𝐷𝐷𝐷𝐷                                              (1) 

Again in the triangle DAE 

                            𝐷𝐷𝐷𝐷2 = 𝐴𝐴𝐷𝐷2 + 𝐴𝐴𝐷𝐷2 − 2𝐴𝐴𝐷𝐷.𝐴𝐴𝐷𝐷. cos𝐷𝐷𝐴𝐴𝐷𝐷                                               (2) 

Subtracting (2) from (1) 

0 = 𝐷𝐷𝐷𝐷2 − 𝐴𝐴𝐷𝐷2 + 𝐷𝐷𝐷𝐷2 − 𝐴𝐴𝐷𝐷2 + 2𝐴𝐴𝐷𝐷.𝐴𝐴𝐷𝐷. cos𝐴𝐴 − 2𝐷𝐷𝐷𝐷.𝐷𝐷𝐷𝐷. cos 𝑎𝑎 

Also the angles OAD and OAE  are right angles, so that      𝐷𝐷𝐷𝐷2 = 𝐷𝐷𝐴𝐴2 + 𝐴𝐴𝐷𝐷2  and 𝐷𝐷𝐷𝐷2 = 𝐷𝐷𝐴𝐴2 + 𝐴𝐴𝐷𝐷2. 

 Hence we have 

0 = 𝐷𝐷𝐴𝐴2 + 𝐷𝐷𝐴𝐴2 + 2𝐴𝐴𝐷𝐷.𝐴𝐴𝐷𝐷. cos𝐴𝐴 − 2𝐷𝐷𝐷𝐷.𝐷𝐷𝐷𝐷. cos𝑎𝑎 

Changing sides 

cos 𝑎𝑎 =
𝐷𝐷𝐴𝐴
𝐷𝐷𝐷𝐷

⋅
𝐷𝐷𝐴𝐴
𝐷𝐷𝐷𝐷

+
𝐴𝐴𝐷𝐷
𝐷𝐷𝐷𝐷

⋅
𝐴𝐴𝐷𝐷
𝐷𝐷𝐷𝐷

cos 𝑎𝑎 

Therefore 

    cos 𝑎𝑎 = cos𝑏𝑏 cos 𝑐𝑐 + sin 𝑏𝑏 sin 𝑐𝑐 cos𝐴𝐴 

Similarly we can prove that 

   cos 𝑏𝑏 = cos 𝑐𝑐 cos 𝑎𝑎 + sin 𝑐𝑐 sin𝑎𝑎 cos𝐵𝐵 

and                                                              cos 𝑐𝑐 = cos𝑎𝑎 cos𝑏𝑏 + sin𝑎𝑎 sin 𝑏𝑏 cos𝐶𝐶 . 

2.2. Sine Formula 

To prove that the angles of a spherical triangle are proportional to the sines of the opposite sides [3]. 

 That is, 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2018) Volume 49, No  1, pp 51-60 

 

54 
 

 

Figure 3 

sin𝐴𝐴
sin𝑎𝑎

=
sin𝐵𝐵
sin 𝑏𝑏

=
sin𝐶𝐶
sin 𝑐𝑐

 

Proof. We already proved in cosine formula that 

cos𝐴𝐴 =
cos 𝑎𝑎 − cos𝑏𝑏 cos 𝑐𝑐

sin 𝑏𝑏 sin 𝑐𝑐
 

∴  sin2 𝐴𝐴 = 1 − cos2 𝐴𝐴 = 1 −
(cos 𝑎𝑎 − cos𝑏𝑏 cos 𝑐𝑐)2

sin2 𝑏𝑏 sin2 𝑐𝑐
 

𝑜𝑜𝑜𝑜,  sin2𝐴𝐴 =
(1 − cos2𝑏𝑏)(1 − cos2𝑐𝑐) − (cos2𝑎𝑎 + cos2𝑏𝑏cos2𝑐𝑐 − 2 cos𝑎𝑎 cos𝑏𝑏 cos 𝑐𝑐)

sin2 𝑏𝑏 sin2 𝑐𝑐
 

                              𝑜𝑜𝑜𝑜,   sin2𝐴𝐴 = �1−cos
2𝑏𝑏−cos2𝑐𝑐+cos2𝑏𝑏cos2𝑐𝑐�−(cos2𝑎𝑎+cos2𝑏𝑏cos2𝑐𝑐−2cos 𝑎𝑎 cos 𝑏𝑏 cos 𝑐𝑐)

