Mathematical Analysis of Spherical Triangle (Spherical Trigonometry by H.C. Rajpoot)

“Mathematical Analysis of Spherical Triangle (Spherical Trigonometry by HCR)”
Application of “HCR’s Theory of Polygon” proposed by H. C. Rajpoot (2014)
©All rights reserved
Mr Harish Chandra Rajpoot Jan, 2015
M.M.M. University of Technology, Gorakhpur-273010 (UP), India
1. Introduction: We very well know that a spherical triangle is a triangle having all its three vertices at the
spherical surface & each of its sides as a great circle arc. It mainly differs from a plane triangle by having the
sum of all its interior angles greater than ( ). (See figure 1 below)
2. Analysis of spherical triangle (when all of its sides are known): Consider any spherical having
all its sides (each as a great circle arc) of lengths ( ) on a spherical surface with a radius
such that its interior angles are ( ) (as shown in the figure 1)
Interior angles of spherical triangle: We know
that each interior angle of a spherical triangle is the angle
between the planes of great circle arcs representing any two
of its consecutive sides. Now, join the vertices A, B & C by
straight lines to obtain a corresponding plane (as
shown by the dotted lines AB, BC & CA). Similarly, we can
extend the straight lines OA, OB & OC to obtain a
plane which is the base of tetrahedron OA’B’C’.
Now, consider the tetrahedron OA’B’C’ having angles
between its consecutive lateral edges OB’ & OC’,
OA’ & OC’ and OA’ & OB’ respectively. Now the angles
are the angles subtended by the sides (each as a
great circle arc) of spherical triangle at the centre of sphere
which are determined as follows
Now the interior angles of spherical triangle that
are also the angles between consecutive lateral triangular
faces of the tetrahedron OA’B’C’ meeting at the vertex O (i.e.
the centre of sphere), are determined/calculated by using
HCR’s Inverse Cosine Formula according to which if
are the angles between consecutive lateral edges meeting
at any of four vertices of a tetrahedron then the angle
(opposite to ) between two lateral faces is given as follows
( *
( * ( )
Figure 1: A spherical triangle ABC having its sides (each as a
great circle arc) of lengths 𝒂 𝒃 𝒄 & its interior angles
𝑨 𝑩 𝑪 . A plane 𝑨𝑩𝑪 corresponding to the spherical
triangle ABC is obtained by joining the vertices A, B & C by
the straight lines.

( * ( )
( * ( )
Area of spherical triangle: In order to calculate area covered by spherical triangle ABC, let’s first calculate
the solid angle subtended by it at the centre of sphere. But if we join the vertices A, B & C of the spherical
triangle by straight lines then we obtain a corresponding plane which exerts a solid angle equal to that
subtended by the spherical triangle at the centre of sphere. Thus we would calculate the solid angle subtended
by the corresponding plane at the centre of sphere by two methods 1) Analytic & 2) Graphical as given
below.
1. Analytic method for calculation of solid angle:
Sides of corresponding plane : Let the sides of corresponding plane be
( )
In isosceles
⇒
( )
⇒
( *
⇒ ( )
Now from HCR’s Axiom-2, we know that the perpendicular drawn from the centre of the sphere always
passes through circumscribed centre of the plane triangle (in this case plane ) obtained by joining the
vertices of a spherical triangle to the centre of sphere (See the figure 2)
Hence, the circumscribed radius ( ) of plane having its
sides ( ) is given as follows
Where,
√ ( )( )( )
Hence, the normal height ( ) of plane from the centre O of
the sphere is given as follows
In right
√( ) ( )
Figure 2: The perpendicular OO’ drawn from the
centre O of the sphere to the plane 𝑨𝑩𝑪 always
passes through its circumscribed centre O’ according
to HCR Axiom-2

√
Now, in right
√( ) ( ) √ ( )
√
( )
Now, from HCR’s Theory of Polygon, the solid angle subtended by the right triangle having its orthogonal
sides at any point lying at a height on the vertical axis passing through the vertex common to the
side & the hypotenuse is given from standard formula as
(
√
* {(
√
* (
√
*}
Hence, the solid angle ( ) subtended by the isosceles at the centre O of the sphere
( ) ( )
Hence, by setting the corresponding values in the above formula, we get
[ (
√( * (
√
)
)
{(
√( * (
√
+
) (
√
√(√ ) (
√
+
)}]
[ (
√
) {(
√
) (
√
√
)}]
[
( )
{
( )
(
√
√ ( ) )}]
* ( ) ,( ) (
√
)-+
[ ( ) (( ) √ ( ) ,]
[ ( ) (( ) √ ( ) ,] ( )

