GEOMETRICAL PROBLEMS. 
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The reason of which is evident, at one view, from tha 
construction. The metho l of solution is also very easy; 
for it will be, as AB is to AD — DB, so is the si ieof 
AC8 to the sine of B — A (see Problem 4.), whence B 
and A, and also BCD and ACD are given. 
PROBLEM XXVIII. 
To divide a given angle ABC i >to two parts, so that 
their secants may obtain a given ratio, 
CONSTRUCT I OX. 
Take BE to BT in the given ratio of the secants* 
join T, E, and let BF be drawn perpendicular to 
ET, and the thing is done. The truth of which is 
manifest, from the construction. 
METHOD OF CALCULATION. 
The angle EBT and the ratio of the sides BE, and 
BT being given, the angles E and T will al-o be given, 
and consequently their complements EBF and FBF. 
PROBLEM XXIX. 
From a given point O, to draw a right line OF, to cut 
tico right lines AC, AB, given by position, so that the■ 
parts thereof, OE, OF, intercepted between that point 
and those lines, may be to one another in a given ratio. 
CONSTRUCTION. 
From O, through A, the point of concourse of BA 
and CA, let OAD be drawn, in which take AD to AO 
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