350' 
THE CONSTRUCTION' OF 
then it will be, as the sine of FCA (TCH) : radius 
(:: TH : CH : : DH : CH) :: the sine of 1ICD : the 
sine CDH ; therefore in the triangle HCD tliere will 
be given all the angles, and the side CD, whence CH 
and HD will be known. 
PROBLEM XLir. 
Having given AB ,and also AD and BG, perpendicular 
to AB; to find a point T in AB, to which if two right- 
lines DT, GT be drawn, the angle DTG, formed by 
those lines, shall be the greatest possible. . 
CONSTRUCTION. 
Describe by the last problem, a circle GDQ, that 
shall pass through G and D and touch AB, and the 
point of contact T will be the point required. 
DEMONSTRATION. 
Join G, T and D, T; and from any other point 
R in the line AB, draw RG 
and RD; also, from the 
point Q where GR cuts the 
circle, draw QD: then, the 
angle GQD, being exter 
nal with regard to the tri 
angle DQR, will be greater 
than GRD; therefore GTD, 
standing inthesame segment 
with GQD, will be also great 
er than ¿RD. Q. E. I). 
METHOD OF CALCULATION. 
Draw DE parallel to AB ; then in the triangle GDE 
will be given DE, EG ( — BG — AD) and the right- 
angle DEG, Whence the other angles EDG, EGD, 
and the side DG will be found; then in the triangle 
CFP, similar to GDE, will be given all the angles and 
the side FP ^ — ^ whence FC will be given; 
from which, by prpeeeding as in the last problem, all: 
the rest will be found.