Half the Sum of the two oppofite Angles to the
Tangent of Half their Difference.*

73. This Theorem ferves for the fame Purpose as the third, only when we have the Logarithms of the Sides given, inftead of the Natural Numbers, as is the Cafe in calculating the Places of the Planets from aftronomical Tables, it is Something readier in Practice.

74. Theorem 5. In any Plane Triangle, as the Bafe is to the Sum of the other two Sides, fo is their Difference to the Difference of the Segments of the Bafe.t

E

75.

* Demonftration. Let AC6 be a right-angled Triangle, rightangled at C, and having two Sides, AC, Cb, refpectively Pl. IT. the two Sides, AC, CB, of the Triangle ABC. Then, by Fig. 16. Theorem 3d, in the right-angled Triangle ACь, As AC +

(but 90° 4 AbC = ▲ CAb, •.• fubftituting for CAb, we

ZA6C-45°. Alfo, by Theorem 3d, in the Triangle ABC,

is the fecond Part of the Theorem. Q. E. D.

Laftly, by the 6th Cafe of Trigonometry, As bC (CB): AC: Radius Tangent of AC; which is the ift Part of the Theorem. Q. E. D.

+ Demonftration. Let ABC be the Triangle, AB the Bafe. About C, as a Center, with the Radius CB, defcribe the Cir

cle

Pl. II.

Fig. 17.

Pl. III.

Fig. 1.

75. Corollary. This Theorem gives alio twice the Distance of the Perpendicular from the Middle of the Base.

For, if we put m to reprefent the Middle of the Bafe, and b Half the Bafe; then AE = b + mE, and BEb-mE; and, confequently, the Difference of the Segments 2mE.

76. The common Cafes of oblique Triangles are, Cafe ift. Given two Angles and one Side, to find the other Sides. This is folved by Theorem 1.

EXAMPLE. Let the Bafe AB be 150 Yards, the
A 30 d. and 4 B 40 d. Quære AC and BC.
Conftruction. Make AB 150 Yards, LA

3od. B40 d. and draw AC and BC to interfect in C; then is the Triangle ABC that required. Calculation. The A 30 d. + B 40 d. — 70 d. and 180 d. 70 d. — 110 d. ≈ 4 C.

To find BC.

As S. 4 C 110° 9.9729858
Is to AB 150 Yards 2.1760913
So is S. A 30° 9.6989700

11.8750613
9.9729858

To find AC.

As Sup. C 70° 9.97 9858

Is to AB 150 Yards 2.1760913 ↑
So is S. B 40° 9.8080675

11.9841588 9.9729858

To BC 79.81 Yards 1.9020755 To AC 102.6 Yards 2.0111730

77.

cle GFBDG; join C, F; bifect FB by the Line CE; then,
CB being CF, FE EB, by Conftruction, and CE com-
mon, the Triangles FEC and BEC are equal in every Re-
fpect, by our Geometry, Art. 95. Therefore ▲ FEC =
Z CEB; or, in other Words, CEL to AB. The two Parts,
into which the Bafe is cut by the CE, are called Segments.
In our Effay on Geometry, Art. 212, it is demonstrated,
that AB × AF AD × AG. Hence this Analogy, As AB :
AD: AG AF. But AB is the Bafe; and it is plain, from
the Construction, that AD is the Sum of the Sides AC and
CB; AG their Difference, viz. — AC — CG (or CB);
and AF AE- EF (or EB) the Difference of the Segments.
The above Analogy, expreffed in Words, will be as in the
Theorem.

:

1

77. Cafe 2. Given two Sides, and an Angle oppofite to one of them, to find the other Side and Angles.

EXAMPLE. Let AB the B 49 d. 30'; Angles A and C.

82 Feet, AC 66 Feet, quære the Side

BC,

and

Pl. III.
Fig. 2.

Conftruction. Make AB 82 Feet; B = 49 d. 30': then, taking a Distance of 66 Feet between the Compaffes, place one Leg on A, and turn the other about to interfect the Line BD, which it will do in two Places, C and E; fo that this Cafe admits of two Solutions; for either of the Triangles, ABC or ABE, has the Conditions required in the Question, and therefore this is called the doubtful Cafe, unless it is known, from fome Circumstance, whether the Angle opposite AB is obtufe or acute. If acute, Theorem 1ft finds the Angle E; if obtufe, its Supplement to 180 d. is the Angle C: for it has been already observed, that any Arc and its Supplement to 180 Degrees have the fame Sine.

In the following Solution we will fuppofe the Angle oppofite to AB to be obtufe, and fo ABC to be our Triangle.

78. Cafe 3. Given two Sides and the Angle contained between them, to find the other Angles and Side.

Pl. III.

The Angles are found by Theorems 3d and 2d; then the other Side by Theorem ift.

EXAMPLE. Let the Angle A 22 d. 30', AC= Fig. 3. 300 Yards, and AB400 Yards: Quære the Angles at B and C, and Side BC.

Conftruction. Make AB 400 Yards, A = 22 d. 30', AC 300 Yards; join B, C ; and the Triangle ABC is that required.

This Cafe may also be folved by Theorem 4th, and by letting fall a Perpendicular, as in Pl. III.

Fig. 4, independently of either of these Theorems; for in the right-angled Triangle ADC will be given. the Hypothenule AC, and Angle A, to find, by Right-Angle Trigonometry, the Bafe AD, and Perpendicular DC. Then ABAD BD; and therefore, in the right-angled Triangle BDC, are given BD and DC, to find the Hypothenufe BC.

79. Cafe 4. Given the three Sides, to find the Angles.

PIII.

Method of Calculation. From the greater Angle let Fig. 4. fall the Perpendicular DC; with the Radius CD defcribe the Arc BE; then are BD and AD called Segments; and, BD being = DE, AE is the Difference of the Segments, which is found by Theorem 4th: Then from AB fubtract AE, and the Remainder is EB; Half of which is ED, or its Equal DB, the fhort Segment.

Now, the Triangle is divided into two rightangled Triangles; in each of which are given the Hypothenufe and Bafe, to find the Angles, by Cafe 4th of right-angled Triangles, or by Theorem ift. Lastly, the ACD, added to the DCB, will give the whole ACB; or the Ls A and B, added together, and fubtracted from 180, are = the ACB.

EXAMPLE. Given AB 350 Feet, AC 240 Feet, BC 200 Feet; to find the Angles.

The Triangle is constructed by the Method fhewn in Art. 224. of the Essay on Geometry.

The Operation by the above Directions will be as under.

AC+ BC 440, and AC- BC — 40.

CANONS.