GEOMETRICAL PROBLEMS, 
319 
to the sine of (COF rr CBG) the difference of the angles 
at the base; whence the angles themselves are given. 
A fter the same manner a segment of a circle may be 
described to contain a given angle, when that angle is 
greater than a right one, if, instead of BO being drawn 
above AB, it be taken on the contrary side. 
problem v. 
Having given the base, the perpendicular, and the-angle 
at the vertex of any plane triangle, to construct the tri 
angle. 
CONSTRUCTION - . 
Upon AB the given base (see the preceding figure) let 
the segment ACGB of a circle be described to contain 
the given angle, as in the last problem; take EF equal 
to the given perpendicular,and draw FC parallel to A B, 
cutting the periphery of the circle in C; join A, C and 
B, C, and the thing is done: the demonstration whereof 
is evident from the last problem. 
METHOD OF CALCULATION. 
In the triangle EBO are given all the angles and the 
side EB, whence EO will be known, and consequently 
OF ( — DC — EO); then it will be as EB : OF :: 
the sine of EOB (the given angle at the vertex) to the 
sine of OCF, the complement of (COF or CBG) the 
difference of the angles at the base ; whence these an 
gles themselves are likewise given.—This calculation 
is adapted to the logarithmic canon ; but by means of a 
table of natural sines, the same result may be brought 
out by one proportion only : for BE being the sine of 
BOE, and OE and OF co-sines of BOE and COF 
(answering to the equal radii OB and OC) it will there 
fore be, BE : EF :: sine BOE (ACB) : co-sine BOE 
+ co-sine COF; from which, by subtracting the co-sine 
of BOE, the co-sine of COF (—CBG) is found. 
PROBLEM vi. 
The angle at the vertex, the sum of the two including 
sides, and the difference of the segments of the base being 
given, to describe the triangle.