Fig. whofe Center fuppofe n. Then take Bλ, equal to BC, equal XXXIV.to B L; and draw the Diameter Bm A B, and D K, parallel to

Bm, parallel to H. Therefore, DB is equal to o B; and m
K, equal to m H; and F, equal to D L. Draw a perpendi-
cular to Dd, then the Triangle λdy is equiangular to the Trian-
gle DLy; and D L K, equiangular to the Triangle dw; there-
fore, Lydy: DL:: And, as DLλ::LK:
w; and therefore Ld, parallel to K w. But the Points & Ho
are in a Circle, whofe Diameter is dλ: Therefore the Angle
λw H, is equal to (the Angled H, equal to the Angle LD,
equal to) Kwy. Therefore the Angle Ko H, equal to yo
equal to a Right Angle: Therefore, m Hmw m K. Al
=
fo, λK: LK:: (wλ: w R::) Tangent of the Angle Dλd:
the Tangent of the Angle D&L; that is,

As the Sine of the Sum of the Sides,

Is to the Sine of the Difference of the Sides,

So is the Tangent of half the Sum of the Sides,

To the Tangent of half the Difference of the Sides.

Therefore,

As the Sine of the Sum of the Angles,

Is to the Sine of the Difference of the Angles;

So is the Tangent of half the Sum of the Angles,
To the Tangent of half the Difference of the Angles.

But,

The Sine of the Sum of the Sides,

Is to the Sine of the Difference of the Sides,

(as the Sine of the Sum of the Angles: Sine of the Difference of the Angle;)

So is the Tangent of half the Sum of the Angles,

To the Tangent of half the Difference of the Angles.

That is, Asλ K:λH:: AP: A O.

Therefore, ^K^H^K^H::÷÷APAO:AP-AO.

That is, m: mw:: Ann G.

Bat

Fig.

But alfo, the Angle mw, is equal to the Angle G An; therefore (by 7 Pr. 6 Lib. Eucl.) the Triangles λm, and A z G, are XXXIV. equiangular; and therefore the Triangles & Hand AO G, and allo the Triangles K w, and AP G, are equiangular. Therefore, λω :Η: : AG. GO, and λω : λΚ: : AG: A P.

But, D: :: DL (LK=) : λH,

And

λλ HAG: AO, and λw: XK :: AG: AP. But Dλ:: DL (LK): λH; and oλ :λw::♪ L λω (LH): λ K.

A. Therefore,
:

Dλ: DL:: (λλH) AG: AO.

And L:: (^: ^K::) AG: AP.

That is,

As the Sine of half the Sum of the Sides,

Is to the Sine of half the Difference of the Sides; So is the Co-tangent of half the Vertical Angle,

To the Tangent of half the Difference of the Angles.

And,

As the Co-fine of half the Sum of the Sides,

Is to the Co-fine of half the Difference of the Sides;
So is the Co-tangent of half the Vertical Angle,
To the Tangent of half the Sum of the Angle.

COROLLARIE S.

In any Spherical Triangle,

I

I

2

I Tan. Bafe: Tan. Sum of the Sides :: Tan. Differ. of the Sides: Tan. Differ. of the Segments of the Base AG, made by a perpendicular Arch falling thereon from E.

A GAO:: AP: Ad.

H. Tan. Bafe: Tan.
fine Dift.'s. A G
EAG= the Arch D
Arch B C.

Sum Sides :: Co-fine.

Sun's: Co

A PλL: For the Angle-
B, and the Angle AGE to the

III. Tan. Bafe: t.Sum Sides: :s. Sum of Ang. :s. Diff. <s

AG: AO:: Dλ: DL.

IV. IF

Fig. IV. If the Angle E G A be fuppofed Right, then A P, AO= XXXIV. A G q. that is, In a Right-angled Spherical Triangle, EGA, the Tang. Hypot. Per. Tan. Hypot. the Square of the Tangent of half the Bafe.

XXXV.

CHA.P. IX:

Per is

Some Problems in Plain Triangles, which come not within the Limits of the Twelve foregoing Cafes.

PROBLEM I.

In the Oblique-angled Triangle C B D, there is given the Angle at
C, (116.20 deg.) the Side D B, oppofite thereunto (1270 Foot :)
And the Two Sides, DC and CB, in One Sum, (1496 Foot:)
To find the other Angles D and B, and the Sides feverally.

CONSTRUCTION.

Xtend the Side of the Triangle D C, to A, making C A

Fig. Equal to CB, and draw the Line B A, fo fhall you have conftituted a new Triangle A O B, in which you have given, (1.) The Side A D, equal to the Sum of the Sides DC and CB, (1497.) And (2.) the Side DB (1270.) And (3.) being you have the Angle D CB, (116 d. 12 m.) you have alfo the Angle BCA in the other Triangle, (63 80 d.) the Complement thereof to a Semicircle, or 180 deg.

Then in the Triangle AC B, having the Angle at (63.80) you have alfo the Sum of the other Two Angles, CA B and A BC, equal to the given Angle D C B, (116.20) the half whereof is the Angle CAB (58.10 d.) to which the Angle ABC is equal, because the Sides C A and C B fubtending thofe Angles, are equal And now in the Triangle DAB you have given, (1.) the Side D A, 1497. (2.) The Side DB, 1270. And (3.) the Angle, DAB (oppofite to D B) (58.10 d.) by which you may find the whole Angle DBA (by Cafe I. of Oblique Triangles: For, As Log. DBs. CAB Log DA; s. ABD 95.53 d.

1270

58.10 d.

1497

From

Fig.

IXV.

Fig. XXVI.

Fig. XXXI

Fig.

XXXI