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From which, the Angle C B A 58.10 d. being fubftracted, there will remain 37.43 d. for the Angle CB D, and then the Angle XXXV. CD B will be 26.37 d. the Remainder of the other Two, Cand B, to 180 d.
And for the Sides, they may be found by the first Cafe of Oblique
As s C: Log. DB::s B: Log. DC.
Ass B Log. DC:: D: Log. C B.
In the Oblique-angled Triangle CDB there is given the Angle at
N the Given Triangle, make C A equal to CB, and draw the Line B A; fo is the Given Triangle CD B, reduced into Two XXXVI. Oblique-angled Triangles CA B and A DB; in which laft is given the Two Sides, A D and D B, but never an Angle; and in the other, only the Angle at C, and never a Side.
But (by Conftruction) the Side C A being made equal to the Side CB, the Angles CAB and C B A are equal, and the Sum of them, equal to the Complement of the Angle at C, (116.20 d.) to 180 d. that is, to 63.80 d. the half whereof 31.90 d. is equal to the Angle CA B, and the Complement thereof to 180 d. is 148.10 d. equal to the Angle DA B.
And now in the Triangle D AB, there is given, the Sides A D and D B, and the Angle at A, to find the other Angles, by Case I. and II. of Oblique Triangles,
As Log. D B: SA:: Log. AD: s. A B D.
- Which added to the Angle CB A, before found, 31.90 d. the
Ass DCB Log. DB::s CBD: Log. CD
Ass DCB: Log. DB:: sCDB: Log. CB
In the Right-angled Plain Triangle ABC, there is given, the Bafe
N the given Triangle ABC, extend the Perpendicular A C to
a new Triangle D AB; in which there is given, (1.) The Right
28 t.45.00 d. 56
Whofe Complement 26.57 d. is the Angle A DB, to which the Angle CBD is equal, because the Sides C B and C D, fubtending them, are equal by Conftruction. Then from the Angle D BA 63.43 d. before found, you substract the Angle D B C, 26.57 d. laft found, the Remainder 36.86 d. is equal to the Angle CBA, and the Complement thereof, 53.14 d. to the Angle C A B.
Otherwife, by this Theorem.
If from the Square of the Sum of the Bafe and Perpendicular D A 56 (3136) you fubftract the Square of the Bafe A B 28, (784) and divide the Remainder (2352) by double the Sum of the Hypotenuse and Perpendicular 56, (viz. 112), the Quotient (21) will give the Perpendicular: Which fubftracted from the Sum (56), leaves (35) for the Hypotenuse.
The Three Sides of a Right-lined Triangle, being given; To find the Area.
DD the Three Sides of the given Triangle together, and take the half thereof: From which half Sum, fubtract each Side of the Triangle feverally, and note the feveral Differences: Then multiply the Half Sum, by any one of the Differences, and that Product multiply by another of the Differences, and that Product multiply by the third Difference: The Square Root of this third Product, fhall be the Area of the given Triangle.
In the Triangle ABC, whofe Three Sides are A B 20, AC Fig. 34, and BC 42: Their Sum is 96, the half whereof is 48: XXXVIII. From which fubtract the feveral Sides, 20, 34, and 42, and there will remain these Differences, 28, 14, 6.Then, multiply 48 (the half Sum) by 28 (the firft Difference) the Product will be 1344: Which multiplied by 14 (the fecond Difference (the Product will be 18816: And this Produ& multiplied by 6 (the third Difference) produceth 112896: The Square Root whereof is 336, for the Area of the Triangle ABC.
To the Logarithm of half the Sum of the Sides, add the Logarithms of the feveral Differences of the Sides from the half Sum: Half the Sum of thofe Logarithms fhall be the Logarithm of the Area of the Triangle.
The Half Sum
Which is the Logarithm of 336, the Area of the Triangle.
By the Three Sides given; To find the Point in the longer Side, where a Perpendicular fhall fall, and the Length of that Per pendicular.
AKE the Sum and Difference of the Two Sides containing. the Angle from whence the Perpendicular is to fall, and multiply them together, and divide the Product by the third Side, upon which the Perpendicular is to fall, the Quotient added to the third Side, or fubitracted from it, fhall be the Double of the.. Greater of Leffer Segment on either Side of the Perpendicular.
In the former Triangle A B C, the Sum of the Sides A B and XXXVIII. A C, is 54, and their Difference is 14; which multiplied together, produce 756, which divided by the Side B C 42, the Quotient is 18; which added to the Side B C 42, gives 60; the halfwhereof 30, is the Greater Segment of the Side B C; or the Quotient 18, fubftracted from the Side BC, leaves 24; the half whereof. 12, is the Leffer Segment B D, where the Perpendicular is to fall.
To the Sum of the Logarithms of the Sum and Difference of the Sides containing the Angle from whence the Perpendicular is. to fall, fubftract the Logarithm of the Side upon which it is to fall. The Remainder fhall be the Logarithm of a Number; which added to, or fubftracted from the Side on which the Perpendicu-lar is to fall, fhall be the Double of the Greater or Lejer Segments of the Side on which the Perpendicular is to fall.
The Sum of A B and A C is 54.
The Logarithm of the third Side B C 42, fubftract
Which is the Logarithm of 18.
Which 18 added to B C 42, makes 60; the half whereof 30, is the Greater Segment CD, or 18 fubftracted from B C 42, leaves 24; the half whereof 12,
is the Leffer Segment B D.
For the Length of the Perpendicular.
Multiply the Sum of A B and B D 32, by the Difference of A B and B D, 8; the Product will be 256, whofe Square Root is 16; for the Length of the Perpendicular A D.
Half the Sum of the Logarithms of 32 and 8, the Sum and
Which is Logarithm of 16, the Length of the Perpendicular AD.
The Bafe BA (or longeft Side) of a Plain Triangle BA D, being 47.8 P. and a Perpendicular DC, let fall from the Angle oppo fite to that Side 17.33 P. being given; To find the Area of that Triangle.
O the Logarithm of half the given Side, add the Logarithm
the Logarithm of the Area of that Triangle.