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In the Triangle A B C, we have this Analogy ;

As Radius
To the Co-tangent of B = 50° 561 9.9094022
So is the Tangent of AC =
of AC = 20° 34'

9.5742761 To the Sine of B A = 17° 43' 9.4836783

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Now the Motion of 17° 43' is performed in i
Hour 10' 56"; Whence the Sun is due East in the
Morning at 7 Hour 10'56", and at 4 Hours 49'04"
in the Afternoon, he will be due West.
In the Triangle C P Z, we have this Analogy;

As 1 CP:R::tZP:cs P = 72° 17'.
This reduced to Time gives 4 Hours 49' 04", the
Time in the Afternoon ; and this subducted from 12
Hours, leaves the Time 7 Hours 10' 561 in the
Morning ; both as before.


The Latitude of the Place, and the Sun's Declination, given ; to find his Altitude when due East or Weft.


In the two Triangles A B C, or C P Z ( fupposing the Time and Place, the same as in the last Problem ) there is given in the first Triangle ABC, the Side AC = 20° 34' the Declination, the Angle B = 50° 56' che Latitude of Chichester ; to find the Side B C, the Altitude required.




Analogy. As the Sine of B = 50° 56' 9.8900929 Is to the Sine of AC = 20° 34' 9 5456745 So is Radius

10.0000000 To the Sine of the Altitude BC=26° 54' 9.6555816

In the Triangle C P Z, there is given C P, the CoDeclination, and Z P the Co-Latitude, to find the Side Z C, the Co-altitude required. For which you have this Analogy; viz. C SZP:R::csCP:cs CZ = 639 061.


The Latitude of the Place, and Declination of the Sun given, to find the Sun's Azimuth at the Hour of Six.

PraEtice. Suppose the Latitude that of Chichester 500 56', and the Sun's Declination 200 34' as it will be on the 12th of May, A. D. 1735. Then, the Diagram being prepared, you

H will have therein formed, the two Rightangled Triangles ABC, and CZP, by either S of which this Problem is fatisfied.

For in the Right Triangle A BC, there is the Side BC = 20° 34' the Declination, and the Angle at B = 509 56' the Latitude of Chichester, both given ;


to find the Side A B, the Azimuth of the Sun from the East or West Points of the Horizon.

The Analogy is As Radius

10.0000000 To the Tangent of BC = 20° 34' 9.5742761 So is Co-sine of B = 50° 561

9.7994951 To the Tang. of the Azim. AB=13° 18' 9.3737712

Wherefore the Azimuth of the Sun at fix in the Morning is about E by N and ; and at six in the Evening, W by N and

Also in the Triangle C Z P, there is given the Side CP = the Co-Declination, and ZP = the Co-Latitude ; to find the Angle at Z = the Azimuth from the North = AO.

The Analogy for this is,
is P.Z:R::tCP:tZ = 768 42'.

Or varied thus,
+ CP:R::SPZ:ctZ = 139 18'.

Or thus,
R:ct CP::sPZ:ct Z.


The Latitude and Declination given, to find the Altitude of the Sun at the Hour of fix.

Practice. In the foregoing Diagram, and in the Triangles A B C and C P Z there are given the same Things as in the last Problem, to find the Side A C in the first, and C.Z in the latter ; either of which answers the Problem.

To find the Altitude AC, this is the Analogy. VOL. II.

Hh 2


As Radius

10.0000000 Is to the Sine of BC= 20° 34' 9.5456745 So is the Sine of B = 50° 56'

9.890og29 To the Sine of the Al. at six AC=15° 49'59.4357675

In the Triangle C P Z to find CZ = Co-altitude, this is the Analogy ;

R:cs P Z::05 CP:csC Z= 74° 11'.

Note, The Altitude in Summer, is equal to the Depresion in Winter.


Given the Latitude of the Place, the Declination, and Hour of the Day, to find the Sun's Altitude.




Case 1. Let the Declination of the Sun be North 200 34' at the Hour of 10 in the Morning, or 2 in the After noon (on the 12th of May, 1735 ) and in the Latitude of Chichester 50° 56', I demand the Height of the Sun at that Time ?

N According to these Data, let the Scheme be constructed as hath been heretofore taught ; then there will be formed the Oblique Spherical Triangle CZP.


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In this Triangle there is given the Side ZP = 390 04', the Co-latitude ; and the Side CP = 69° 26' the Co-declination ; and the included Angle at P= 30° oo', the Time to Noon ; to find the Side CZ the Co-altitude required.

This is found by Cafe 4, in the Synopsis, thus ;

As the Radius
To the Co-sine of P= 300 001
So is the Tangent of ZP=

39° 04'
To the Tangent of PL = 35° 06'
Which substract from C P = 69° 26'
There will remain CL = 349 200

10.0000000 9.9375306 9.9094022 9.8469328

Wherefore fay again ;

As the Co-sine of LP=359 06! Co. Ar. 0.0871672 Is to the Co-fine of ZP=39°04' 9.8900929 So is the Co-line of LC=34° 201 9.9168593 To the Co-sine of CZ=38° 25' 9.8941194

The Complement of which to 90° = 2 B, is CB = 51° 35', which therefore is the Altitude of the Sun the Time specified.

Case 2. Let the Declination of the Sun be South 20° 34'; the Latitude, and Hours as above, to find the Altitude


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