Prob. 106. The Sun being due Eaft at feventeen Minutes paft Eight in the Morning, and his Declination 19°. 57′ North, given, to find the Latitude of the Place.

Conftruction. With the Chord of 60°, draw the Meridian EPQS; draw EOQ the Equinoctial, and POS the Axis at right Angles to it; lay 19°.57 from Æ to a and from Q to b; alfo lay the Half-Tangent of 19°.57′ on the Axis from O to c, and, by Prob. 17, draw the Parallel of Declination a cb; fubtracting eight Hours and feventeen Minutes from Twelve o'Clock,

or Noon, there remains three Hours and forty-three Minutes for the Time the Sun wants of coming to the Meridian EPQS; convert this three Hours and forty-three Minutes into Degrees and Minutes, by allowing 15° to an Hour, and 15 of the Equinox for one Minute of Time, and it is 55°.45; lay, by Prob. 42, Cafe 2, this 55°.45' on the Equinox from Q to D, and, by Prob 17, draw the Meridian PDS, interfecting the Parallel of Declination a cb in O, which is the Sun's Place at the given Time; then thro' the Point draw the right Circle ZOON, which is the Prime Vertical of the Place, and the two Points Z and N are the Zenith and Nadir. In the Triangle OOD, right-angled at D, given the Bafe OD 34°. 15' the Complement of 55°.45 to 90°, and the Perpendicular DO 19°. 57', to find the Angle at O, being the Distance of the Zenith from the Equinox, which is always equal to the Latitude.

This is an Extreme conjunct, and the Angle at O, being one of the Extremes, may be found by Lord Nepier's fecond Rule, and its Note, Page 266,

As the Tangent of DO, 19°. 57',
Is to the Radius,

So is the Sine of O D, 34°. 15',

Co. Ar. 0.440115

10.

9.750358

To the Co-tangent of O OD, 32°. 49 ́s

10.190473

The

The Declination being North, and the Sun being due East after he was above the Herizon, the Latitude is 32°.49′ North.

The Angle at O, being made at the Center by the right Circles EOQ and Z ON, may be measured by Prob. 53.

If a right Circle were drawn at right Angles to the right Circle ZON, it would be the Horizon of the Place, above which P, the North Pole, would be elevated 32°. 49′.

Prob. 107. The Sun due Eaft at forty-fix Minutes after Seven in the Morning, and his Altitude 28°. 53' at the fame Time, given, to find the Latitude of the Plate.

Conftruction. With the Chord of 60°, draw the Meridian PQS; draw OQ the Equinoctial, and POS the Axis at right Angles to it; then converting the Time from Six o Clock, one Hour and forty-fix Minutes, into Degrees and Minutes, it makes 26°. 15', which lay from O to B, by Prob. 42, Cafe 1, and draw the Meridian PBS, by Prob. . 17; then take the Half-Tan

N

P

2

B

Q

Ο

gent of the given Altitude 28°.53 from the Scale, fet one Foot of the Compaffes in the Center O, and with the other interfect the Meridian PBS in the Point ; thro' the Point draw the right Circle Z OON, the Prime Vertical of the Place, which completes the Triangle OO B, rightangled at B; where we have given the Hypothenuse Ŏ O 28.53, and the Bafe O B 26°. 30', to find the Angle at O, the Latitude of the Place.

This is an Extreme conjunct, and the Angle at O, being the middle Part, may be found by Lord Nepier's first Rule, and its Note, Page 265:

As the Radius

Is to the Co-tangent of OO, 28°.53',
So is the Tangent of O B, 26°. 30',

10.

10.258336

9.697736

9.956072

To the Co-fine of OO B, 25°. 20',

Therefore the Latitude is 25°. 20', and may be either

North or South.

Prob.

Prob. 108. The Sun being in the Equinoctial, his Azimuth 24° 2' at Seventeen Minutes after One O'Clock in the Aftermaon given, to find the Latitude of the Place.

