BOOK ONE.

size, namely GFC and GFD, then you will have the external flanking angle CFD.


GFC is	75 degrees.
The double: 150	the Angle CFD.

5. The double of the interior flanking angle (that is to say, the two angles of the same size BCH and FCG together) being subtracted from the angle of the figure BCD, there will remain the flanked angle HCF.


The interior flanking angle FCG	15 degrees.
The double	30.
The angle of the figure BCD	90 degrees.
	30 removed.
The flanked angle HCF remains	60.

A large technical plate containing eleven geometrical diagrams labeled No. 4 through No. 14, plus a small figure labeled Fig: B. The diagrams illustrate various polygons (squares, pentagons, hexagons) and their internal geometric constructions for fortification, showing angles, radii, and intersecting lines within circles and squares. Points are labeled with capital letters (A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R).

TO FIND THE LINES.

1. The side is given as 100 feet, of which the half will be CG, namely 50 original: 50 followed by a circled dot; this symbol likely denotes a specific unit or decimal place in 17th-century mathematical notation..
2. In the right-angled Triangle FGC, with CG being the Radiusoriginal: "Raid." In 17th-century mathematics, the "Raid" (Radius) or "Sinus Totus" was the base value—often 100,000—used in trigonometric tables to calculate proportions., FG will be the Tangent, and CF the Secant, of the interior flanking angle FCG.


To find FG. The Tangent of the angle FCG of 15°	26795.
Multiplied by CG	50000 original: followed by a circled 3, a notation for decimal precision.
Gives the product.	1339750000.
Which being divided by the Radius	100000.
FG will come to	13397 original: followed by a circled 3. FG.
To find CF. The Secant of the angle FCG of 15°	103528.
Multiplied by CG	50000 original: followed by a circled 3.
	Gives