Page images
[blocks in formation]
[blocks in formation]

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.

[blocks in formation]

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 =

[blocks in formation]

sin c

sin a

[ocr errors][merged small][merged small][ocr errors]


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.

[ocr errors]

log cos c + 10.

[blocks in formation]

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

log tan c = 10 + log tan 76°52′

[blocks in formation]

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'.

Ex. 7. In the quadrantal triangle ABC, given the quadrantal side a = 90°, an adjacent angle c = 42° 12′, and the opposite angle A = 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

sin A =✔sin (s—b) . sin (țs—c) .


sin b. sinc


Or, log sin A = log sin (s—b)+log sin (s−c)+ar. comp. log sin bar. comp. log sin c: where s = a+b+c.

log sin (5-6)= log sin log sin (sc) = log sin A.c.log sin b = A. c. log sin A.C. log sin c A. c. log sin

Sum of the four logs.

43°58' = 9.8415749

= 9.3418385

83°13′ = 0.0030508 114°30′ = 0.0409771

Half sum log sin A= log sin 24°15'4=


[merged small][merged small][ocr errors][merged small][merged small][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][merged small][ocr errors][merged small][ocr errors][ocr errors]



log = 9.9219401 log 9.8745679


sin 62°56'

[ocr errors]
[ocr errors]


[ocr errors]

sin 114°30' log

[ocr errors]
[ocr errors]



= 9.9495770

= 9.9590229

sin 125°19' log = 9.9116507.

[ocr errors]

So that the remaining angles are, B=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

have tan diff. seg. of c =

This in logarithms, is

tan (b+a). tan 1(b−a)

tan c

log tan (b+a) = log tan 69°56′ = 10·4375601
log tan (b-a) = log tan 13°16′ = 9.3727819

Their sum = 19.8103420

Subtract log tan c = log tan 57°15′ = 10-1916394

Rem. log eos dif. seg.= log cos 22°34′ =


Hence, the segments of the base are 79°49′ and 34°41′.


Therefore, since cos A = tan 79°49′ x cot 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 log cos A = log cos 48°32'



= 9.8209820

The pre

The other two angles may be found as before. ference 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°56′ or 156°14'. Side betw. giv. angles, either 119°4' or 152°14'. either 79°12′ or 100°48′.

Side opp. 58 8',

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 7957 miles?

Ans. 76-75299, or nearly 764 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′′. between the base and each face 66°35′12′′.

Solid angle at the vertex 114.49768 The max. angle
Each ditto at the base 222-34298 being 1000.






General Account of this kind of Surveying.

ART. 1. In the treatise on Land Surveying in the second 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 county maps being so erroneous, as in some cases to exhibit blunders of three miles in distances of less than twenty.

3. The

« PreviousContinue »