H

P

N

R

S

Conftruction. With the Chord of 60, draw the Meridian EPQS; draw the quinoctial EOQ, and POS the Axis at right Angles to it; by Prob. 42, Cafe 2, lay 190.15 (that is, one Hour and feventeen Minutes, the Time from Noon) on the Equinoctial from Q to Q, and, by Prob. 17, draw the oblique Circle POS; then, by Prob. 59, draw an oblique Circle thro' the Point, to make an Angle of 24.2, the given Azimuth, with the Meridian, interfecting it in the two Points Z and N, the Zenith and Nadir of the Place; draw ZON the Prime Vertical, and the Horizon HOR at right Angles to it, interfecting the oblique Circle ZON in a, and forming the right-angled fpherical Triangle Oa, right-angled at a; in which is given the Hypothenufe OO 70°.45', the Complement of Q, the Time from Noon, and the Bafe Oa 65°. 58′, the Complement of the given Azimuth, to find the Angle OO a, the Complement of ZOQ, the Latitude of the Place fought.

This is an Extreme conjunct, and the Angle at O, being the middle Part, may be found by Lord Nepier's first Rule, and its Note, Page 265:

As the Radius

Is to the Co-tangent of OO, 70°.45′

So is the Tangent of Oa, 65°58',

To the Co-fine of Oa, 38°.27', the Complement of Latitude,

-10.

9-543094

10.350737

i 9.893831

Therefore the Latitude is 51°. 33' either North, or South,

as the Obfervation is made either on the North or South

Side of the Equinox ; but the Figure is here made for North Latitude.

The Angle at O, being made at the Pole of the primitive Circle, may be measured by Prob. 47. If it were required to find the Altitude, the fame Parts being given, the Perpendicular a may be found by Lord Nepier's fourth Rule, and its Note, Page 266, as the required Perpendicular is one of the Extremes disjunct.

Having gone thro' the principal Aftronomical Problems folvable by right-angled spherical Trigonometry, I fhall now add fome useful Problems in oblique Trigonometry.

Prob. 109. The Latitude of the Place 51°. 32′ North, the Suu's Declination 22°. 47′ North, and his Altitude 49°. 29′, given, to find his Azimuth.

Conftruction. With the Chord of 60°, draw the Meridian H-Z RN; draw HOR the Horizon, and at right Angles to it ZON the Prime Vertical; lay 51°.32', the Latitude, from N to E, and from Z to Q; then draw EOQ the Equinoctial, and at right Angles to it POS the Axis; by Prob. 61, draw the Parallel of Declination acb, at 22°47? Distance from the Æquinoctial;

H

b

Z

S

R

and the Parallel of Altitude d'ef, at 49°. 29′ Distance from the Horizon, interfecting the Parallel of Declination acb in O, the Sun's true Place; through which, by Prob. 17, draw the Meridian PS, and the Azimuth-Circle Z ON at right Angles to it, which will complete the oblique-angled Triangle ZOP; in which is given the Side ZO 40°. 31, the Complement of the given Altitude, the Side P Z 38°. 28, the Complement of the given Latitude, and the Side PC 67°.13′, the Distance of the Sun from the elevated Pole, which, in this Cafe, is the Complement of the given Declination, to find the Angle PZ O, the Sun's Azimuth from the North Part of the Meridian.

*

Here are the three Sides given to find an Angle, which may be done by Prob. 90, where we have fhewn the Solution of

Tt

this

this Problem, both by letting fall a Perpendicular, and allo by a Method without a Perpendicular, as follows:

Sine of 60. 29

9.939638

Which, being doubled, gives 120° 58′ for the Angle PZO, the Sun's Azimuth from the North Part of the Meridian; and this, being fubtracted from 180°, the Remainder, 59.2, is the Sun's Azimuth from the South, being the Angle RZO, which is made at the Periphery, and may be measured by Prob. 54:

In the fame Manner may the Angle at P be found, which is the Time from Noon. The two containing Sides in that Cafe will be PZ and PO, the Arithmetical Complements of their Sines added to the Sines of the Half-Sum and Remainder, as before: The Half of that Sum will be the Sine of 18°.35'; whole Double, 37°. 10, is the Angle Z PO or QPO, the Time from Noon; which, being converted into Time, is two Hours, twenty-eight Minutes, and forty Seconds; therefore, if it were in the Forenoon, it is thirty-one Minutes and twenty Seconds after Nine o'Clock; but, if in the Afternoon, twenty-eight Minutes and forty Seconds after Two.

Thus this Triangle contains the Complement of the Latirude in the Side P Z, and the Complement of the Sun's Declination in the Side PO, when the Declination and Latitude