XXVI. As cs. AB:cs,AC;;cs.BD:cs.DC.

CASE IV. Two Sides AC and DC, with the Angle ADC, opposite to AC given; To find the third Side A D.

In this Cafe, Let the Perpendicular fall from the Concourse of the given Sides, upon the Side enquired, continued, if need be.

(1.) As ct. CD: Rad.::cs. ADC: t. BD.

(2.) As cs. BD:cs.CD:: Rad.: cs. CВ.

(3.) As cs. CB: Rad.::csAC:csAB.

Therefore,

cs.CD:cs.BD:;cs. AC:cs. AB.
And, BD+AB = AD in the first
And, BD-AB = AD in the second 5

} Triangle.

CASE V. Two Sides C A and DA, and their contained Angle
CDAgiven; To find one of the other Angles.

In this Cafe the Perpendicular may fall from the Extremity of either of the Sides, oppofite to the Angle given.

(1.) Asct. AC: Rad.:: Cs. DAC: 1. AB.
And AD - AB=BD in the first
And ADAB = BD in the second

(2.) Asct. CAD: S.AB: Rad.: t. B С.
As t. B C. Rad. :: s. BD: ct. A D C.

Therefore,

} Triangle.

AssAB: ct. CAD: : SBD: ct. AD C.

CASE VI. Two Angles, ADC and C A D, and the Side AC, opposite to one of them being given, To find the Side between them. If it be known whether the Side fought, or the Side oppofite to the other given Angle, be Acute or Obtuse.

In this Cafe, Let the Perpendicular fall from the Extremity of the given Side, on the Side enquired, continued, if need be. Analogies.

Analogies.

(1.) As ct. AC: Rad. :: cs. DAC: t. A B.
(2.) Asct. CAD: S.AB:: Rad. : t. BC.
(3.) As Rad.: t.BC::ct.ADC: s. B D.

Therefore,

As ct. CAD:S.AB::ct.ADC: s. B D.
And, AB BD = AD in the ist

But, D B --- AB = A D in the 2d } Triangle.

Fig. XXVI.

CASE VII. Two Angles D AC and ADC, and the Side Fig.
A C, opposite to one of them, (viz. D) being given; To find XXVI.
the other Angle at C. If it be known, whether the Angle en- Cafe VII.
quired, or the Side opposite to the other given Angle, be Acute
or Obtufe.

In this Cafe, Let the Perpendicular fall from the Angle en-
quired.

Analogies.

(1.) As ct. CAD: Rad.::cs. AC:ct. ACB.
(2.) As s. ABC:cs.CAB :: Rad. :cs. BC.

(3.) As cs. BC: Rad.::cs.BDC:s.B C D.

Therefore,

Ascs. CAB:s. ACB::cs.BDC:s. BCD.
And, ACB+BCD = ACD in the ift
But, BCD-ACB=ACD in the 2d

} Triangle.

CASE VIII. Two Angles, ACD and CAD, with the Side Fig. between them, AC, being given; To find the third Angle D.

XXVL

In this Cafe, Let the Perpendicular fall from the Extremity of CajeVIII.

the given Side, and oppofite to the Angle enquired.

Analogies.

ACB=BCD in the ist

(1.) As ct. CAB: Rad.cs. AC: ct. ACB.

And, ACD --- A C B = B C D

But, ACD+ACB=BCD in the 2d

(2.) As s. ACB:cs.CAB:: Rad. :cs. В С.

(3.) As Rad.: cs.BC::s.BCD:cs. CDB.

L

Triangle.

There

Fig. XXVI.

Cafe IX.

Therefore,

As S. ACB:cs.CAB::s.BCD:cs. CDB.

CASE IX. Two Sides, A C and CD, with the Angle ADC, opposite to one of them, (viz. A.C) being given; To find their contained Angle AC D. If it be known, whether the enquired Angle, or the Angle opposite to the other given Side, be Acute or Obtufe.

In this Cafe, Let the Perpendicular fall from the Angle ens quired.

Analogies.

(1.) As ct. CDB: Rad. ::cs.CD:ct. BCD.

(2.) As ct. CD:cs.BCD: : Rad. : t. B С.