sin2 𝑏𝑏 sin2 𝑐𝑐
 

𝑜𝑜𝑜𝑜, sin𝐴𝐴 =
�(1 − cos2𝑎𝑎 − cos2𝑏𝑏 + cos2𝑐𝑐 + 2 cos 𝑎𝑎 cos 𝑏𝑏 cos 𝑐𝑐)

sin 𝑏𝑏 sin 𝑐𝑐
 

We  have  taken  only  positive  sign, with  the  radical  as we know that the angle and sides of a spherical 

triangle are each less than two right angles and such sin𝐴𝐴 , sin𝐵𝐵 , sin𝐶𝐶 are all positive. 

∴
sin𝐴𝐴
sin𝑎𝑎

=
�(1 − cos2𝑎𝑎 − cos2𝑏𝑏 + cos2𝑐𝑐 + 2 cos𝑎𝑎 cos𝑏𝑏 cos 𝑐𝑐)

sin𝑎𝑎 sin 𝑏𝑏 sin 𝑐𝑐
 

The symmetry of the result shows that 

∴
sin𝐵𝐵
sin 𝑏𝑏

=
�(1 − cos2𝑎𝑎 − cos2𝑏𝑏 + cos2𝑐𝑐 + 2 cos 𝑎𝑎 cos 𝑏𝑏 cos 𝑐𝑐)

sin 𝑎𝑎 sin 𝑏𝑏 sin 𝑐𝑐
 

𝑎𝑎𝑎𝑎𝑎𝑎  
sin𝐶𝐶
sin 𝑐𝑐

=
�(1 − cos2𝑎𝑎 − cos2𝑏𝑏 + cos2𝑐𝑐 + 2 cos 𝑎𝑎 cos 𝑏𝑏 cos 𝑐𝑐)

sin𝑎𝑎 sin 𝑏𝑏 sin 𝑐𝑐
 

So we have 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2018) Volume 49, No  1, pp 51-60 

 

55 
 

sin𝐴𝐴
sin𝑎𝑎

=
sin𝐵𝐵
sin 𝑏𝑏

=
sin𝐶𝐶
sin 𝑐𝑐

=
�(1 − cos2𝑎𝑎 − cos2𝑏𝑏 + cos2𝑐𝑐 + 2 cos 𝑎𝑎 cos 𝑏𝑏 cos 𝑐𝑐)

sin 𝑎𝑎 sin 𝑏𝑏 sin 𝑐𝑐
 

∴  
sin𝐴𝐴
sin 𝑎𝑎

=
sin𝐵𝐵
sin 𝑏𝑏

=
sin𝐶𝐶
sin 𝑐𝑐

 

 

Figure 4 

3. Shortest Distance and Direction Between Places 

Let NAB be the spherical triangle where A and B the two places on the sphere of the earth.  𝐴𝐴 ≡ (𝑥𝑥1°𝑁𝑁  , 𝑦𝑦1°𝐷𝐷) 

𝐵𝐵 ≡ (𝑥𝑥2°𝑁𝑁 ,𝑦𝑦2°𝐷𝐷).  

N be the north pole. “O” be the center of the sphere of  the  earth. Let,  ∠𝐴𝐴𝐷𝐷𝐵𝐵 = 𝑎𝑎,   ∠𝐵𝐵𝐷𝐷𝑁𝑁 = 𝑎𝑎,   ∠𝐴𝐴𝐷𝐷𝑁𝑁 = 𝑏𝑏. 

We have to find out the shortest distance between the given two places and the direction of the two places 

towards North pole. 

3.1. Calculation for the shortest distance between A and B towards North pole 

 We have from cosine rule of spherical trigonometry  

cos𝑎𝑎 = cos 𝑎𝑎 cos 𝑏𝑏 + sin𝑎𝑎 sin 𝑏𝑏 cos𝑁𝑁 

= cos(90°−x2°) cos(90° − x2°) + sin(90° − x2°) sin(90° − x2°) cos(y2° − y1°)

= sin x2° sin x1° + cos x2° cos x1° cos(y2° − y1°) 

∴  𝑎𝑎 = cos−1 {sin x2° sin x1° + cos x2° cos x1° cos(y2° − y1°)}                                (1) 

∴  The distance between  A  and  B = the  length of the arc of the great circle passing through A and B = 𝑎𝑎 =

𝑅𝑅.𝑎𝑎   (𝑅𝑅 =Radius of the earth). 