Similarly, we have
[ ( ) (( ) √ ( ) ,] ( )
[ ( ) (( ) √ ( ) ,] ( )
Now, we must check out the nature of plane whether it is an acute, a right or an obtuse triangle.
Since the largest side is hence we can determine the largest angle of plane using
cosine formula as follows
Thus, there arise two cases to calculate the solid angle subtended by the plane at the centre of
sphere & so by the spherical triangle ABC as follows
Case 1: Corresponding plane is an acute or a right triangle ( ):
In this case, the foot point O’ of the perpendicular drawn from the centre of sphere to the acute plane
lies within or on the boundary of this triangle. All the values of solid angles corresponding to all
the sides respectively of acute plane are taken as positive. Hence, the solid angle ( )
subtended by the acute plane at the centre of sphere is given as the sum of magnitudes of solid angles
as follows
( )
Case 2: Corresponding plane is an obtuse triangle ( ):
In this case, the foot point O’ of the perpendicular drawn from the centre of sphere to the obtuse plane
lies outside the boundary of this triangle. (See the figure 3 below). In this case, solid angles
corresponding to the sides respectively are taken as positive while solid angle corresponding to
the largest side of obtuse plane is taken as negative. Hence, the solid angle ( ) subtended by
the obtuse plane at the centre of sphere is given as the algebraic sum of solid angles as follows
( )
2. Graphical method for calculation of solid angle:
In this method, we first plot the diagram of corresponding plane having known sides & then
specify the location of foot of perpendicular (F.O.P.) i.e. the circumscribed centre of plane then draw
the perpendiculars from circumscribed centre to all the opposite sides to divide it (i.e. plane ) into
elementary right triangles & use standard formula-1 of right triangle for calculating the solid angle subtended
by each of the elementary right triangles at the centre of sphere which is given as follows

(
√
* {(
√
* (
√
*}
Then find out the algebraic sum ( ) of the solid angles subtended by the elementary right triangles at the
centre of the sphere & hence the area covered by the spherical triangle ABC
3. Analysis of spherical triangle (when two of its sides & an
interior angle between them are known): Consider any
spherical triangle , having its two sides (each as a great
circle arc) of lengths and an interior angle between them,
on a spherical surface with a radius . Now we can easily
determine all its unknown parameters i.e. unknown side ( ), two
interior angles and area covered by it.
Now the angles are the angles subtended by the sides
(each as a great circle arc) of spherical triangle at the centre of
sphere which are determined as follows (See the figure 2 above)
( )
Now, apply HCR’s Inverse cosine formula for known interior
angle as follows
( * ( )
⇒ ⇒
( * ( *
Again by applying HCR’s Inverse cosine formula for calculating the unknown interior angle as follows
( * ( )
( * ( )
Area of spherical triangle: In order to calculate area covered by the spherical triangle ABC, let’s first
calculate the solid angle subtended by it at the centre of sphere. But if we join the vertices A, B & C of spherical
triangle by the straight lines then we obtain a corresponding plane which exerts a solid angle equal to
that subtended by the spherical triangle ABC at the centre of sphere. Now all the sides of the plane
can be calculated by following the previous method (as mentioned above) as follows
Figure 3: Corresponding plane 𝑨𝑩𝑪 is an obtuse triangle
𝒄 𝒃 𝒂 𝑪 𝟗𝟎 𝒐
. Centre O (𝟎 𝟎 𝐡) of the
sphere is lying at a height h perpendicular outwards to
the plane of paper

Thus we can calculate the solid angle subtended by the corresponding plane & so by the spherical
triangle ABC at the centre of sphere by following the previous two methods 1) Analytic & 2) Graphical (See the
above procedures). Hence we can calculate the area covered by the given spherical triangle.
These examples are based on all above articles which are very practical and directly & simply applicable to
calculate the different parameters of a spherical triangle. For ease of understanding & the calculations, the
value of side of the spherical triangle ABC is taken as the largest one).
Example 1: Calculate the area & each of the interior angles of a spherical triangle, having its sides (each as a
great circle arc) of lengths 12, 18 & 20 units, on the spherical surface with a radius 50 units.
Sol. Here, we have
⇒
Now, all the interior angles of spherical triangle can be easily calculated by using inverse cosine formula as
follows
⇒ ( ) ( )
( ) ( )
( ) ( )
⇒ ( )
Now, the sides of corresponding plane are calculated as follows
( )
( )
( )
Area of plane is given as

