Note Note [34 — Literal and other Errors proper to be examined, and corrected with the Pen, before the Book is read.
Reading edition
[34]
To find the Length of the Shadow at half past III.
(See Section 84, Note [25].)
Given Lat. of Chester, 53° 12′ To find Sun’s Alt.
Sun’s Dec. 5 29
Hour III, 30M. 52 30
This is the Case of an oblique spheric Triangle, wherein are two Sides and one Angle between them given, to find the Sun’s Azimuth, and the Sun’s Co-Alt.
Side 84. 31 Sum of Sides 121. 19
Side 36. 48 Diff. of Sides 47. 43
(31⁄2 Hour) Angle contained 52. 30
Half ditto 26. 15 Co. 63. 45
Half Sum of Sides 60. 39 29. 22
Half Difference ditto 23. 51 66. 9
THE FIRST PREPARATIVE PROPORTION.
As Sine of 1⁄2 Sum of Sides 60. 39 0.05966 Co-Ar.
To Sine of 1⁄2 Difference of Sides 23. 51 9.60675
So Co-Tangent 1⁄2 contained Angle 63. 45 10.30703
——— ————
To T. of 1⁄2 Diff. of the other two Angles 43. 15 9.97344
SECOND PREPARATIVE PROPORTION.
As Co-Sine 1⁄2 Sum of Sides 29. 21 0.30968 Co-Ar.
To Co-Sine 1⁄2 Diff. 66. 9 9.96123
So Co-Tangent 1⁄2 contained Angle 63. 45 10.30703
————
To T. 1⁄2 Sum of other Angles 75. 11 10.57794
Half Diff. before found 43. 15
———
Sum, is greater Angle 118. 26 = Sun’s Azim.
Diff. is lesser Angle 31. 56 = S’s right Asc.
Then by first Axiom in Trigonometry, to know the Sun’s Altitude say,
As Sine Sun’s right Asc. 31. 56 0.27659
To Sine Co-Lat. 36. 48 9.77744
So Sine of the contained Angle 52. 30 9.89947
————
To Co-Sine of the Sun’s Alt. 63. 57 9.95350
from 90.
———
Sun’s Alt. 26. 3
Having Sun’s Alt. to find the Shadow,
As Sine Sun’s Alt. 26. 3 0.35738 Co-Ar.
To Person’s Height, 66 Inches, 1.81954
So Co-Sine of the Sun’s Alt. 63. 57 9.95350
————
To Length of Shadow, 135 Inches, 2.13042
Then 6(66⁄135 = 11⁄22 − | − 3⁄6 or 1⁄2, i. e. as 22 to 45: supposing the Length of the Shadow divided into 45 Parts; the Height of the Object woud be 22 of those Parts; or not quite half the Length of the Shadow, at half past III.