(3​1⁄2 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.