Spherical Trigonometry, for the Use of Colleges and Schools
Area of a Spherical Triangle. Spherical Excess
Excerpts
Area of a Spherical Triangle. Spherical Excess
A *Lune* is that portion of the surface of a sphere which is comprised between two great semicircles.
Area of a Spherical Triangle. Spherical Excess
The expression $A+B+C-\pi$ is called the *spherical excess* of the triangle;
Area of a Spherical Triangle. Spherical Excess
*the area of a spherical triangle is the same fraction of half the surface of the sphere as the spherical excess is of four right angles.*
Area of a Spherical Triangle. Spherical Excess
Hence since the whole surface of a sphere may be considered as a lune with an angle equal to four right angles, we have for a lune with an angle of which the circular measure is $A$,
Area of a Spherical Triangle. Spherical Excess
The triangles are, however, not absolutely equal, but *symmetrically* equal (Art. 57), so that one cannot be made to coincide with the other by superposition.
Area of a Spherical Triangle. Spherical Excess
This expression is true even when the polygon has some of its angles greater than two right angles, provided it can be decomposed into triangles, of which each of the angles is less than two right angles.
Equations
Area of a Spherical Triangle. Spherical Excess
\dfrac{\text{area of lune}}{\text{surface of sphere}} = \dfrac{A}{2\pi}\,A lune's area is to the whole sphere's surface as its angle (in circular measure) is to 2π, since lunes are proportional to their angles.
Area of a Spherical Triangle. Spherical Excess
\text{area of lune } = \dfrac{A}{2\pi} 4\pi r^2 = 2Ar^2.The area of a lune of angle A on a sphere of radius r is 2Ar².
Area of a Spherical Triangle. Spherical Excess
\text{triangle } ABC=(A+B+C-\pi)r^2.The area of a spherical triangle equals its spherical excess times r².
Area of a Spherical Triangle. Spherical Excess
E=A+B+C-\piThe spherical excess E of a triangle is the sum of its angles minus π.
Area of a Spherical Triangle. Spherical Excess
\text{area of polygon} = \Bigl\{\Sigma - (n-2)\pi \Bigr\} r^2.The area of a spherical polygon with n sides and angle sum Σ is (Σ − (n−2)π) r².
Area of a Spherical Triangle. Spherical Excess
\sin\tfrac{1}{2}E = \dfrac{\surd\{\sin s \sin(s-a) \sin(s-b) \sin(s-c) \} x} {2\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b \cos\tfrac{1}{2}c }The sine of half the spherical excess equals the square root of sin s sin(s−a) sin(s−b) sin(s−c), divided by 2 cos½a cos½b cos½c. The stated form carries a factor x that the proof in Art. 101 does not produce; this is flagged as a discrepancy between the statement and its derivation (possible transcription error or erratum), not silently corrected.
Area of a Spherical Triangle. Spherical Excess
\tan\tfrac{1}{4}E = \surd\{\tan\tfrac{1}{2}s \tan\tfrac{1}{2}(s-a) \tan\tfrac{1}{2}(s-b) \tan\tfrac{1}{2}(s-c) \}The tangent of a quarter of the spherical excess equals the square root of the product of the tangents of half of s and of half of s−a, s−b, s−c.
Area of a Spherical Triangle. Spherical Excess
\sin\tfrac{1}{2}E = \sin C \sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \sec\tfrac{1}{2}c;Half the spherical excess has sine equal to sin C times sin½a sin½b divided by cos½c.
Area of a Spherical Triangle. Spherical Excess
\tan\tfrac{1}{2}E = \frac{\sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \sin C } {\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b + \sin\tfrac{1}{2}a \sin\tfrac{1}{2}b \cos C }The tangent of half the spherical excess expressed in the two sides a, b and the angle C between them.
Area of a Spherical Triangle. Spherical Excess
\frac{\cos^2\tfrac{1}{2}a + \cos^2\tfrac{1}{2}b + \cos^2\tfrac{1}{2}c-1 } {2\cos\tfrac{1}{2}a \cos\tfrac{1}{2}b \cos\tfrac{1}{2}c }Right-hand side of the result for cos½E in terms of the half-sides only (equation (3)); the chapter's chain writes cos½E equal to this expression.
