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CHAP. XV. Having both sides of a rectangle Triangle, to finde the two angles and the base.
AT the intersection of the contrary parallel of the greater side, with the right parallel of the lesser side, there place the threed; which being so placed giveth the lesser angle required. And if to the intersection you place the bead and then apply the threed to the side of the quadrat, it shall there shew the base required.
As in the rectangled Triangle ABC in the 11 Chapter, where the two given sides is AB 80, and BC 60: Wherefore I place the threed at the intersection of the contrary parallel of 80, with the right parallel of 60, so doth the threed give me the lesser angle at A, which is 36 degrees 52 minutes, which being known, the angle ACB will be found by complement, to be 53 degrees 8 minutes, thus have I both the angles required.
The threed lying in the former position, I place the bead to the intersection of the two sides; and then apply∣ing the threed to the side of the quadrat, it giveth me 100 for the base AC, which was required.