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In plain trigonometry we have formulas that relate the measures of sides and the trigonometric ratios of interior angles of any triangle. Hence, with minimal information we can solve a triangle, meaning we can find the measures of all the sides and all the internal angles. However, in case of plane geometry, the sides are all straight line segments. But what happens if a triangle has curved sides? In such case, the triangle laws using trigonometric ratios cannot be applicable. In case of triangles with curved sides, there are a set of special formulas to relate half the measure of sides with the internal angles. The set as such, is called as halfside formula. It may be clearly noted that the sides in these cases are arc of circles and hence measured in angular measure of radians.

These formulas deal with the relationships between the side lengths and angles of a spherical triangle. Half side formulas are alternative ways of solving triangles. For example,
If semiperimeter of the triangle having sides a, b and c is $S$ = $\frac{1}{2}$ $(a + b + c)$, then half angle formula for sines can be written as
$Sin(A/2)$ = $\sqrt{\frac{sin(Sb)(Sc)}{sin\ b sin\ c} }$
$Sin(B/2)$ = $\sqrt{\frac{sin(Sa)(Sc)}{sin\ a sin\ c} }$
$Sin(C/2)$ = $\sqrt{\frac{sin(Sa)(Sb)}{sin\ a sin\ b} }$
The above simple diagram shows a spherical triangle with vertices at $U,\ V$ and $R$. $A,\ B$ and $C$ are the respective internal angles of the triangle. The corresponding curved sides are denoted as $'a',\ 'b'$ and $'c'$ which are the arc measures in radians. The radius of the sphere is R. This triangle will not obey the law of sine or the cosine rule and also the sum property of internal angles (that is sum of $180^{\circ}$) will not be applicable. The formulas that are applicable are,
$tan$ $(\frac{a}{2})$ = $R\ cos(S  A)$
$tan$ $(\frac{b}{2})$ = $R\ cos(S  B)$
$tan$ $(\frac{c}{2})$ = $R\ cos(S  C)$
where, $S$ = $\frac{(A + B + C)}{2}$
It may be noted that all the three formulas are same excepting for the permutation of the variables. Hence, as single set it is referred as halfside formula. In addition, there is one more formula which is better called as spherical law of cosines. It states,
$cos (c)$ = $cos(a)cos(b) + sin(a)sin(b)cos(C)$
Here again we can permute the variables and obtain two more formulas.
The concept and the applications of these formulas can be better understood with the following examples.
Let us first study the situations where the sides of a triangle could be curved. Suppose you draw a triangle on a plain sheet of paper, the sides are straight line segments. We can extend the same concept to show three closer places on earth but it is only an approximation. The earth is of a spherical shape and hence the path between any two places shall be along the periphery of earthâ€™s spherical surface. Means it is arc of a great circle with center as center of the earth. However, because of the huge radius of the earth, that arc length of the great circle is almost same as the length of chord of the same circle connecting the same places, provided the paces are very close. But in cases of two very long distant places the path between them cannot be approximated as a straight line segment. The true distance between them is along the arc of the great circle passing through them. Now if you consider three places of considerable distance between each pair, those three places form a triangle of curved sides. Such a type of triangle is called as a spherical triangle, since the triangle lies on the surface of a sphere and not on a plane surface.
In the above picture, the shaded part is a spherical triangle.
Example 1:
The angles between the sides of a spherical triangle are $70^{\circ},\ 80^{\circ}$ and $100^{\circ}$. The radius of the sphere is $30$ in. Find, to the nearest inch, the curve length of the side opposite to the angle of $70^{\circ}$.
Solution:
Let us label the given data as $A$ = $70^{\circ},\ B$ = $80^{\circ},\ C$ = $100^{\circ}$. Hence $S$ = $\frac{(70^{\circ} + 80^{\circ} + 100^{\circ})}{2}$ = $125^{\circ}$. Also $R$ = $30$. Let us now find the radian measure of '$a$'.
$tan$ $(\frac{a}{2})$ = $Rcos(S  A)$ = $30cos(125^{\circ}\ \ 70^{\circ})$ = $30 cos(55^{\circ})$
Therefore, $(\frac{a}{2})$ = arc $tan\ [30 cos(55^{\circ})]$ or $a$ = $2 \times$ arc $tan\ [30\ cos(55^{\circ})]$ radians.
Hence, the curve length of '$a$' = $R \times a$
= $R\ \times 2\ \times$ arc $tan\ [30 cos(55^{\circ})]$
= $30\ \times 2\ \times$ arc $tan\ [30 cos(55^{\circ})]$
$\approx\ 91 in$.
Example 2:
A space craft flies on a spherical orbit with a radius of $4000$ miles. It starts from point $U$ to point $V$ and then it turns towards a point $W$ at angle of $50^{\circ}$ from the path $UV$. After reaching the point $W$ it returns to point $U$ making an angle of $80^{\circ}$ from the path of $VW$. It is noticed that the path $WU$ is at angle of $60^{\circ}$ from the original path $UV$. What is the distance flown by the spacecraft in the final path $WU$?
Solution:
In this case, $A$ = $60^{\circ},\ B$ = $50^{\circ},\ C$ = $80^{\circ}$. Hence $S$ = $\frac{(60^{\circ} + 50^{\circ} + 80^{\circ})}{2}$ = $95^{\circ}$. Also $R$ = $4000$.
First we need to find the arc measure of $b$ in radians. As per the half side formula,
tan $(\frac{b}{2})$ = $Rcos(S  B)$ = $4000cos(95^{\circ}  50^{\circ})$ = $4000\ cos(45^{\circ})$
Therefore, $(\frac{b}{2})$ = arc $tan\ [4000\ cos(45^{\circ})]$ or $b$ = $2 \times\ arc\ tan\ [4000 cos(45^{\circ})]$ radians.
Hence, the curve length of $WU$ = $R \times b$
= $4000 \times 2 \times\ arc\ tan\ [4000 cos(45^{\circ})]$
$\approx\ 12564\ miles$.
Example 3:
Two cars leave from the same point. The first one travels towards north at a speed $80$ mph and the second one heads towards east at a speed of $90$ mph. What is the distance between the two cars after $5$ hours? (Consider the shape of earth to be spherical with a radius of $3959$ miles).
Solution:
Let us assume the cars start from the point $U$. Let the first car reaches $W$ in the north and the second car reaches $V$ in the east, after $5$ hours. Obviously, angle $A$ = $90^{\circ}$. The distance $UW$ along the surface of the earth is $80 \times 5$ = $400$ miles and the same from $U$ to $V$ is $90 \times 5$ = $450$ miles. Then, $b$ = $(\frac{400}{3959})$ and $c$ = $(\frac{450}{3959})$, both in radians.
Using spherical law of cosines,
$cos (a)$ = $cos(b)cos(c)\ +\ sin(b)sin(c)cos(90^{\circ})$
= $cos$ $(\frac{400}{3959})$ $cos$ $(\frac{450}{3959})$ $+$ $cos$ $(\frac{400}{3959})$ $cos$ $(\frac{450}{3959})$ $cos(90^{\circ})$
= $cos$ $(\frac{400}{3959})$ $cos$ $(\frac{450}{3959})$ $+\ 0$
It gives $a$ = $0.152$ radians (approximately)
Therefore the distance between the cars along the earth surface = $3959 \times 0.152\ \approx\ 601.5\ miles.$