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HAving given for a rigbt-angled Triangle one segment of the base and the side adjacent to the other segment, to find rest and construct the Triangle.
As if the lesser segment of the base DB be given (Fig. 39.1.) and the side AC adjacent to the other segment; to find greater segment of the base, which being found the rest easily obtain'd, and consequently the whole Triangle.
Make the greater segment = b, the given side = c, the segment sought = x. Now if we suppose the triangle ABC to be already found, it is evident, 1. If from the square of AC you substract the □ AD, you'l have the □CD=cc−xx. 2. The same □ CD may also be otherwise hence obtain'd, be∣cause, the angle at C being a right one, CD is a mean pro∣portional between BD and DA, i. e. between b and x; whence the rectangle of the extremes bx is = □ of the mean CD. Wherefore now it follows, 3. that cc−xx=bx; and adding xx, cc=bx+xx; and substracting bx,−bx+cc=xx. Therefore according to case 2. x=−½b+√¼bb+cc.
Geometrical Construction. Join EF=½b (n. 2. Fig. 39.) and FA=c at right angles, and having described a Circle from E thro' F draw AEB; so you'l have DA the greater segment and DB the less; having erected therefore a perpendicular from D, and described a semi-circle upon AB, you'l have C the vertex of the triangle sought, whence you are to draw the sides AC and BC.
The Arithmetical Rule. Join the □ of half the given seg∣ment, and the □ of the given side into one sum; and having extracted the root of it, if you thence take half the given seg∣ment, you'l have the segment sought.