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TABLE III.

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

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Values of the terms required.

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

Tan of its half ==

tan

diff. giv. sides X sin sum opp. angles

tan

sin diff. of those angles sum giv, sides X cossum opp. angles cos diff. of those angles

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Tan their diff. = cot giv. ang. X sin † diff. giv. sides

Tan their sum =

cot

sinsum of those sides giv. ang. X cos diff. giv. sides cos sum of those sides

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The other two sides.

Tan their sum =

Third angle.

tangiv. 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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=

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Cos angle sought tan adj. seg. X cot adja. side.

Will be obtained by finding its corresponding 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.

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

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Or, log sin clog sin clog sin a + 10.
log cos B = log tan c

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log tan a + 10.
log cos c + 10.

log sin 76°52′ = 19.9884894
log sin 78°20′ = 9·9909338

log sin 83°56′ = 9.9975556

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

Again, 10+ log tan c = 10+ log tan 76°52' =20-6320468

Remains,

log tan a =
log cos B =

B is here acute, because a Lastly, 10+ log cos a = 10

Remains,

log cos c =

log cos b =

where b is less than 90°,

log tan 78 20' = 10.6851149 log cos 27°45′ = 9-9469319 and c are of like affection. + log cos 78 20′ = 19-3058189 log cos 76°52′ = 9.3564426 log cos 27° 8' = 9.9493763 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 b=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', or its supplement, c=54'44', or its supplement, c64°35', or its supplement.

Ex. 6. Given в = 48°, c = 64°35'; required the other Aas. b = 42°12′, c = 54°44', a = 64°40'1.

parts?

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Ex. 7. In the quadrantal triangle ABC, given the quadrantal side a = 90°, an adjacent angle c = 42°12', and the opposite angle = 64°40'; required the other parts of the triangle?

Ex. 8. In an oblique-angled spherical triangle are given the three sides, viz. a = 56°40', b = 83°13', c = 114°30'; to find the angles.

Here, by the fifth case of table 2, we have

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=

sin (†s—b) . sin (†s—c) ̧

sin b. sin c

Or, 2 log sin log sin(4s-b)+log sin (sc)+ar. comp. log sin bar. comp. log. sin c: where s=a+b+c. = log sin 43 589.8415749 c) log sin 120419.3418385 c. log sin 83°13′ =0.0030508 A. c. log sin 114°30′ = 0.0409771 Sum of the four logs.

log sin (sb) : log sin (s A. c. log sin b = A A. C. log sin c

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19-2274413

Half sum = log sina log sin 24°15' =96137206

Consequently the angle A is 4831'.

Then, by common analogy,
Ás, sin a... sin
To, sin a... sin

So is, sin b... sin

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56°40′... tog = 9.9219401
48'31'... log = 9.8745679
3'13'... log =9.9969492

To, sin B sin 62'56...log = 9.9495770
And so is, sin c.. sin 114'30'... log 9-9590229

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To, sin c... sin 125°19′... log = 9·9116507

So that the remaining angles are, в=62°56′, and c=125°19′.

2dly. By way of comparison of methods, let us find the angle A, by the analogies of Napier, according to case 5 table 3. In order to which, suppose a perpendicular demitted from the angle c on the opposite side c. Then shall we

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Subtract log tan c = log tan 57°15′ = 10·1916394
Rem. log cos dif. =
seg log cos 22°34′ = 9.6187026
Hence, the segments of the base are 79°49' and 34°41'.

Therefore, since cos Atan 79°49′ × cct b:

To log tan adja. seg. = log tan 79°49′ = 10-7456257 Add log tan side b log tan 83°13′ = 9.0753563 The sum, rejecting 10 from the index

=

=3 9.8209620

= log cos a = log cos 48°32' The other two angles may be found as before. The preference is, in this case, manifestly due to the former method.

Ex. 9. In an oblique-angled spherical triangle, are given two sides, equal to 114°30′ and 56°40′ respectively, and the angle opposite the former equal to 125°20'; to find the other parts. Ans. Angles 48-30' and 62°55'; side, 83°12'.

Ex. 10. Given, in a spherical triangle, two angles, equal to 48°30′ and 125°20′, and the side opposite the latter; to find the other parts.

Ans. Side opposite first angle, 56°40′; other side, 83°12′; third angle, 62°54'.

Ex. 11. Given two sides, equal 114°30′ and 56°40′; and their included angle 62°54: to find the rest.

Ex. 12. Given two angles, 125°20′ and 48°30′, and the side comprehended between them 83°12′: to find the other parts.

Ex. 13. In a spherical triangle, the angles are 48°31', 62 56, and 125°20′; required the sides?

Ex. 14. Given two angles, 50°12′, and 58°8′; and a side opposite the former, 62°42′; to find the other parts. Ans. The third angle is either 130°54′33′′ or 156°16′32′′. Side betw. giv. angles, either 119°3′32′′ or 152°14′14′′. Side opp. 58°8′, either 72°12′13" or 100°47 37".

Ex. 15. The excess of the three angles of a triangle, measured on the earth's surface, above two right angles, is 1 second; what is its area, taking the earth's diameter at 79574 miles?

Ans. 76-75299, or nearly 763 square miles.

Ex. 16. Determine the solid angles of a regular pyramid with hexagonal base, the altitude of the pyramid being to each side of the base, as 2 to 1.

Ans. Plane angle between each two lateral faces 125°22′35′′. between the base and each face 66°35'12". 89-60648 The max. angle 218-19367) being 1000.

Solid angle at the vertex
Each ditto at the base

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