328 
TIfE CONSTRUCTION OF 
METHOD OF CALCULATION. 
Let CA ami CD be expressed by the numbers exliF 
biting the given ratio of the sides’: then in the triangle 
ACD will !>e given two sides and the included angle 
AC1); whence the angle CAL (CGF) and CEA (CFG) 
will be given, and from thence the sides CG and CF. 
PROBLEM XV. 
The base, the perpendicular, and the difference of the 
angles at,the base, being given, to construct the triangle. 
CONSTRUCTION. 
Bisect the given base AB by the perpendicular DF, 
in which take DF equal to the given height of the 
triangle; draw CFGH parallel to AB, and make the 
angle EDH equal 
to the given dif 
ference of the an 
gles at the base ; 
draw EAQ, and 
take Q therein, 
so that QD =r 
£)H; and paral- 
p lei to QD, draw 
■k AO, meeting DE 
in O; upon O, as 
a centre, with the 
* radius OA, describe the circle AGFCB, and from the 
point G, where it cuts the right-line CI1, draw GA and 
GB; then will AGB be the triangle required. 
DEMONSTRATION. 
Let OG and BC be drawn. Bv reason of the paral 
lel lines QD and AO, it will be QD (DH) : AO (OG) 
:: ED ; EO; therefore the two triangles EHD, EGO, 
having one angle E, common, and the sides about the 
other angles D and O proportional are equiangular 
(Euc. 7. C.) and consequently EOG = EDH. More 
over, because DOEF is perpendicular both to AB and 
GC, and AD equal to CD, it is evident that the circle 
passes through the point B, and that the arches FC,l G, 
F