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PART II.

THE ELEMENTS OF

Spherical Trigonometry.

SECTION IV.

ON THE COMPUTATION OF SPHERICAL SURFACES.

PROP. I.

(265.) Problem. To express the surface of a spherical triangle, in terms of its three angles, and of the radius of the sphere.

Put r for the sphere's radius ; a for the circumference of a circle, of which the diameter is unity : then (Archim. or Legendre Geom, and E. 2. 12.) the surface of the hemisphere is expressed by 2r? ; and, if T be put for the surface of the spherical triangle, of which the angles are A, A', A", T: 2r? :: (A + A +A") - 180° : 360° (Art. 222.)

282 T
T= .(A+A+A" – 180°)

360

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PROP. II. (266.) Problem. To express the surface of a spherical triangle, in terms of any two of its sides and of the included angle.

If sin 1".pa be taken for an unit, then, the same notation being used as in the preceding article, and S, S', S", being put for the sides of the spherical triangle,

T = A+ A' + A" – 180 (Art. 265.)
- cot į T=tan Į (A+ A' + A") (Introd. 13. 19.)

tan į A+tan Ž (A' + A"').
1 - tan į A tan 1 (A' + A")

(Introd. 29.) tan 1 A cos } (S'+S") + cot Ž A cosį (S' –S")

cos į (S' +S") - cos } (S' - S") (Art. 244.) ... cot IT =

cos į S' cos { S" +sin S sin S" cos A

sin I S' sins" sin A (Introd. 29.) cot ; S'cot S"+cos A

(Introd. 17.) sin A

(267.) Cor. Hence, two spherical triangles are equal to one another, if their vertical angles be equal, and if the tangents of the halves of the sides containing the equal angles, be reciprocally proportionals. And, by

* By consulting a book of logarithms, it will be found that the logarithm of sin 1" is, also, the logarithm of

180

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means of this property, on a given sphere a spherical triangle may be described, on a given arch, that shall be equal to a given triangle, of the sphere, and shall have one of its angles equal to the angle of the given triangle. Also, an isosceles spherical triangle may be described, which shall be equal to a given triangle, and shall have its vertical angle, equal to the vertical angle of that given triangle.

PROP. III. (268.) Problem. To express the surface of a spherical triangle, in terms of its sides.

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Let the same notation be used as in Art. 265. 266. and let P be put for the semi-şum of the three sides of the triangle.

2 sin P sin (P-S) Then, 1+cos A =

sin S sin S"

(Art. 263. Case 1.) 2 sin (P-S') sin (P-S") And, 1 - Ços A =

sin S" sin S" If, therefore, the square root be taken of the product of the two last equations,

2 /[sin P sin (P-S) sin (P-S') sin (P-S“)] sin A

sin Sr sin 3" (Introd. 15.)

cos S-cos S'cos S" Again, since cos A=

(Art. 230. sin s' sin S"

1+cos S 1 +cos S" and that cot Scot į S" (Introd. 30.)

1+cos S+cos S'+cos S" .. cot{S'cot S"+cos. A =

sin s' sin S"

sin s

sin S"

TE;

cot I S'cot IS" + cos A ::. cot. {T=

(Art. 266.)

sin A

1.+cos Scoś $'+cos S" 2/[sin P sin (

PS) sin (P-S”) sin (P-S")]

(269.) Cor. If the spherical triangle (T) be equilateral,

1 + 3 cos s cot įT=

3 S 2 sin

(270.) SCHOLIUM. The following remarkable expression for the surface (T) of a spherical triangle, which may easily be deduced from what has been premised, was first given by Lhuillier;

P P-S P-S P-S" tanţT=V (tan

.tan

tan

tan

PS).

2

2

2

This form is manifestly analogous to the equation which exhibits the surface of a plane rectilineal triangle, in terms of its sides, and which might be deduced from the spherical form, by supposing the sphere's radius to be infinitely great. It serves, also, to shew, that when the sides of a spherical triangle are very small, compared with the radius of the sphere, its surface is nearly equal to that of a plane rectilineal triangle, contained by sides that are equal, in length, to the sides of the spherical triangle, each to each. For (Introd. 35.) a circular arch, when it is very small, in comparison of the radius, may be taken for its tangent, without any considerable error: and, in that case, Lhuillier's equation becomes

4 T = V[P.(P-S).(P-S).(P-S")],

or T = [P (P-S'). (P-S).(P-S")]; which also (Introd. 34.) expresses the surface of a plane rectilineal triangle, of which the sides are S, S' and S".

PROP. IV. (271.) Problem. Two sides of a quadrilateral spherical figure being at right angles to its base, to express in terms of those two sides, and of the base, the surface of the figure; all its sides being arches of great circles of the sphere.

Let the two sides BD, CE, of the quadrilateral spherical figure BDEC, which is bounded by arches of great circles of the sphere, be at right angles to the base DE:

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It is required to express the surface of the figure BDEC, in terms of the arches BD, CE and DE.

Produce BD and EC, until they meet in A: then (Art. 51. 36.) DA and EA are quadrants ; and (Art. 54.) the side DE is the measure of the spherical angle A.

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