Stèb ' ' 
THE CON.'TJitJCTION OF 
V> 
Prob. £G.) by saying, as the sum of the said sines is to 
their difference, so is the tangent of half BAG to the 
tangent of half BA Q — CAQ. 
Agaui, it will he as sine QAH : sine QUA (: : Qll 
: QA :: QP : QÀ) :: sine QAP : sine QPA; there 
fore, in the triangle AQP, are given ail the angles and 
one side AP, whence AQ and PQ will be found. 
PROBLEM XLVII. 
Having the three perpendiculars, let fall from the angles 
of a plan eTnangle on the opposite sides, cjual to three gi 
ven right-lines kk, LI, and Mw?, to describe the triangle. 
CONSTRUCTION. 
Draw thè indefinite right-liue RS, in which take 
AB equal to K/r : find a fourth proportional to Mw, 
LI, and K/r, with which as a radius, Irom the centre A, 
let. an arch rCs be described; and from B, with the 
> radius LI, let, another arch be described intersecting the 
former jn C ; join A, C and B, C, and upon ItS let 
fall the perpendicular QC, in which produced, take 
QP zz LI, and draw PF parallel to US, meeting AC, 
produced, in F, draw FG parallel to CB, and AFG 
will be the triangle required. 
DEMONSTRATION. 
Draw FE, Gq and Av, perpendicular to the three 
sides of the triangle.