By equation (1) we can easily find out the value of n for any two places and hence we can find out shortest 

distance d. 

3.2. Calculation for the direction 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2018) Volume 49, No  1, pp 51-60 

 

56 
 

 From spherical  triangle using sine rule 

sin𝐴𝐴
sin𝑎𝑎

=
sin𝐵𝐵
sin 𝑏𝑏

=
sin𝑁𝑁
sin 𝑎𝑎

 

or,   sin𝐴𝐴 =
sin𝑎𝑎 sin𝑁𝑁

sin 𝑎𝑎
 

or,   sin𝐴𝐴 =
sin(90° − x2°) sin(y2° − y1°)

sin 𝑎𝑎
 

                                      or,   sin𝐴𝐴 =
cos x2° sin(y2° − y1°)

sin 𝑎𝑎
                                               (2) 

Similarly,   

                                         sin𝐵𝐵 =
cos x1° sin(y2° − y1°)

sin 𝑎𝑎
                                                    (3) 

By, these two equations we can easily find out the direction of any place with respect to another place towards 

North Pole. 

4. Examples 

4.1. Example 

To find the shortest distance and direction from one towards the other between  London (U.K) and Bangkok 

(Thailand) towards North Pole. 

Here,                                       𝐴𝐴 ≡ 𝐿𝐿𝑜𝑜𝑎𝑎𝑎𝑎𝑜𝑜𝑎𝑎;                     𝑥𝑥1 = 51.30°𝑁𝑁 

                                              𝑦𝑦1 = 0.10°𝑊𝑊 

𝐵𝐵 ≡ 𝐵𝐵𝑎𝑎𝑎𝑎𝐵𝐵𝐵𝐵𝑜𝑜𝐵𝐵;                   𝑥𝑥2 = 13.44°𝑁𝑁 

                                                 𝑦𝑦2 = 100.30°𝐷𝐷 

N = y2° − y1° = 100.30 ° − (−0.10°) 

      = 100.30° + 0.10° = 100.40° 

We know from Spherical Trigonometry, 

𝑎𝑎 = cos−1 {sin x2° sin x1° + cos x2° cos x1° cos(y2° − y1°)} 

          = cos−1 {sin 51.30° sin 13.44° + cos 51.30° cos 13.44° cos 100.40} 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2018) Volume 49, No  1, pp 51-60 

 

57 
 

= cos−1(0.1813931 − 0.1097772) 

= cos−1(0.0716159) 

= 85.8931980° 

                             = 85.8931980 ×
𝜋𝜋

180
  𝑜𝑜𝑎𝑎𝑎𝑎𝑟𝑟𝑎𝑎𝑎𝑎 

              = 1.4991191𝑜𝑜𝑎𝑎𝑎𝑎𝑟𝑟𝑎𝑎𝑎𝑎 

Calculation for the shortest distance: 

Radius of the earth, 𝑅𝑅 = 6378.388  𝐵𝐵𝑘𝑘𝑘𝑘 

∴  Shortest Distance between 𝐿𝐿𝑜𝑜𝑎𝑎𝑎𝑎𝑜𝑜𝑎𝑎 and 𝐵𝐵𝑎𝑎𝑎𝑎𝐵𝐵𝐵𝐵𝑜𝑜𝐵𝐵 is, 

                                                  𝑎𝑎 = 𝑅𝑅 ⋅ 𝑎𝑎   

                   = 6378.388 × 1.4991191 = 9561.96  𝐵𝐵𝑘𝑘𝑘𝑘 

So, the distance between London and Bangkok is  9561.96  𝐵𝐵𝑘𝑘𝑘𝑘. 

