RULE*

As the Sum of the two Sides
Is to their Difference

So is the Tang, of half the Sum of the
unknown Angles

To Tang, of half their Difference.

* DEMONSTRATION.

Suppofe A B C the Triangle, make BD= B C, and `BG=AB, then is GD= the Dif ference of the 2 Sides, and CBD the Sum~ of the two unknown Angles (Prop. 5 Sect.

2.)

Therefore, let

A

B

C

F

E

}

To

b D

fall the Perpendicular B E. and with that as a Radius, defcribe the Arch, a Eb and draw CD, then will ED be - the Tangent of half the Sum of the two unknown Angles. Draw BF parallel to AC. Then per fimilar Triangles Props 12 Sect. z.)

QED.

As AG: GD:: DE: EF. But half the Difference fubtracted from half the Sum gives the leffer Angle, and added to half the Sum gives the greater. Suppofe a Eb the Sum of the two Angles ac the lefs, and cb the greater. Now the Line BE cuts a Eb into two equal Parts a E and Eb, make Edc E, then will FH be the whole Difference, and therefore if half the Difference c E be fubtracted from the half Sum a E, it will give the leffer a c, and if added c E to Eb, gives the greater cb. QE D.

As the Bafe or largeft Side

Is to the Sum of the other two Sides
So is the Difference of thofe Sides>

To the Difference of the Segments of the Bafe, made by a Perpendicular let fall from the vertical Angle upon the faid Base. Then half the Difference added to half the longer Side, gives the greater Segment, and fubtracted from it, gives the leffer.

In

DEMONSTRATIO N,

From one end B of the least Side A B of the Triangle ABC as a Centre, and Radius BA, defcribe a Circle cutting the other two Sides in E and F; and let fall the Perpendicular BD. Then it is evident that G B = BF = AB, and by (Prop. 13. Sect. 2)

AD DE, and confequently EC-CD-DA, FC=CBBA, and GC CB+BA. Bur (Prop. 9 Sect. 2.) A C: CG:: FC: CE that, is, AC: CBBA:: CB-BA: CD-DA. QED.

And that the half Sum of two Quantities increased and deminished by their half Difference, gives the greater and lefs Quantities refpectively, was proved in the last Cafe.67

In the oblique angled plain Triangle ABC are given A B 120, AC 1687 and BC 70, to find the Angles A, B and C.

B

12.0

168.7

GEOMETRICALLY.

Draw the Line AC and lay off thereon 1687 from a Scale of equal Parts. From the fame Seale take 70 in your Compaffes and one Foot in A defcribe an Arch, then take 120 and with one Leg in C defcribe with the other, as fecond Arch cutting the Former in B, which completes the Triangle.

i,

The Angles may be meafured as before, e, from a Line of Chords.

Ad Sides 190

So is the Diff. of thofe Sides 50 1.698970

お

To the Diff of the Seg. of the } 1758609

Base 56'31

Now

Oblique Plain Trigonometry.

T I

N Definition 5th. it is faid, that an oblique angled plain "Triangle is fuch as has no right. Angle. As in right angled Triangles there ares 6 Cafes, fo in oblique there are 4, which It fhall now proceed to explain. Boys) bus

N. B. Oblique angled plain Triangles, as well as right angled ones, have their Sides proportional to each other, as the Sines of their oppofite Angles, but when there is not a proper Data given for fuch a Proportion, other Rules are obferved for bringing out the Solutions, which I fhall point out in their proper Places

CAS E I.

In the oblique angled plain Triangle ABC, there are given the Side BC 120, the Side AB 70 7༠ and the Angle A 36°40'; required the other Side and Angles.

GEOMETRICALLY.

Draw B A and affume any Point thereon, as A, with the Chord of 60o in your Compaffes, and one Foot in A, draw the Arch ba, and fet off thereon 36°40' through a, draw A B and lay off A B equal 70, then with BC equal 120 in your Compaffes and one foot in B, cross, with the other, the Line A C, as in C; draw CB, and you will have the Triangle required.

The Angles may be found by taking the Chord of 60 in your Compaffes, and with one Foot in C and B defcribe the Arches a b and cd, which take in your Compaffes and apply to the Line of Chords, give their refpective measure.

The Side BC being taken in your Compaffes and applied to a Line of equal Parts gives its Length.

LOGARITHMETICALLY,

As each Side is the Sine of its oppofite Angle, the Operation will be fimilar to thofe in the ft. Variety of right angled Triangles.

As