Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[merged small][merged small][ocr errors][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]
[blocks in formation]

74 35 sec = 10.575385 cosec = 10015915 cos = 9'424615

13 45 cos = 9.987372

sin =

9.376003

cot 37 45 cot = 10.111100

= 10.111100 sin = 9.786906

[blocks in formation]
[merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

Ans. X 60° 22′ 20′′; Z = 95° 41′ 56′′; XZ = 76° 32′ 40′′.

ART. 22.-To express the cosine of an angle of a spherical triangle in terms of the sines and cosines of the sides.

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

Again, in triangle TOS—

ST2

SO2 + OT2 - 2SO. OT cos SOT = SP2+PO2+PT2 + PO2 - 2SO.OT.cos RQ

since OPS and OPT are right angles

=2PO2 + SP2 + PT2

-2SO. OT cos RQ II.

Subtract I. from II.

Hence cos RQ

.. 0 = 2PO2 + 2SP. PT cos P-2SO. OT cos RQ PO PO PT SP

=

+

OT' SO OT SO

COS P

=cos TOP.cos POS+ sin TOP. sin POS. cos P = cos PR. cos PQ + sin PR. sin PQ. cos P. This is the fundamental formula, and, where ABC is the triangle and sides denoted by a, b, c, is expressed

cos a = cos b. cos c + sin b. sin c.cos A.

The formula used in Case I. is derived from this in the following

[blocks in formation]

82° 26', AC = 92° 2′, AB = 104° 28'. Solve.

2. If BC = 140°, AC = 70°, AB = 80°. Solve.
3. If BC = 113° 2′, AC = 82° 40′, and C = 138° 50'.

4. If AC 60°, AB = 84° 22′, and A
=

Solve.

= 95°. Find BC.

5. If BC= 115° 20', AB = 84° 42', and B = 65° 30′. Find A and C.

6. If BC = 90°, AC = 75° 29', AB = 84° 13'. Find A.

7. Sun's R.A. and Dec. being 1h 30m 10", and 9° 26′ 30′′ N. Find his long.

8. The R.A. of a star is 16h 49m 48", and its Dec. is 7° 23′ 10′′ N.; find its long. and lat., the obliquity of the ecliptic being taken to be 23° 27′ 30′′.

9. In a right-angled spherical triangle C = 90°, show tan a tan c. cos B = tan A. sin b; also if B = b, show triangle must be isosceles.

10. Two ports A and B, both in northern hemisphere, are in long. 90° W. and 10° W. Two ships leave A and B respectively, proceeding each upon the great circle which touches the parallel of the port at the point of departure.

Ship from A steams to the eastward at uniform rate of 12 knots, that from B to the westward at 17 knots; they meet each other in 5 days. Find lat. and long. of point of meeting.

CHAPTER VI.

THE SAILINGS.

P

ART. 23. Parallel Sailing.-In parallel sailing, the ship's course is supposed to continue on the same parallel of latitude. There are three things concerned, viz. the Latitude, the Departure or Meridian Distance, and the Difference of Longitude.

Definitions.-(1) Latitude: the distance of the ship north or south from the Equator.

(2) Departure (or Meridian Distance): the distance in geographical miles between two places on the same parallel of latitude.

(3) Difference of Longitude: the arc of the Equator corresponding to the Meridian Dis

tance.

Illustration.-AB and CD are Departures at different distances from the Equator; EQ is the D. long., or corresponding Arc of the Equator.

FIG. 70.

B

Remark. It is seen that the Meridian Distance decreases as the Latitude increases, and vanishes altogether at the Poles.

The formula which express the relation between the three terms are

[blocks in formation]

These are the three cases in Parallel Sailing, and show how the D. long. or Dep. or Lat. may be found when the other two terms are given.

Let P be the pole, PE and PQ meridians of longitude, EQ

an arc of the Equator, DL an arc of a parallel of latitude, and C the centre of the Earth.

E

FIG. 71.

P

Arc DL

Then QL or ED will represent the lat. of the parallel DL, EQ is the diff. long., and DL the departure.

PC.

Join CL, and draw FL perpendicular to

It is evident that FL is a radius of the small circle of which DL is a part, and CQ a radius of the great circle of which EQ is a part; that is, a radius of the Earth. And since the arcs are proportional to the radii of the two circles

are EQ :: rad. FL: rad. CQ

therefore DL: EQ:: FL: CL

because CL is also a radius of the Earth;

[blocks in formation]

because CFL is a plane right-angled triangle, and QCL is the complement of FCL.

Now, the angle QCL is measured by the arc QL, which is the latitude of the parallel DL

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

dep.
D. long.
D. long.
dep.

= cos lat.

[ocr errors]
[ocr errors]
[ocr errors]

= sec lat.

[ocr errors]
[ocr errors]

(2)

(1)

From (1) dep. = D. long. x cos lat. (3) from (2) D. long. = dep. x sec lat.. (4)

which are the same as those derived from the sphere.

« ΠροηγούμενηΣυνέχεια »