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

SCHOLIUM I.

FRrom this Proposition flows first of all the chiefest part of Trigonometry for the Resolution of Right Angled ▵▵: For since in any Right Angled Triangle, if one side, e. g. AB (Fig. 64.) be put for the whole Sine, the other BC will be the Tangent of the opposite Angle at A (and in like manner if CB be the whole Sine, BA will be the Tangent of the Angle C;) but if the Hypothenuse AC be made Radius or whole Sine, then the Side BC will be the Right Sine of the Angle A, or the Arch CD described from the Center A, and AB the Right Sine of the Angle C, or the Arch AE, described from the Center C, (we will omit mentioning the Secants, because the business may be done without them) which all follow from Def. 10. Wherefore you may find,

• I. The Angles.
  • 1. From the Sides by inferring As one leg to the other, so the whole Sine to the Tangent of the Angle opposite to the other Leg.
  • 2. From the Hy∣poth. & one side, by inferring As the Hyp. to the W.S. (whole sine) so the given leg to the S. of the opp. angle
• II. The Sides.
  • 1. From the Hy∣poth. and Angles: As the W. S. to the Hypoth. so the Sine of the Angle, opposite to the Leg sought, to the Leg it self.
  • 2. From one Leg and the Angles: As the W. S. to the given Leg, so the Tan∣gent of the Angle adjacent to it, to the Leg sought.
  • 3. From the Hypoth. and one of the Sides: Having first found the Angles, it's done by the 2, 1. or by the Pythagorick Theorem.

• III. The Hypothenuse.
  • 1. From the An∣gles and one of the Legs. As the S. of the Angle, opposite to the gi¦ven Leg, to that Leg, so the W. S. to the Hypoth.
  • 2. From the Legs given; Having first found the Angles its done by the 1. or by the Pythagorick Theorem.

III. Inversly also, if two Triangles ABC and ABC (〈◊〉〈◊〉 the Figure of the present Proposition) have one Angle of o•••• equal to one Angle of the other (e. g. B and B) and the Sid•••• that contain these equal Angles proportional (viz. as AB to B•• so AB to BC) then the other Angles (A and A, C and C will be also equal, and the Triangles similar(α) 1.1 for to 〈◊〉〈◊〉 like Chords AB and AB, BC and BC, there answer by t•••• Hypoth. like or similar Arches, i. e. equal in the number 〈◊〉〈◊〉 Degrees and Minutes; and to these also there answer equal A••¦gles both at the Periphery and Center.

IV. (Fig. 65. N^o 1.) If(β) 1.2 the Sides of the Angle BA•• are cut by a Line DE, parallel to the Base BC, the Segments 〈◊〉〈◊〉 those, Sides will be proportional, viz. AE to EC as AD to BD•• for by reason of the Parallelism of the Lines BE and BC, th•• Triangles ADE and ABC are Equiangular: Therefore as th•• whole BA to the whole AC, so the part AD to the part A•• and consequently also the remainder EC to the remainder D•• as the part EA to the part AD, by Prop. 26. and alternative•••• by Prop. 24. EC will be to EA as BD to AD.

Notes

• (α) 1.1

  Eucl prop. 6. lib. 6.
• (β) 1.2

  Eucl. 2. lib. 6.