Fig. XXX.

Fig.

As LO:DC::OK:D X,
NE:EC::OB:Ο Κ.

It will alfo be

As LONE: DC * EC :: OK*OB:DX *OK:

Or rejecting the common Altitude O K, it will be
As LO*NE: DC*EC::OB: DX,

The Versed Sine of the Angle fought. Which was to be Demon-
Strated.

THEOREM V.

In all Spherical Triangles,

As the Rectangle of the Sines of the Sides containing the enquired
Angle,

Is to the Square of the Radius;

So is the Rectangle of the Sines of the half Sum, and half Difference of the Base, and Difference of the Legs,

To the Rectangle made of the Radius, and half the Verfed Sine of the Angle enquired.

DEMONSTRATION..

It is already proved by the last Theorem,
XXXI. That, As LONE: to the Square of the Radius:

So is OB: DX.

And therefore also,

LONE: Sq. of Rad. :: half OB: half D X.

And now in the following Diagram, OEGH, the Sum of ES and OE, is the double Measure of the Angle BSO; and the Arch OS is the Difference between the Base ES and E O, the Difference of the Sides A K and A E; And,

As R: half OS:: OH: half OB. And,
Half OS, and half OH, is equal to O B * Rad.
Therefore,

LONE: Sq. Rad. :: half OB Rad.: half D X * Rad.
And

LONE: Sq. R.:: half OS * half O H *: half D X * R. XXXI.

Which was to be Demonstrated.

THEOREM VI.

In all Spherical Triangles,

As the Rectangle of the Sines of the Sides containing the enquired

Angle,

Is to the Square of Radius;

So is the Rectangle of the Sines of the half Sum, and half Difference

of the Bafe, and Difference of the Legs,

To the Square of the Sine of half the Angle enquired.

DEMONSTRATION.

Fig.

It is already proved by the last Theorem, That
LO*NE: Sp. R.:. half OS * half OH: balf D X R. XXΧΙ.

But the Rectangle made of Radius, and half the Versed Sine of

an Arch, is equal to the Square of the Sine of half the Arch:

As in the foregoing Diagram; let the Arch given be D T, then is DX the Versed Sine of that Arch; and D F the Right Sine of half the Arch; and the Triangles DFR and DTX are like.

Therefore,

DR:DF:: half DT (=DF): half D X:
And, DR * half DX, is equal to the Square of D F.

Therefore,

LO*NA: Rq.:: half O S * halfO H: the Square of D F.
Which was to be Demonstrated.

Fig. XXXII.

CHAP. VII.

The Solution of the Twelve Cases of Oblique-angled Spherical Triangles, without any Regard had to a Perpendicular let fall, whereby to reduce it into Two Right-angled Triangles.

A

ND the Spherical Triangle which I shall make use of is that noted with ZSP, whose Sides and Angles, both in Sexagenary Degrees and Minutes, and Decimal Parts also, are as in this Table is expressed.

Fig.

CASE I. Two Sides, ZS and ZP, with the Angle P, oppo

XXXII. fite to one of them, being given; To find the Angle S, opposite

Cafe I.

to the other.

Fig.

ΧΧΧΙΙ.

Cafe II.

CASE II. Two Angles, S and P, with the Side ZP, oppofite 10 one of them, being given; To find the Side Z S, opposite to the other

Analogy.

As the Sine of the Angle S, 36 Deg. 8 Min. Co-Ar.

Is to the Sine of ZP, 24 Deg. 4 Min.

So is the Sine of the Angle P, 46 Deg. 18 Min.

To the Sine of Z S, 30 Deg.

0.2293936

9.6104465

98591180

Fig. ΧΧΧΗ. Cafe III.

)

८

A

3

2