Spherical Triangle

Any section made by a cutting plane that passes through a sphere is circle. A great circle is formed when the cutting plane passes through the center of the sphere. Spherical triangle is a triangle bounded by arc of great circles of a sphere.
 

000-spherical-triangle-definition.gif

 

Note that for spherical triangles, sides a, b, and c are usually in angular units. And like plane triangles, angles A, B, and C are also in angular units.
 

Sum of interior angles of spherical triangle
The sum of the interior angles of a spherical triangle is greater than 180° and less than 540°.

$180^\circ \lt (A + B + C) \lt 540^\circ$

 

Area of spherical triangle
The area of a spherical triangle on the surface of the sphere of radius R is given by the formula

$A = \dfrac{\pi R^2E}{180^\circ}$

Where E is the spherical excess in degrees.
 

Spherical excess

$E = A + B + C - 180^\circ$

or

$\tan \frac{1}{4}E = \sqrt{\tan \frac{1}{2}s~\tan \frac{1}{2}(s - a)~\tan \frac{1}{2}(s - b)~\tan \frac{1}{2}(s - c)}$

Where $s = \frac{1}{2}(a + b + c)$
 

Spherical defect

$D = 360^\circ - (a + b + c)$

 

Note:
In spherical trigonometry, earth is assumed to be a perfect sphere. One minute (0° 1') of arc from the center of the earth has a distance equivalent to one (1) nautical mile (6080 feet) on the arc of great circle on the surface of the earth.

1 minute of arc = 1 nautical mile
1 nautical mile = 6080 feet
1 statute mile = 5280 feet
1 knot = 1 nautical mile per hour

 

Right Spherical Triangle

Solution of right spherical triangle
With any two quantities given (three quantities if the right angle is counted), any right spherical triangle can be solved by following the Napier’s rules. The rules are aided with the Napier’s circle. In Napier’s circle, the sides and angle of the triangle are written in consecutive order (not including the right angle), and complimentary angles are taken for quantities opposite the right angle.
 

000-right-spherical-triangle-napiers-circle.gif
 

$\bar{A} = 90^\circ - A$

$\bar{B} = 90^\circ - B$

$\bar{c} = 90^\circ - c$

 

Napier’s Rules

SIN-COOP Rule
In the Napier’s circle, the sine of any middle part is equal to product of the cosines of its opposite parts.
 

If we take $a$ as the middle part, its opposite parts are $\bar{c}$ and $\bar{A}$, then by sin-coop rule
$\sin a = \cos \bar{c}~\cos \bar{A}$

$\sin a = \cos (90^\circ - c)~\cos (90^\circ - A)$

$\sin a = \sin c~\sin A$
 

SIN-TAAD Rule
In the Napier’s circle, the sine of any middle part is equal to the product of the tangents of its adjacent parts.
 

If we take $a$ as the middle part, the adjacent parts are $b$ and $\bar{B}$, then by sin-taad rule
$\sin a = \tan b~\tan \bar{B}$

$\sin a = \tan b~\tan (90^\circ - B)$
 

Spherical triangle can have one or two or three 90° interior angle. Spherical triangle is said to be right if only one of its included angle is equal to 90°. Triangles with more than one 90° angle are oblique.
 

Oblique Spherical Triangle

Definition of oblique spherical triangle
Spherical triangles are said to be oblique if none of its included angle is 90° or two or three of its included angles are 90°. Spherical triangle with only one included angle equal to 90° is a right triangle.
 

Sine law
$\dfrac{\sin a}{\sin A} = \dfrac{\sin b}{\sin B} = \dfrac{\sin c}{\sin C}$
 

000-oblique-spherical-triangle.gif

 

Cosine law for sides
$\cos a = \cos b ~ \cos c + \sin b ~ \sin c ~ \cos A$

$\cos b = \cos a ~ \cos c + \sin a ~ \sin c ~ \cos B$

$\cos c = \cos a ~ \cos b + \sin a ~ \sin b ~ \cos C$
 

Cosine law for angles
$\cos A = -\cos B ~ \cos C + \sin B ~ \sin C ~ \cos a$

$\cos B = -\cos A ~ \cos C + \sin A ~ \sin C ~ \cos b$

$\cos C = -\cos A ~ \cos B + \sin A ~ \sin B ~ \cos c$
 

Napier's analogies
$\dfrac{\sin \frac{1}{2}(A - B)}{\sin \frac{1}{2}(A + B)} = \dfrac{\tan \frac{1}{2}(a - b)}{\tan \frac{1}{2}C}$

$\dfrac{\cos \frac{1}{2}(A - B)}{\cos \frac{1}{2}(A + B)} = \dfrac{\tan \frac{1}{2}(a + b)}{\tan \frac{1}{2}C}$

$\dfrac{\sin \frac{1}{2}(a - b)}{\sin \frac{1}{2}(a + b)} = \dfrac{\tan \frac{1}{2}(A - B)}{\cot \frac{1}{2}c}$

$\dfrac{\cos \frac{1}{2}(a - b)}{\cos \frac{1}{2}(a + b)} = \dfrac{\tan \frac{1}{2}(A + B)}{\cot \frac{1}{2}c}$
 

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