Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

Another Solution of the same Problem.

This Problem may be more easily solved, and will give a far more simple Equation, if you are to find not FD but AE. Make therefore (n. 1. Fig. 53.) = x, and the rest as above; AC will be = x+b, and its □xx+2bx+bb; therefore the □BC=xx+2bx+bb−aa; therefore the □ of the tan∣gent HC=xx+2bbx+bb−2aa. But the rectangle ACE will be bb+bx. Therefore by Prop. 47. Lib. 1. Math. E∣nucl.

〈 math 〉〈 math 〉; and turning all over to the right hand, 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉. Therefore by case 2 of affected quadraticks, 〈 math 〉〈 math 〉.

The Geometrical Construction may be performed according to the rules of quadraticks Fig. 54. n. 1. as will be evident to any attentive Reader. Having therefore described a circle at the interval BA, whether it be done from any arbitrary cen∣ter (see n. 4. Fig. 53.) or upon AE found in the pres. Fig. making an intersection at the said interval in B; and apply∣ing AE, and producing it until EC become equal to the gi∣ven quantity q, and at length having drawn BC you'l have the triangle right-angled at B, and also the difference of the sides FC. But to make it more short and elegant; having determined AE by a little circle (Fig 54 n. 3.) prolong it to the opposite part of the circumference in C, and draw AB and CB; for as the radius of the little circle is ½b, so EC is =b.

Now if you would construct the Equation above by Baker's rule (that its universality may also be confirm'd by an Exam∣ple in quadraticks) viz.

〈 math 〉〈 math 〉;

The centtal rule will be (taking a for 1 and the L. R.) 〈 math 〉〈 math 〉 = AD

[illustration]

LIV

LV

LVI

LVII

〈 math 〉〈 math 〉 = DH. i. e. by our Reduction, 〈 math 〉〈 math 〉 = AD 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

The Construction therefore will consist in these: LM (n. 2. ••g. 54)=a, MO=¼b, LP=½b, therefore PR=〈 math 〉〈 math 〉: ••N=PR viz. 〈 math 〉〈 math 〉, therefore Q=〈 math 〉〈 math 〉. In Fig. 53. n. 3. ••=¼b, A1=½LM, 1, 2, or 1D=PR; D1=LP MO, 1, 2, or 1H=PQ Having drawn a circle from 〈◊〉〈◊〉 thro' A you'l have the true root RO= to AE sought; ••d MO the false root.

NB. Hence it is evident that one Problem may have seve∣••l Solutions and Constructions, some more easy and simple, ••hers more compound and laborious; viz. according as the ••••known quantity is assumed more or less commodious to the ••urpose: which may not be amiss here to note for the sake 〈◊〉〈◊〉 Learners.