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


sum X tan diff. of the sides tan base

Tan 1 sum or 1 diff. of the seg. of the base. S

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.

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Questions for Exercise in Spherical Trigonometry.

Ex. 1. In the right-angled spherical triangle BAC, right-
angled at A, the hypothenuse a≈ 78°20′, and one leg.c
76°52′, are given to find the angles в, and c, and the other
leg b.


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Here, by table 1 case 1, sin c


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Or, log sin c = log sin c-log sin a + 10.
log cos B = log tan c-log tan a + 10.
log cos blog cos a-log cos c + 10.
Hence, 10+ log sin c = 10 + log sin 76°52′
log sin 78°20′ —


log sin a =


log sin c =

log sin 83°56′ = 9.9985556

Here c is acute, because the given leg is less than 90°.
Again, 10+ log tan c = 10 + log tan 76°52′


log tan α =

log tan 78°20′


log cos B =

log cos 27°45'


в 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 9.3564426 log cos 76°52′ :


log cos b =

log cos 27°8′


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

Ans. a



Ex. 6. Given в ≈ 48o, c = 64935'; required the other
Ans. b = 42°12′, e = 54°44′, a = 64°40'1.


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tan c

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tan a

cos b =

sin c

sin a

Cos a

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Ex. 7. In the quadrantal triangle ABC, given the quadrantal side a = 90o, an adjacent angle c=42° 12′ and the opposite angle A64° 40'; required the other parts of the triangle?

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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 the table 2, we have sin (1s-b). sin (s—c)

sin A

sin b. sin c



Or, log. sin ¦ ▲ — log sin (s—b)+log sin (1s—c) + ar. comp.
log. sinar. comp. log. sin c: where s=a+b+c.
log sin (sb) = log sin 43° 58′ = 9·8415749
log sin (sc) log sin 12° 41′
A. c. log sin b
c. log sin 83° 13′
A. c. log sin c c. log sin 114° 30′
Sum of the four logs



9.3418385 0.0030508 0-0409771 19-2274413


Half sum
log sin a=log sin 24° 15′
Consequently the angle a is 48° 31′

Then, by the common analogy,

As, sin a

sin 24° 15′9:6137206

sin 56°40′...log9-9219401 . sin 48°31'


sin 83°13′...log


sin 62°56′

log sin 114°30 ..log

And so is, sin c

To, sin c

sin 125°19′... log 9.9116507. So that the remaining angles are, B=62°56′, and c

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To, sin A..
So is, sin b
To, sin в


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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 have tan
tan (b-a) tan (b-a)
diff. seg, of c
tan c
This in logarithms, is

log tan 1 (b+a) = log tan 69°56′
log tan 1 (b-α) = log tan 13°16′



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

Their sum = 19·8103420




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tan 79° 49′Xcot b:
log tan 79° 49′
log tan 83° 13′

Therefore, since cos a
To log tan adja. seg. =
Add log tan side b



The other two angles may be found as before. The preference is, in this case, manifestly due to the former method. triangle, are given

Ex. 9. In an oblique-angled sphericalectively, and the


two sides equal to 114° 40' and 56° 30′
angle opposite the former equal to 125o 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'.

The sum rejecting 10 from the index
log cos a log cos 48° 32′ (

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 125o 20': required the sides?

opposite the former, 62° 42′ and 58° 8; and a side

Ex. 14. Given two angles, 50° 12′,

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to find the other parts.

Ans. The third angle is either 130°56′ or 156°14'.

Side betw. giv. angles,
Side opp. 5898',

either 119°4′ or 152°14'. either 79°12′ or 100°48′.

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 79573 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 126°52′11′′1 between the base and each face 66°35′12′′i.

The max. angle being 1000.

Solid angle at the vertex 114-49768 )
Each ditto at the base 222·34298


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General Account of this kind of Surveying.


ART. 1. In the treatise on Land Surveying in the first volume of this Course of Mathematics, the directions were restricted to the necessary operations for surveying fields, farms, lordships, or at most, counties; these being the only operations in which the generality of persons, who practise this kind of measurement, are likely to be engaged: but there are especial occasions when it is requisite to apply the principles of plane and spherical geometry, and the practices of surveying, to much more extensive portions of the earth's surface; and when of course much care and judgment are called into exercise, both with regard to the direction of the practical operations, and the management of the computations. The extensive processes which we are now about to consider, and which are characterised by the terms Geodesic Operations and Trigonometrical Surveying, are usually undertaken for the accomplishment of one of these three objects. 1. The finding the difference of longitude, between two moderately distant and noted meridians; as the meridians of the observatories at Greenwich and Oxford, or of those at Greenwich and Paris. 2. The accurate determination of the geographical positions of the principal places, whether on the coast or inland, in an island or kingdom; with a view to give greater accuracy to maps, and to accommodate the navigator with the actual position, as to latitude and longitude, of the principal promontories, havens, and ports. These have, till lately, been desiderata, even in this country: the position of some important points, as the Lizard, not being known within seven minutes of a degree; and, until the publication of the board of Ordnance maps, the best country maps being so erroneous, as in some cases to exhibit blunders of three miles in distances of less than twenty.


3. The

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