1. The Distance between them. Suppofe Cittadella at A, and Peckin at B; draw proper great Circles through these two Places, and there will be formed the Ifofceles Triangles AP B, in which each Side A P or P B is equal to the Complement of the Latitude, viz. 50° 00', and the Angle at P= 104° 30', the Difference of Longitude. But the Perpendicular PC divides it into two Right Angled Triangles APC and CPB equal to each other; by either of which this Problem is fatisfied.

Therefore in the Right-angled Triangle AP C, there is given AP = 50° 00′, and the Angle AP C 52° 15'; to find AC = Half the Distance required, thus ;

As Radius

To the Sine of AP 50° 00'
So is the Sine of APC 52° 15′
To the Sine of AC = 37° 17′

The Double whereof is 74° 34'

10.0000000

9.8842540 9.8980060

9.7822600

the Distance re

quired in the Arch of a great Circle, viz. A B = 5182.2 English Miles nearly; and 4674 Italian ones.

2. To find how they bear from each other; fay,

As Radius

10.0000000

9.8080675

To the Co-fine of 50° 00' AP So is the Tangent of 52° 15' = APC 10.1111004 To the Co-tangent of 50° 18' CAP 9.9191679 That is, Peckin bears N. E. and 5° 18' eafterly of Cittadella; confequently Cittadella bears S. W. and 5° 18 more wefterly of Peckin. Whereas according to the Rhumbs, they lie directly Eaft and Weft of each other. Such a Difference there is between the Pofition

of

of a Rhumb and a great Circle on the Surface of the Globe.

PROBLEM IV.

To find the Distance and Bearing of two Places, the one lying under the Equinoctial, the other in any given Latitude; their Difference of Longitude being given also.

First draw the proper great Circles (one of which LIO, paffeth through London and the faid Island, ) and you will have formed the Right-angled Triangle 102; in which there is given the Side O 2

ÆL

51° 32′ the Latitude of London; and the Side 12 80', the Complement of the Difference of Longitude Æ I to 180°, or a Semicircle.

1. To find the Side IO, the Complement of the Distance from London to 180°, or a Semicircle. Thus

To the Co-fine of O I= 83° 48'

Which fubftract from

180° 00'

There remains LI= 96° 12' the Arch of Diftance between the two Places; which reduced is 6685 English Miles; and fo far is Sumatra from London the nearest Way that even a Bird could fly to it.

2. To find the Angle at O = PLI, the Point on which Sumatra bears from the N towards the E of London. Thus,

As the Sine of O I= 83° 48' Is to Radius

So is the Sine of 12,= 80° oo!

9.9974523

10.0000000

9.9933515

To the Sine of the Angle at 0 = 82° 08′ 9.9958992 Thus Sumatra lies E by N, and 3° 23' to the East of London.

3. To find the Angle at I, the Point on which London bears from the W to the N of Sumatra. Thus,

As the Sine of O I= 83° 48′

Is to Radius

So is the Sine of 20 =
= 51° 321

9.9974523 10.0000000

9.8937452

To the Sine of the Angle at I= 51° 57′ 9.8962929

Thus Sumatra hath London bearing from it NW, and 6° 57' to the North.

PRO

PROBLEM Y.

Given the Difference of Latitude and Longitude of any two Places (on one Side the Equator,) to find their Diftance and Pofition from each other.

don be 51° 321, the Latitude of Jerufalem 32° 44′; and their Difference

of Longitude to be 37° 24'; I demand their Diftance from

each other?

Draw the Circles and prepare the Scheme as here you fee it, and you will have formed the Oblique Triangle LP I, in which there is given the Side LP

38° 28', the Side IP 57° 16'; and the Angle at P37° 24' the Difference of Longitude; to find the Side IL. In order to this, let fall the Perpendicular AI, then in the Right-angled Triangle API, find AP, thus;

VOL. II.

Q9 2

As

As Radius

10.0000000

Is to the Co-fine of P

So is the Tangent of IP = 57° 16′

37° 24'

9.9000472

10.1919171

51° 01' 10.0919643

Then AP-LP = AL 12° 33'. Now fay;

As the Co-fine of AP = 51 or C. Ar. 0.2012842 Is to the Co-fine of AL = 12 33

= 57 16

9.9894973

So is the Co-fine of IP

9.7329803

To the Co-fine of IL

32 58

9.9237618

This 32° 58' reduced, make 2291 English Miles; and fo far is Jerufalem diftant from London, according to the latest Obfervations.

Note, The Difference of Longitude by many Authors and Maps is made 50° 24', and then the Arch IL would be found 40° 49' 2837 Miles; 546 more than it is now known to be; for this Reaton I did not follow the old Account of the Longitude of Jerufalem; though in moft other Places I was obliged to do it for want of better.

2. To find the Pofition of Jerufalem from London or the Angle ALI. Thus ;

As