Area of a Spherical Triangle. Spherical Excess
\sin^2\tfrac{1}{4}E = \frac{\sin\frac{1}{2}s \sin\frac{1}{2}(s-a) \sin\frac{1}{2}(s-b) \sin\frac{1}{2}(s-c) } {\cos\frac{1}{2}a \cos\frac{1}{2}b \cos\frac{1}{2}c }The square of the sine of a quarter of the spherical excess equals the product of sines of half s and of half (s−a), (s−b), (s−c), over the product of cosines of half the sides.
Area of a Spherical Triangle. Spherical Excess
\cos^2\tfrac{1}{4}E = \frac{\cos\frac{1}{2}s \cos\frac{1}{2}(s-a) \cos\frac{1}{2}(s-b) \cos\frac{1}{2}(s-c) } {\cos\frac{1}{2}a \cos\frac{1}{2}b \cos\frac{1}{2}c }The square of the cosine of a quarter of the spherical excess equals the product of cosines of half s and of half (s−a), (s−b), (s−c), over the product of cosines of half the sides.
Area of a Spherical Triangle. Spherical Excess
\sin(C-\tfrac12E) = \frac{\surd\{\sin s \sin(s-a) \sin(s-b) \sin(s-c) \} } {2\sin\frac12a \sin\frac12b \cos\frac12c }The sine of (C − ½E) equals the square root of sin s sin(s−a) sin(s−b) sin(s−c), over 2 sin½a sin½b cos½c.
Problems
Exercise VIII
Exercise VIII, problem 1, p. 083
Find the angles and sides of an equilateral triangle whose area is one-fourth of that of the sphere on which it is described.
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Exercise VIII, problem 10, p. 083
If the angles of a spherical triangle be together equal to four right angles ^212a + ^212b + ^212c = 1.
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Exercise VIII, problem 11, p. 083
If $r_1$, $r_2$, $r_3$ be the radii of three small circles of a sphere of radius $r$ which touch one another at $P$, $Q$, $R$, and $A$, $B$, $C$ be the angles of the spherical triangle formed by joining their centres, areaPQR = (Ar_1 + Br_2 + Cr_3 - )r^2.
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Exercise VIII, problem 12, p. 083
Shew that s = 12E (A-12E) (B-12E) (C-12E) ^12 212A 12B 12C .
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Exercise VIII, problem 13, p. 083
Given two sides of a spherical triangle, determine when the area is a maximum.
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Exercise VIII, problem 14, p. 083
Find the area of a regular polygon of a given number of sides formed by arcs of great circles on the surface of a sphere; and hence deduce that, if $\alpha$ be the angular radius of a small circle, its area is to that of the whole surface of the sphere as $\operatorname{versin}\alpha$ is to 2.
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Exercise VIII, problem 15, p. 083
$A$, $B$, $C$ are the angular points of a spherical triangle; $A'$, $B'$, $C'$ are the middle points of the respectively opposite sides. If $E$ be the spherical excess of the triangle, shew that 12E = A’B’12c = B’C’12a = C’A’12b .
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Exercise VIII, problem 16, p. 083
If one of the arcs of great circles which join the middle points of the sides of a spherical triangle be a quadrant, shew that the other two are also quadrants.
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Exercise VIII, problem 2, p. 083
Find the surface of an equilateral and equiangular spherical polygon of $n$ sides, and determine the value of each of the angles when the surface equals half the surface of the sphere.
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Exercise VIII, problem 3, p. 083
If $a=b=\dfrac{\pi}{3}$, and $c=\dfrac{\pi}{2}$, shew that $E=\cos^{-1}\dfrac{7}{9}$.
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Exercise VIII, problem 4, p. 083
If the angle $C$ of a spherical triangle be a right angle, shew that 12 E= 12 a 12 b 12 c, 12 E= 12 a 12 b 12 c.
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Exercise VIII, problem 5, p. 083
If the angle $C$ be a right angle, shew that ^2 ccE= ^2 aa+^2 bb.
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Exercise VIII, problem 6, p. 083
If $a=b$ and $C=\dfrac{\pi}{2}$, shew that $\tan E=\dfrac{\sin^2 a}{2\cos a}$.
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Exercise VIII, problem 7, p. 083
The sum of the angles in a right-angled triangle is less than four right angles.
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Exercise VIII, problem 8, p. 083
Draw through a given point in the side of a spherical triangle an arc of a great circle cutting off a given part of the triangle.
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Exercise VIII, problem 9, p. 083
In a spherical triangle if $\cos C=-\tan\dfrac{a}{2}\tan\dfrac{b}{2}$, then $C=A+B$.
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