(3.) As Rad.: t. BC::ct. AC:cs. ACB..

Therefore,

As ct. CD:cs.BCD::ct.AC: cs. ACB.
And BCD+ACB=ACD in the ist

But BCD ACB ACD in the 2d } Triangle..

Fig. CASEX. Two Angles, DAC and ACD, with the Side beXXVI. tween them, A C, being given; To find either of the other Cafe X. Sides, DC or AD.

In this Cafe, Let the Perpendicular fall from the Concourse of the Side given and fought, on the third Side, continued, if need be.

Analogies.

(1.) As ct. CAB: Rad. :: cs.AC:ct. A C B.
And, ACD-ACB=BCD in the ist
But, ACD+ACB=BCD in the 2d
(2) Asct. AC:cs.ACB:: Rad. :t. B C.
(3.) Ast. BC: Rad.::cs. BCD:ct. C D..

Therefore,

} Triangle.

As cs. BCA: ct. AC::cs. BCD:ct. CD.

CASE

*

CASE XI. The Three Sides being given ; To find an Angle. Fig.
XXVI.

For the refolving of this Problem, there must be fome Prepara- Cafe XI.
tion made; for that the Univerfal Propofition of the Lord Nepier's
is not, of it self, fufficient for the Solution of this or the follow-
ing Cafe. And therefore, the said Lord Nepier's, to bring these
Cafes within fome Compass of this his Universal Propofition, he
first finds the Difference of the Segments of that Side; which be-
ing made the Bafse of the Triangle, is divided into Two Parts, by
letting fall of a Perpendicular; and by help of this

ANALOGY.

As the Tangent of half the Base,

Is to the Tangent of half the Sum of the other Two Sides;

So is the Tangent of half the Difference of those Sides,

To the Tangent of half the Difference of the Segments of the Base.

Thus then,

In the Oblique-angled Spherical Triangle ACD, there is given,
the Two Sides (or Legs) AC and CD, together with the
Base AD; to find the Angle CAD.

Analogies.

As the Tangent of half A D,

Is to the Tangent of half A C and AD;

So is the Tangent of half the Difference of A C and A D,

To the Tangent of half A E.

And half AD+ half AE = AB in the ist

But half A E-half AD = BE or B D in the 2d

Hence, to find the Angle at A.

As Rad.: ct. AC::t. AB:cs.CA В.

Fig.

}Triangle. XXVII.

CASE XII. The Three Angles being given; To find a Side. Fig.

XXVI.

This Cafe is but the Converse of the last beforegoing, and is to Cafe XII.

be Solved after the fame manner: If fo be that we convert the An-
gles into Sides. Which, how to perform, is shewed in the 12th
Cafe of the following Traft of the Solution of Oblique-angled Sphe-
rical Triangles, without letting fall a Perpendicular.

L2

And

Fig.

And for the Proportions or Analogies for the Resolving of this XXVII. Cafe there are also Two Ways proposed, to which I referr you.

CHAP. VI.

The foregoing XII Cafes, of Oblique-angled Spherical Triangles, Geometrically Demonstrated, and Resolved, with... out letting fall a Perpendicular.

F

:

OR the Performance of what is here promised, these Six
Theorems following, being demonstrated, will clear.

THEOREM I.

In any Spherical Triangle, whether Right or Oblique-angled..

The Sines of the Angles are Proportional, to the Sines of their op-pofite Sides: & contra.

This is demonftrated in Seft. 1. Chap. 4. of Right-lined Triangles, and is the fame in Spherical Triangles, whether Right or Oblique Angular.

THEOREM II.

In all Oblique-angled Spherical Triangles, whose Three Sides together are less than a Semicircle, or 180 Degrees.

As the Sine of half the Sum of the Angles at the Base,

Is to the Sine of the half Difference of those Angles;

So is the Tangent of half the Base,

To the Tangent of half the Difference of the Sides..

And also,

As the Co-fine of half the Sum of the Angles at the Bafe,
Is to the Co-fine of half the Difference of those Angles

So is the Tangent of half the Base,

To the Tangent of half the Sum of the Sides.

DE