are both the fame, as in the prefent Cafe; but, if the Declination and Latitude are, one North, the other South, then the Side PO will be the Sun's Declination added to 90°, as. may be seen by drawing a Parallel of Declination on the contrary Side of the Equinox, as srt; for, in that Cafe, the Sun's Place will be at n, and the Side Pr is the whole Quadrant Pu more than the Declination u r. And, as the Angles at Z and P are the Sun's Azimuth and the Hour of the Day, this one Triangle is fufficient to folve all the feveral Problems wherein three of these five Parts are given to find the other two, The Method of projecting the Triangle and finding the required Parts, in all the different Cafes that can be put, may be found in the Problems of oblique-angled spherical Triangles, from Prob. 80 to Prob. 91 inclufive,

But, as fome of thofe Cafes require the primitive Circle to be the Azimuth or Meridian-Circle, on which the Sun is, and not the Meridian of the Place, it will be proper to give an Example in each of thofe Cafes; for, when a given Side adjoins to a given Angle, that Side must always be laid on the primitive Circle.

Prob. 110. The Sun's Declination 22°.54′ North, and bis Altitude 14° 36' at three Minutes paft Two o'clock in the Afternoon, given, to find the Latitude of the Place: (Suppose South Latitude).

Here are two Sides, viz. the Declination added to 90°, (the Latitude and Declination being contrary Ways) 112°. 54'; and the Complement of Altitude, 75°.24', with the Time from Noon, two Hours and three Minutes; which, being converted into Degrees, is 30°. 45', the Angle oppofite one of the Sides given, to find the third Side, being the Complement of the Latitude.

Conftruction. With the Chord of 60°, draw the Meridian SPQ; draw A OQ the Equinox, and SOP the Axis at right Angles to it; as the Latitude is South and Declination North, lay 22°. 54′ from. Qto O, which will be the Sun's Place, and the Side SO will be. the Sun's Distance from the vifible Pole, or his Declination added to go; by Prob. 48, draw an

Tt2

B

oblique

oblique Circle SZ BP, to make an Angle with the primitive Circle of 30°.45' at the Point S, and SZ BP will be the Meridian of the Place; lay 75°. 24', the Complement of Altitude, from to a, and from O to b; then draw the Parallel of the Sun's Altitude az b, at 75°. 24′ from the Point O, interfecting the Meridian in z, the Zenith of the Place; and an oblique Circle, being drawn from the Point thro' the Point z, will be the Azimuth-Circle on which the Sun is, and completes the oblique-angled Triangle SO z; in which we have given the Side S 112°. 54', the Sun's Declination added to 90°, or his Distance from the elevated Pole, the Side Oz 75°. 24', the Complement of the given Altitude, and the Angle at S 30° 45', the Time from Noon converted into Degrees, &c. to find the Side S z, the Distance of the Zenith from the vifible Pole, being the Complement of the Latitude of the Place.

Here being two Sides, with an Angle oppofite one of them, given, to find the third Side, it may be done by Prob. 87: For, if a Perpendicular be let fall from the End of the given Side S on the Side Sz, by drawing an oblique Circle from the Angle at thro' the Pole of the Meridian Szp, it will interfect the Circle S z P at right Angles in B, and thereby form the two right-angled Triangles SBO and z B O, both right-angled at B; in the former of which are given the Side SO 112°54', and the Angle at S 30°-45', to find the Perpendicular BO; which is an Extreme disjunct, and the required Perpendicular is the middle Part; therefore it may be found by Lord Nepier's third Rule, and its Note, Page 266: But, as the Side S is more than a Quadrant, we must take the Supplement of it to 180°, which is 67°. 6', and the Proportion will be,

As the Radius

Is to the Sine of the Supplement of SO, 67°.6',
So is the Sine of BSO, 30°. 45',

To the Sine of BO, 28°.6',

By the fame Parts given, to find the Side SB, is an Extreme conjunct, and one of the Extremes required; which may be found by Lord Nepier's fecond Rule, and its Note, Page 266: Only obferve, that, in this Cafe, it will be the Supplement of the Side S B, which will be found by Calculation:

As

As the Co-tangent of the Supplement of Co. Ar. 0.374259

SO, 67°.6′,

Is to the Radius,

So is the Co-fine of BSO, 30°.45',

}Co.

10.

