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2 Probl. Two sides with the angle comprehended by them given, to find ei∣ther of the other angles.
In the Oblique angled plain Tri∣angle BDC.
The angle DBC is inquired.
The given Sides
The given Angle DCB
Subtract the angle given, out of 180d. the remainer is the sum of the other angles, and ½ Z c••ur. DC and BC. ½ X crur ∷ t ½ Z ang. t ½ X ang. by the 3 Ax. To the half sum of the other angles, adde the half difference found, and you will have the greater angle, sub∣duct it & you will have the lesser, in our example ½ Z ang. + ½ Xang. is = DBC sought, because that is the greater ang.
3 Probl. The angles and a side gi∣ven, to find either of the other sides.
In the Oblique angled plain triangle CBD,
The side DB is inquired.
The given Angles
The given Side CB
s. CDB, CB ∷ s. DCB. DB. by the second Axiome.
4 Probl. Two sides with the angle comprehended by them given, to find the third side.
In the oblique angled plain triangle DCB
The side CB is inquired.
The given Sides
The given Angle CBD
First, find the angle BCD by the third Axiome.
Secondly, find the side CB by the se∣cond Axiome.
5 Probl. The three sides given to find an angle.
In the Oblique angled plain trian∣gle CBD.
The angle CDB is inquired.
The three sides be∣ing given.
The resolution of this Problem doth require two operations, first, for the segment of the base GB.
DB.Z crur. DC & CB ∷ X crur. GB. ••y the fourth axiome.
And DB − GB = DG, and ½ DG =DA or AG.
And DC. Rad ∷ AD. s ACD, by the second axiome whose complement ••s ADC.
And AG + GB = AB, therefore ang. ABC may be also found in like manner.