√ ( )( )( )
√ ( )( )( )
√
Since, the largest side of plane is hence the largest angle of the plane is
which is calculated by using cosine formula as follows
⇒ ( )
(
( ) ( ) ( )
( )( )
)
Hence, the plane is an acute angled triangle.
Note: If all the interior angles of any spherical triangle are acute then definitely the corresponding
plane will also be an acute angled triangle. It is not required to check it out by calculating the largest
angle of plane . (As in above example 1, we need not calculate the largest angle to check out the
nature of the plane we can directly say on the basis of values of interior angles A, B & C of the spherical
surface that the plane is an acute if each of A, B & C is an acute angle)
Hence the foot of perpendicular (F.O.P.) drawn from the centre of sphere to the plane will lie within
the boundary of plane (See the figure 2 above) hence, the solid angle subtended by it at the centre of
sphere is calculated as follows
[ ( ) (( ) √ ( ) ,]
[ (
( )
*
(
(
( )
*
( )
√ ( *
)
]
[ ( ) (( ) √ ( ) ,]
[ (
( )
*
(
(
( )
*
( )
√ ( *
)
]
[ ( ) (( ) √ ( ) ,]

[ (
( )
*
(
(
( )
*
( )
√ ( *
)
]
Note: In this case, all the values of solid angles corresponding to all the sides
respectively of the acute plane are taken as positive.
Hence, the solid angle ( ) subtended by the acute plane or spherical triangle ABC at the centre
of sphere is given as the sum of magnitudes of solid angles as follows
The above value of area implies that the given spherical triangle covers of the total
surface area ( ) & subtends a solid angle at the centre
of the sphere with a radius 50 units.
Example 2: A spherical triangle, having its two sides (each as a great circle arc) of lengths 25 & 38 units and
an interior angle included by them, on the spherical surface with a radius 200 units. Calculate the
unknown side, interior angles & the area covered by it.
Sol. Here, we have
⇒
Now in order to calculate unknown side c, apply HCR’s Inverse cosine formula for known interior angle as
follows
( ) ⇒ ( *
( *
Again by applying HCR’s Inverse cosine formula for calculating the unknown interior angle as follows
( ) ( )
( ) ( )
⇒ ( )
Now, the sides of corresponding plane are calculated as follows

( )
( )
( )
Area of plane is given as
√ ( )( )( )
√ ( )( )( )
√
Since, the largest side of plane is hence the largest angle of the plane is
which is calculated by using cosine formula as follows
⇒ ( )
(
( ) ( ) ( )
( )( )
)
Hence, the plane is an obtuse angled triangle.
Hence the foot of perpendicular (F.O.P.) drawn from the centre of sphere to the plane will lie outside
the boundary of plane (See the figure 3 above) hence, the solid angle subtended by it at the centre of
sphere is calculated as follows
[ ( ) (( ) √ ( ) ,]
[ (
( )
*
(
(
( )
*
( )
√ ( *
)
]
[ ( ) (( ) √ ( ) ,]

[ (
( )
*
(
(
( )
*
( )
√ ( *
)
]
[ ( ) (( ) √ ( ) ,]
[ (
( )
*
(
(
( )
*
( )
√ ( *
)
]
Note: In this case, solid angles corresponding to the sides respectively are taken as positive
while solid angle corresponding to the largest side of obtuse plane is taken as negative.
Hence, the solid angle ( ) subtended by the obtuse plane or spherical triangle ABC at the centre
of sphere is given as the algebraic sum of solid angles as follows
The above value of area implies that the given spherical triangle covers of the total
surface area ( ) & subtends a solid angle at the
centre of the sphere with a radius 200 units.
Conclusion: All the articles above have been derived by Mr H.C. Rajpoot by using simple geometry &
trigonometry. All above articles (formula) are very practical & simple to apply in case of a spherical triangle to
calculate all its important parameters such as solid angle, surface area covered, interior angles etc. & also
useful for calculating all the parameters of the corresponding plane triangle obtained by joining all the vertices
of a spherical triangle by the straight lines. These formulae can also be used to calculate all the parameters of
the right pyramid obtained by joining all the vertices of a spherical triangle to the centre of sphere such as
normal height, angle between the consecutive lateral edges, area of plane triangular base etc.
Note: Above articles had been derived & illustrated by Mr H.C. Rajpoot (B Tech, Mechanical Engineering)
M.M.M. University of Technology, Gorakhpur-273010 (UP) India Jan, 2015
Email:rajpootharishchandra@gmail.com
Author’s Home Page: https://notionpress.com/author/HarishChandraRajpoot

Mathematical Analysis of Spherical Triangle (Spherical Trigonometry by H.C. Rajpoot)

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