9.934199

To the Tang. of the Supplement of S B, 63°. 49', 10.308458

Which, being fubtracted from 180°, gives S B 116°. 11. Then, in order to find S z, the Complement of Latitude, we muft find the Side B z of the right-angled Triangle z BO; in which we have given the Hypothenufe Oz 75°. 24 ́, and the Perpendicular BO 28°.6', by which either of the oblique Angles may be found, as the Angle at z; which is an Extreme disjunct, and one of the Extremes required; therefore, by Lord Nepier's fourth Rule, and its Note, Page 266,

As the Sine of %, 75°. 24',
Is to the Radius,

So is the Sine of BO, 28°.6",

To the Sine of B z O, 29°. 8',

Having found the Angle at z, Hypothenufez, the Side B z

Co. Ar. 0.014255

10.

9.673032

9.687287

with that, and the given may be found by Lord

Nepier's fecond Rule, and its Note, Page 266; being an Extreme conjunct, and one of the Extremes required:

Which, being fubtracted from the before-found SB 116°. 11', leaves Sz 42° 47', the Complement of Latitude; and that, being fubtracted from 90°, leaves 17°. 13' South Latitude, fought.

If the Sun's Azimuth had been required, the Angle at z is the Azimuth from the South, which we have found to be 29°.8'; whofe Supplement to 180° is 151°.52', the Sun's Azimuth from the North Part of the Meridian.

Prob.

Prob. 111. The Altitude of the Sun 11°.45, bis Declina tion 23°. 26' North, and his Azimuth from the South 62°. 16', given, to find the Latitude of the Place.

Here the Side reprefenting the Complement of the Altitude is adjoining to the Angle of the Azimuth; therefore the primitive Circle will be the Azimuth, on which the Sun is at that Time.

Construction. With the Chord of 60°, draw the Azimuth-Circle HZRN; thro' which draw HOR and ZON at right Angles, for the Horizon and Prime Vertical; by Prob. 48, draw an oblique Circle, Z DN, to make an Angle of 62°. 16', equal to the given Azimuth, with the primitive Circle in the Point Z, and Z DN is the Meridian of the Place; lay 78°. 15', the Complement of

R

H

Ꮓ

D

R

P

N

the given Altitude, from Z to O, which will be the Sun's Place at that Time; then lay 66°. 34, the Complement of the Sun's Declination, both Ways, from to a and b, and draw the Parallel of the Sun's Declination a Pb, interfecting the Meridian in P and p; thro' either of thefe Points and ZN, draw the oblique Circle H PO, (or Hp) which will complete the Triangle Z OP, (or Zp); in either of thefe are given the Side Z.78°. 15', the Complement of Altitude, the Side PO (or p O) 66°34', the Complement of Declination, and the Angle at Z 62°. 16', the Sun's Azimuth from the South, to find the Side Z P, (or Zp) the Complement of the Latitude of the Place.

This may alfo be done by Prob. 87: A Perpendicular being let fall from the End of the Side Z O on the Side Z P, (or Zp) by drawing an oblique Circle from O, thro' r, the Pole of the Circle Z PpN, will interfect it at right Angles in the Point D, and form the right-angled Triangles Z OD and POD, (as alfo pO D) all right-angled at D; in the first of which we have the Side ZO 78°. 15', and the Angle at Z 62°. 16', to find the Perpendicular D; which is an Extreme disjuna, and the middle Part required; whence, by Lord Nepier's third Rule, and its Note, Page 266,

A

Then, by the fame Parts given, to find the Bafe Z D, is an Extreme conjunct, and one of the Extremes required; whence, by Lord Nepier's second Rule, and its Note, Page 266,

As the Co-tangent of ZO, 78°: 15",
Is to the Radius,

So is the Co-fine of DZ O, 62°.16',

To the Tangent of DZ, 65'. 55 ́,

Co. Ar. 0.681936

-

10.

9.667786

10.349722

Then, in order to find the Side Z P, the Complement of Latitude, we must first find the Side PD, the Bafe of the right-angled Triangle POD; in which we have the Hypothenufe PO 66°. 34', and the Perpendicular OD 64°.4', by which either of the oblique Angles may be found, as the Angle at P; which is an Extreme disjunct, and the required Angle is one of the Extremes; therefore, by Lord Nepier's fourth Rule, and its Note, Page 266,

Now, by the Angle at P, and the Hypothenufe PO, given, to find the Bafe PD, is an Extreme conjunct, and the required Bafe is one of the Extremes; therefore, by Lord Nepier's fecond Rule, and its Note, Page 266,

As the Co-tangent of P O, 66°. 34',
Is to the Radius,

So is the Co-fine of OPD, 70°.49′,

Co. Ar. 0.363081

10.

9.516657

9.879738