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XXXVIII. The half Sum of the Sides- -48

1.6812412 S

A B-28(whose Loga. 1.4471580 The Difference of


rithm is 1.1461280 ? BCBC-6

0.7781512 Their Sum

5.0526784 The Half Sum

2.5263392 Which is the Logarithm of 336, the Area of the Triangle.


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By the Three Sides given; To find the point in the longer Side,

where a Perpendicular shall fall, and i he Length of ihat Per-

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 Produ&t by the third Side, upon which the Perpendicular is to fall, the Quotient added to the third Side, or fubitra£ted from it, shall be the Double of the Greater or Lesser Segment on either Side of the Perpendicular.

Example Fig.

In the former Triangle A B C, the Sum of the Sides A B and xxxy11. A C, is 54, and their Difference is 14; which multiplied together, XXXVIIC

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 half whereof

32, is the Greater Segment of the Side B C; or the Quotient 18, substracted from the Side BC, leaves 24; the half whereof 12, is the Leser Segment. B D, where the Perpendicular. is to fall.

By Logarithms. To the Sum of the Logarithms of the Sum and Difference of the Sides containing the Angle from whence the perpendicular is. to fall; subftra& the Logarithm of the Side upon which it is to fall. The Remainder shall be the Logarithm of a Number; which added to, or fubftracted from the Side on which the Perpendicu-lar is to fall, shall be the Doul te of the Greater or Lejjer Seg. ments of the Side on which the Perpendicular is to fall.




XXXVII. The Sum of A B and A C is 54

1.7323937 The Difference of A B and A C is 14

1.1461281 Their Sum

2.878;218 The Logarithm of the third Side B C 42, substract 1.6232493

Remains 1.2552725 Which is the Logarithm of 18. Which 18 added to B C 42, makes 60 ; the half whereof 30, is the Greater Segment C D or 18 subftracted from B C 42, leaves 24; the half whereof 1-2, is the Leser 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, whose Square Root is 16;
for the Length of the Perpendicular A D.

By Logarithms.
Half the Sum of the Logarithms of 32 and 8, the Sum and
Difference of A B and B D is the Logarithm of the Perpendicular..

The Logar. of the Sum of A B and B D 32, 1.50515
The Logar. of the Difference of A B and A D 8. 090309
Their Sum

The half

1.20412 Which is Logarithm of 16, the Length of the. Perpendicular AD.


The Base B A (or longest Side) of a Plain Triangle B A D, being'

4.7.8 P. and a Perpendicular DC, let fall from the Angle opposite to that Şide 17.33 P. being given ; To find the Area of that Triangle.

O the Logarithm of half the given Side, add the Logarithm Fig.

of the Perpendicular; the Sum of those Logarithms shall be XXXIX.the Logarithm of the Area of that Triangle.

T .



42, and


The Base B A, is 47.8
The half Base

23.9 - Log.-1.3783978 The Perpendicular D C 17.33 Log.-1.2387986 The Logarithm of 414.2

2.6171965 Which is the Area of the Triangle.

After the fame manner, if the Base of a Triangle were
the Perpendicular 16, the Area will be found to be 336.

There is an Oblique-angled Plain Triangle A B C, ane of whose An-

gles at the Base B C is 53.48 d. And the Sum of the Two Sides
opposite to the Base B C, viz. A B and A C, is 129, and the
Base CB, is longer than the longer side A C, by 16.8. The re-
Spective Sides and Angles of the Triangle are required.


one end thereof, at B, make, an Angle A B C, to contain 53.84 d. Then the sum of the Two Sides A B and AC being together 120, break it into any Two equal or unequal Parts, as into 43.2 and 76.8, then (by the Proposition) must the Base BC, be 93.6, which is greater than A C by 16.8.-So in the Obliqueangled Triangle A B C, you have given, (1.) The Side AB 43.2. (2.) The Side A C 76.8. (3.) The Angle opposite thereto, ABC 53 d. 50 m. by which you may find the Angle A C B. For,

As A C 76.8 : Isto s ABC 53.84 d.
So is A B 43.2 : To s ACB 27.02 d.

