geometrical problems. 
373 
construction) as KL<7 : ACq ( : AC : Iv) :: the sine 
of AOC or AEB, the given difference of the angles at 
the base, to the sine of SOI, which added to AOS, gives 
AOI, whose supplement, divided by 2, will be GIG; 
from whence OGI and its supplement OGA are given, 
and consequently ANM (equal to AGE); then in'the 
triangle ANM will be given AN, NM, and the in 
cluded angle ANM, whence the angles M, A r P, will 
3E0 be given. . , 
PROBLEM LX. 
The perpendicular, the angle at the vertex, and the 
sum of the three sides of a triangle being given; to de 
scribe the triangle. 
CONSTRUCTION. 
"Make AB equal to the sum of the sides, which bi 
sect in P, making PO perpendicular to AB, and the 
angle PAO equal to half the given angle at the ver 
tex; from the cen 
tre O with the 
radius OA de 
scribe the circle 
AHB, and in OP, 
produced, take PK 
equal to the given 
perpendicular,and 
draw KH parallel 
to BA, cutting the 
circle in H; join A, II and B, H, and make the angles 
BHE and AHE equal to HBF and HAE respectively ; 
then will EHF be the triangle required. 
DEMONSTRATION. 
Join O, B and O, H, and draw HQ perpendicular to 
r The angle EFH is = BHF + HBF = 2HBF {by 
construction) rr HO A (Euc. 20. 3.): and, in the same 
manner is-FEH = IIOB; hence it follows that EFH 
4- FEH ( = HOA + HOB) ~ AOB; and, by taking 
each of these equal quantities from two right, angles, 
b b 3