Calculation for the direction of  London and Bangkok towards North Pole: 

From the sine rule on spherical triangle we have, 

sin𝐴𝐴 =  
cos x2° sin(y2° − y1°)

sin 𝑎𝑎
 

=
cos 13.44° sin 100.40°

sin 85.8931980°
 

= 0.9590979 

∴ ∠𝐴𝐴 = sin−1(0.9590979) 

= 73.56° 

Again,   sin𝐵𝐵 =  cos x1°sin(y2°−y1°)
sin 𝑛𝑛

 

=
cos 51.30° sin 100.40°

sin85.8931980 °
 

= 0.6165539 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2018) Volume 49, No  1, pp 51-60 

 

58 
 

∴ ∠𝐵𝐵 = sin−1(0.6165539) 

= 38.07° 

The direction of Bangkok with respect to London towards graphical north pole is 

∠𝐴𝐴 = 73.56° 

The direction of London with respect to Bangkok towards graphical  north pole is 

∠𝐵𝐵 = 38.07° 

4.2. Example 

To find the shortest distance and direction for Dhaka and Munshiganj  towards North Pole. 

Here,                                    𝐴𝐴 ≡ 𝐷𝐷ℎ𝑎𝑎𝐵𝐵𝑎𝑎;                             𝑥𝑥1 = 23.42°𝑁𝑁 

                                                    𝑦𝑦1 = 90.22°𝐷𝐷 

𝐵𝐵 ≡ 𝑀𝑀𝑀𝑀𝑎𝑎𝑘𝑘ℎ𝑟𝑟𝐵𝐵𝑎𝑎𝑎𝑎𝐵𝐵;                  𝑥𝑥2 = 23.32°𝑁𝑁 

                                                      𝑦𝑦2 = 90.32°𝐷𝐷 

N = y2° − y1° =  90.32° − 90.22° = 0.10° 

We know from Spherical Trigonometry, 

𝑎𝑎 = cos−1 {sin x2° sin x1° + cos x2° cos x1° cos(y2° − y1°)} 

          = cos−1 {sin 23.42° sin 23.32° + cos 23.42° cos 23.32° cos(90.32° − 90.22°)} 

= cos−1(0.157344 − 0.842653) 

= cos−1(0.999997) 

= 0.1403345° 

= 0.1403345 ×
𝜋𝜋

180
  𝑜𝑜𝑎𝑎𝑎𝑎𝑟𝑟𝑎𝑎𝑎𝑎 

= 0.002449  𝑜𝑜𝑎𝑎𝑎𝑎𝑟𝑟𝑎𝑎𝑎𝑎 

Calculation for the shortest distance: 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2018) Volume 49, No  1, pp 51-60 

 

59 
 

Radius of the earth [3], 𝑅𝑅 = 6378.388  𝐵𝐵𝑘𝑘𝑘𝑘 

∴  Shortest Distance between Dhaka and  Munshiganj is, 

𝑎𝑎 = 𝑅𝑅 ⋅ 𝑎𝑎 

= 6378.388 × 0.002449 

= 15.6237  𝐵𝐵𝑘𝑘𝑘𝑘 

So, the distance between Dhaka and Munshiganj is   15.6237  𝐵𝐵𝑘𝑘𝑘𝑘. 

Calculation for the direction of  Dhaka and Munshiganj towards North Pole: 

From the sine rule on spherical triangle we have, 

sin𝐴𝐴 =  
cos x2° sin(y2° − y1°)

sin 𝑎𝑎
 

=
cos 23.32° sin(0.10°)

sin 0.140345°
= 0.654553 

 

∴ ∠𝐴𝐴 = sin−1(0.654553) 

= 40.89° 

Again,    sin𝐵𝐵 =  cos x1°sin(y2°−y1°)
sin 𝑛𝑛

 

=
cos 23.42° sin(0.10°)

sin 0.140345°
 

= 0.653829 

∴ ∠𝐵𝐵 = sin−1(0.653829) = 40.83° 

The  direction  of  Munshiganj  with  respect to Dhaka towards graphical north pole is 

∠𝐴𝐴 = 40.89° 

The  direction  of  Dhaka  with  respect to Munshiganj towards graphical north pole is 

∠𝐵𝐵 = 40.83° 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2018) Volume 49, No  1, pp 51-60 

 

60 
 

5. Conclusion 

Differential Geometry is related to Astronomy and Geographical problems. Some practical examples are 

discussed in this paper. The aim of this paper is to find a mathematical method of finding distance and direction 

between two places on the surfaces of the earth with respect to the geographical north pole or south pole. Since 

the radius of the earth is not perfect, so the distance and directions found here are approximate. 

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