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TABLE II.—For the Solution of Oblique-Angled Spherical Triangles.

An angle or a side being divided by a perpendicular, the first and second segments are denoted by 1 seg, and 2 seg.

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Values of the Ju

ities required!.

Sines of angles are as sines of oppos. sides.


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Cot 1 seg. of this ang. = cos giv. side x tan adj. angle. sin 1 seg. x cos ang. opp. given side

Sin 2 seg. =



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between the


cos ang. adj. given side

Sines of sides are as sines of their opposite angles.

Let fall a perpen-Cot 1 seg. ang. req dicular from the

included angle. Cos 2 seg =

= tan given ang. × cos adj. side.

cos 1 seg. x tan giv, side adj. giv. angle tan side opp. given angle


= cos given ang. × tan adj. side.

cos 1 seg. x cos side opp. given angle

cos si e aj. given angle

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Tan 1 seg. of div. side = cos giv. ang. x tan side opp. ang. sought.

from the third


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Cos side sought

cos side not div. x cos 2 seg.

cos 1 seg. of side divided


Third angle.

Let fall a perpen.Tan 1 seg. of div. side = cos giv. ang. x tan other given side. on one of the

giv. sides.

dicular on the third side.


Cot 1 seg. of div. ang. = cos giv. side x tan ang. opp. side sought.

Tan side sought


tan giv. side x cos 1 seg. div. ang.

cos 2 seg. of divided angle

Let fall a perpen.) Cot 1 seg. div. ang. = cos giv. side x tan other giv. angle. giv. angles. from one of the Cos angle sought

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Let a, b, c, be the sides; A, B, C, the angles, b and c including the angle sought,

A side and the

to one of the A side opposite given angles.

Let fall a perpen

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sine or cosine


of its half.

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of its half.


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An angle by the and s = a+b+c. Then,

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Let s be the sum of the angles A, B, and C; and let B and C be adjacent to a the side required.


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For the Solution of all the cases of Oblique-Angled Spherical Triangles, by the Analogies of Napier.



Values of the Terms required.

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By the common analogy, sines of angles as sines of opp. sides.

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{By the common analogy.

By the common analogy.

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giv. side x sin diff. giv. angles sinsum of those angles

tangiv. side x cos diff. giv. angles cos sum of those angles

{By the common analogy.


Let fall a perpen. on the side adjacent to the angle sought.

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Cos angle sought = tan adj. seg. x cot adja. side.

Will be obtained by finding its correspondent angle, in a triangle which has all its parts supplemental to those of the triangle whose three angles are given.

Questions for Exercise in Spherical Trigonometry.

Er. 1. In the right-angled spherical triangle BAC, rightangled at A, the hypothenuse a = 78° 20′, and one leg = 76°52′, are given; to find the angles B, and C, and the other leg b.

Here, by table 1 case 1, sin c =

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sin c

sin a

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log sin clog sin a + 10.

log tan c

log cos B =
log cos blog cos a

Hence, 10+ log sin c = 10 + log sin 76°52′ :


log sin a =

log sin c =

log tan a + 10.

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log cos c + 10.

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Here c is acute, because the given leg is less than 90°.

log tan c = 10 + log tan 76°52′

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B is here acute, because a and c are of like affection.

Lastly, 10+ log cos a = 10+ log cos 78°20′ = 19.3058189


log cos c =

log cos b =

log cos 76°52′ = 9.3564426

log cos 27° 8′ = 9·9493763

where b is less than 90°, because a and c both are so.

Ex. 2. In a right-angled spherical triangle, denoted as above, are given a = 78°20′, B = 27°45′; to find the other sides and angle.

Ans. b = 27° 8', c = 76°52′, c = 83°56′. Ex. 3. In a spherical triangle, with A a right angle, given = 117°34′, c = 31°51′; to find the other parts.

Ans. a 113°55′, c = 28°51′, B = 104° 8'. Ex. 4. Given b = 27°6′, c = 76°52′; to find the other parts. Ans. a 78° 20′, B = 27°45', c = 83°56'. Ex. 5. Given b= 42°12′, B=48°; to find the other parts. Ans. a 64°40'1, or its supplement, C= 54°44', or its supplement, c = 64°35', or its supplement.


Ex. 6. Given B = 48°, c = 64° 35'; required the other Ans. b 42°12', c 54°44', a 64°40'.

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