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In the Triangle APZ, there is given the Side* 1.1 ZP 38° 30', the Side PA 70°, and the Angle P, let be 31° 34, and the Side AZ is required.
The Resolution of this Case depends on the Catholike proposition of the Lord of Marchiston, by supposing the Oblique-Triangle to be divided (by a supposed Perpendicular falling either within or without the Triangle) into two Rect∣angulars.
Now in the Triangle AZP, let fall the Per∣perpendicular ZR; so is the Triangle AZP divi∣ded into two Rectangulars ARZ and ZRP. Now the Side AZ may be found at two Opera∣tions thus: say,
As the Radius or S. of 90° 00'
To Sc. of the included V, P. 31 34.
So is T. of the lesser Side PZ. 38 30,
To T. of a fourth Arch. 34 08.
If the contained Angle be less than 90°, take this fourth Arch from the greater Side; but if it be greater than 90°, from its Complement unto 180°, the Remainder is the Residual Arch: Now again say,
As Sc. of the fourth Arch. 34° 08'
To Sc. Residual Arch. 35 52
So Sc. of the lesser Side PZ. 38 30
To Sc. AZ the Side required. 40 00