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a BOC, b

each side has the same numerical measure
subtended angle whose vertex is at
the center of the sphere. Thus,

AOC, c = = AOB.

An angle of a spherical triangle

is the angle included by the planes of its sides which is measured by the angle included by two lines, one line in one plane, the other in the other, both perpendicular to the common intersection of the planes at the same point.

=

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E

90-P

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B

90° h

as the

D

Thus, if BE, in the plane AOB, is perpendicular to OA, and if ED, in the plane AOC, is perpendicular to OA, then the angle BED will measure the inclination of the planes AOB and AOC, and will be equal to the angle A of the spherical triangle.

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A

RIGHT TRIANGLES.

124. Napier's Circular Parts.

Napier's circular parts are the two sides adjacent to the right angle, the complements of their opposite angles, and the complement of the hypotenuse.

Thus, if HBP is a spherical triangle, right-angled at H, the circular parts are b, p, 90°-B, 90° P, and 90'-h.

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с

90°- B

B

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H

Adjacent parts are those which

are not separated by an intervening circular part. Thus, b and 90° P, 90°-P and 90°-h, 90' h and 90° - B, 90°- B and p, p and b are adjacent parts.

The right angle H is not regarded as a circular part, nor as separating the parts b and p.

Opposite parts are those which are separated by an intervening circular part.

Thus, b and 90° — h, 90°- P and 90° B, 90° —— h and p, 90° —B and b, p and 90°-P are opposite parts. Any one of these five circular parts is adjacent to two of the remaining parts, and opposite the other two parts.

Of any three circular parts, one part is either adjacent to both the others or opposite both.

A middle part is that which is adjacent to two other parts, or opposite two other parts.

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125. Exercises.

Tell which is the middle part, and whether the other parts are adjacent to, or opposite, the middle in the following:

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126. Napier's Principles.

1. The sine of the middle part is equal to the product of the tangents of the adjacent parts.

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Draw BD and DE, respectively perpendicular to OH and OP, and draw BE. BDE is a right angle, since the plane BOH is perpendicular to the plane POH, and BD is perpendicular to OH. The angle BED is equal to P.

B

E

D

P

p

H b

EB sin h, OE:

=

= cos h, DB = sin p, and OD

ED OE ED
X
EB EB OE'

ED

OD OD

(1) sin (90°-P) = tan (90°- h) tan b.

ᎠᏴ ED
X

ᎠᏴ

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or cos Pcot h tan b.

=

" or sin b

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tan p tan (90° — P).

tan p cot P.

By changing P, b, p into B, p, b, (1) and (2) become

(3) sin (90°-B)

tan (90'-h) tan p.

sin p=

(4)

tan b tan (90° — B).

=

cos P.

Multiplying (2) by (4), member by member, we have sin b sin p tan b tan p tan (90°-B) tan (90°-P).

Dividing by tan b tan p, and reducing, we have cos b cos ptan (90°—B) tan (90°—P). cos b cos p=

= cos EOD ×OD=OE= cos h=sin (90°— h). ... (5) sin (90'— h) tan (90°-B) tan (90°-P).

2. The sine of the middle part is equal to the product of the co-sines of the opposite parts.

OE cos EOD × OD, or cos h =cos b cos p.

... (6) sin (90°-- h) =cos b cos p.

DB EB sin DEB, or sin p= sin h sin P.
•'. (7)
sin p
(3) gives sin (90°-B)

cos (90'-h) cos (90°—P).

sin (90-h) sin p.
cos (90°-h) cos p

This, by substituting cos b cos p for sin (90°-h), cos (90°-h) cos (90'-P) for sin p, and reducing, gives (8) sin (90°—B) = cos b cos (90°—P).

By changing p, P, B, b into b, B, P, p, (7) and (8) become

(9) sin b = cos (90°-h) cos (90’—B). (10) sin (90°-P) = cos p cos (90'--B).

These ten formulas are thus reduced to two principles, from which the formulas can be written.

The memory will be further aided by observing the common vowel a in the first syllables of the words tangent and adjacent of the first principle, and the common vowel o in the first syllables of the words co-sine and opposite of the second principle; that is, we take the product of the tangents of the parts adjacent to the middle, and the product of the co-sines of the parts opposite the middle.

127. Mauduit's Principles.

If we take, as circular parts, the complements of the two sides adjacent to the right angles, their opposite angles, and the hypotenuse, we can readily deduce from the diagram, or from Napier's principles, the following principles:

1. The co-sine of the middle part is equal to the product of the co-tangents of the adjacent parts.

2. The co-sine of the middle part is equal to the product of the sines of the opposite parts.

Let the ten formulas be written and compared with those of the last article.

128. Analogies of Plane and Spherical Triangles.

The formulas which demonstrate Napier's principles may be placed under forms which will exhibit the analogies existing between Plane and Spherical Triangles, as in the subjoined table.

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These formulas can be committed and applied instead of Napier's principles by those who prefer to do so. The analogies will assist the memory.

S. N. 10.

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