D ;

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Fig. XLI. T be this:

of the Menfuration of the Area of a Spherical Triangle. Lemma I. The Lunary Superficies of the Hemisphericks, are as the

Angles of the Same Superficies.
CHE Proof of this Lemma (amongst many other Ways) may

Let the Meridian Semicircle A E D be imagined to be equally moved over the Longitude of the Equator BEC, upon the Polés A and D. Therefore the Angles on the other fide of A and D


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(to wit, F and G) shall be as the Times. The Superficies alfo, Fig. XLI.
K and M, thall be as the Times, therefore F shall be to G, as
the Superficies K, to the Superficies M. And G unto M, as the
Superficies M, to the Superficies K, &c. howsoever they can be
taken. For,

Those Things that agree to a Third,
Agree among themselves.

Therefore, by Compofition,
As F + G:G:: K + M: M,

Or, As F + G:F::R + M: K.

That is,
As the Two Right. Angles are unto G,
So is half the Spherical Superficies, to M.
And, As Two Right Angles, are to F;

So is half the Spherical Superficies, to K.
Lemma II. The Triangle G, is equal to the Triangle H, because Fig.XLII.-

the Angles and Sides of one, are equal to the Angles and Sides of the other: To wit, A = D B


- E, C= F. Also, O, M=P, N=Q. Therefore they are Congruous and Equal.

The Excess of the Three Angles, over and above Two Right An-

gles, divided by 720, Shews what the Area of the Triangle, is, in
refpe& of the whole Spherick.

180 : A :: Sph. : G +R

' For by Lemma I. 180 : B::Sph. : G+S

180 : C:: Sph.: GHT=H+T. (by the Second Lemma.) Therefore, As 180 : A + B + C :: half the Spherick : to 3 , ( 3G+R+SHT, (by 24 El. 5 Ev.) As 180 : A+B+C780 ::

: :: So half the Spherick : 3 G+R+S +T-half the Sph.. But G +R+S + T is equal to half the Spherick;






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} thews what part the Triangle is of the

Fig. Therefore, 3 G +R+S+T - half the Sph. 2 G.
XLII. So that, As 180 : A B+C - 180 : :

: : So is half the Spherick : to 2 G.
And the Antecedent Terms being Quadropled, it shall be,
As 720.: A +B+C- 180 :: 2 Sphericks : 2 G.

And so the Spherick to G. Therefore,

whole Spherick.
These Things are likewise true in all Spherical Polygons, of what

Ordinate Figure foever they be; or In-ordinote, to all the An-
gles be given. And the Reason is, because all Polygons may be
resolved into Triangles. . Therefore, this Rule shall hold in
these Multangles also.

Multiply 180 d. by the Number of the Angles; subduct the
Product out of the Aggregate of all the Angles increased by 360d.
The Residue divided

by 720 d. gives the Area of the Polygon.

Of the Completion of a Solid Body. FR "Rom the foregoing Mensuration of the Area of a Spherical Tri

angle, this Fruit ariseth. If the Radius of the Sphere be 100000.0), the side of an inscribed Icosaedrum shall be 105146.22, equal to the Subtense of 63 deg. 26 min. 10 sec. Therefore the Plain Equilateral Triangle FEO (in the lcosaedrum) answers to the Equilateral Spherical Triangle in the Sphere; whose Three Spherical Triangles are connected in the Plain Angles, in the same Points, F, E, 0.

And the Sides of this Spherical Triangle are separately taken 63 deg. 26 min. 10 sec. to wit, because their Subtenses F E, EO,

OF, in the Plain Triangle, are equal to one another. Fig. Let fall now the Perpendicular E P, the Spherical Triangle XLIII. EPO shall be Rectangled; where, over and above the Right

Angle at P, are given, E O and PO, equal to half E 0; wherefore the Vertical Angle PE O shall be 36 deg. juft, and the whole Angle at.E, 72 deg. And the Sum of the Three equal Angles, E, F, O, Thall be 216 deg. from whence taking Two Right


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