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reason_n angle_n equal_a side_n 2,221 5 9.5367 5 false
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A52262 Geometrical trigonometry, or, The explanation of such geometrical problems as are most useful & necessary, either for the construction of the canons of triangles, or for the solution of them together with the proportions themselves suteable unto every case both in plain and spherical triangles ... / by J. Newton ... Newton, John, 1622-1678. 1659 (1659) Wing N1059; ESTC R17311 30,729 123

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right sine of an arch and the secant of its complement Demonst. In the preceding diagram the triangles AEF AHG are like therefore AF. AE ∷ AG. AH 31 As the sine of an arch or angle is to Rad. so is the tangent of the same arch to the secant thereof Demonst. In the preceding diagram the triangles AEF AGH are like therefore EF. AE ∷ HG AH 32 As Radius is to the secant of an arch so is the co-tangent of the same arch to the co-secant thereof Demonst. In the preceding diagram the triangles ALH and ACK are like therefore LH AH ∷ CK AK Other more easie and expeditious wayes of making the Tangents and Secants you may see in the first Chap. of my Trigonometria Britannica but the Canons being now already made these Rules we deeme sufficient The construction of the Artificial Sines and Tangents we have purposely omitted they being nothing els but the Logarithmes of the Natural of which Logarithmes we have shewed the construction in a former Institution by the extraction of roots and in my Trigonometria Britannica by multiplication and therefore shall now proceed to the use of the Canon of Sines Tangents and Secants in the solution of all Triangles whether plaine or Spherical CHAP. IV. Of the Calculation of plain Triangles A Plain Triangle is contained under three right lines and is either Right-angled or Oblique 2 In all plain Triangles two angles being given the third is also given and one angle being given the sum of the other two is given because the three angles together are equal to two right by the 9th of the second Therefore in a plain right angled triang. one of the acute angles is the complement of the other 3 In the resolution of plain Triangles the angles onely being given the sides cannot be found but onely the the reason of the sides It is therefore necessary that one of the sides be known 4 In a Right-angled Triangle two termes besides the right-angle will serve to find the third so the one of them be a side 5 In Oblique angled Triangles there must be three things given to find a fourth 6 In right-angled plain Triangles there are seven cases and five in Oblique for the solution of which the four Axiomes following are sufficient 1 Axiome In a right-angled plain Triangle The rectangle made of Radius one of the sides containing the right-angle is equal to the rect-angle made of the other containing side and the tangent of the angle thereunto adjacent Dem. In the right angled plain triang. BED draw the periphery FE then is BE Radius DE the Tangent of the angle at B make CA parallel to DE then are the Triangles ABC and EBD like because of their rightangles at A and E and their common angle at B therefore BA BE ∷ AC ED. and BA × ED BE × AC that is BA × tB Rad. × AC as was to be proved 2 Axiome In all plain Triangles The sides are proportional to the sines of their opposite angles Demonst. In the ●plain triangle BCD extend BC to F making BF DC and draw the arches FG CH then are the perpendiculars FE CA the sines of the angles at B D by the 7th of the third and the triangles BEF and BAC are like because of their right angles at E and A and their common angle at B. Therefore C. C A ∷ BF FE that is BC. sine D ∷ DC BF sine B. as was to be proved 3 Axiome In all plain Triangles As the halfe sum of the sides is to their halfe difference so is the tangent of the half sum of their opposite angles to the tangent if their half difference Demonst. In the triangle BCD let the sides be CB and CD and CG CB. wherefore ½ Z crur. EG ½ × crur. EC draw CH bi-secting BG at right angles and make the angle GCI D then will the angle GCH ½ Z angle B and D whose tangent is HG and the angle ICH ½ × ang. B and D whose tangent is HI But EG EC ∷ HG HI that is ½ Z crur. ½ × crur. ∷ t ½ Z ang t ½ X ang. 4 Axiome In all plain triangles As the base is to the sum of the other sides so is the difference of those sides to the difference of the segments of the base DB. BF ∷ HB GA That is DB. Z crur. ∷ × crur. × seg base These things premised we will now set down the several Problems or cases in all plaine triangles right-angled and oblique with the proportions by which they may be solved manner of solving them both by natural and Artificial numbers Of right angled Plain Triangles IN right angled plain triangles the sides comprehending the right-angle we call the Legs and the side subtending the right angle we call the Hypotenuse 1 Probl. The legs given to find an Angle The given legs AB 230 AC 143.72 AC Rad ∷ AB tA CB by the 1 Axiome That the quantity of this angle or any other term required may be expressed in numbers if the solution be to be made in natural numbers multiply the second term given by the third and their product divide by the first the Quotient is the fourth proportional sought But if the solution be to be made in artificial numbers from the sum of the Logarithmes of the second and third terms given subtract the Logarithme of the first the remainer shall be the Logarithme of the fourth proportional required Illustration by natural numbers As the Leg AC 143.72 Is to the Radius AC 10000000 So is the Leg. AB 235 To the tang of ACB grad. 58.55 The product of the second and third termes is 2350000000 which being divided by 143.72 the first term given the quotient is 16351238 the tangent of the angle ACB which being sought in the Table the neerest less is 16350527 and the arch answering thereto is Grad. 58. 55 parts Illustration by Logarithmes   Logarithmes As the leg AC 143.72 2.157517 Is to the rad. IC 10.0000 So is the leg AB 235 2.371068 To the tang of C. gr. 58.55 10.213551 2 Probl. The angles and one leg given to find the other leg. In the right angled plain Triangle ABC the leg AC is inquired The given Angle ABC Leg. AB Rad. AB ∷ t ABC AC by the first Axiome 3 Prob. The Hypotenuse and a leg given to find an angle In the right angled plain Triangle ABC the angle ACB is inquired The given Hypoth BC. Leg AB BC. Rad ∷ AB s. ACB by 2 Ax. 4 Probl. The Hypotenuse and angles given to find either leg. In the right angled plain Triangle ABC the leg. AB is inquired The given Hypot. BC. Angle ACB Rad. BC ∷ s. ACB AB by 2 Ax. 5 Probl. The angles and a leg given to find the Hypotenuse In the right-angled plain Triangle ABC the Hypotenuse BC is inquired The given Angle ABC The given Leg AC s. ABC AC ∷ Rad. BC. by