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OBLIQUE ANGLED TRIGONOMETRY,

THE folution of all plane triangles, may be deduced from

the three following theorems.

THEOREM I.

In ang plane triangle, the sides are in the same proportion, as the kines of the opposite anglese Fig. 53. plate 4.

Dem. From the angles A and B, draw BE and AD perpena dicular, to the opposite fides, BC and AC produced if necessary. Because the triangles ADB, AEB, are right angled triangles, the fide AD becomes the sine of the angle ABD, and BE the fine of the angle BAE; if AB the hypothenuse, common to both the triangles, be made the radius ; but the two triangles ADC, BEC, have each a right angle at D and E, likewise the common angle ACB, therefore, they are similar, and consequently, BC is to CA, as BE is to AD; that is, the fides are in the fame proportion as the fines of the opposite anglesa

THEO

THEOREM II.

In any plane triangle, the sum of any two sides is to their difference ?

as the tangent of half the sum of the angles at the base, is to the tangent of half their difference. Fig 54. plate 4.

Dem. Let ABC be a plane triangle, AB+BC:AB_BC :: tan. ang. A+Ang. C:tan. ang. C-ang. A, upon A as cena

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tre with AB the longest side for a radius, describe a circles meeting AC produced in E and F; produce BC to D, join DA, FB, EB, and draw FG parallel to BC, meeting EB in G.

The angle EAB is equal to the sum of the angles at the base,, and the angle EAB at the centre is double the angle EFB at the circumference, therefore, EFB is half the sum of the angles at the base; but the angle ACB, is equal to the angles CAD, and ADC, or ABC together, therefore, FAD is the difference of the angles at the base, and FBD is half that difference, but FBD is equal to the alternate angle BFG; since the angle FBE, in a semi-circle, is a right angle, FB being radius, BE, BG will be tangents of the angles EFB, BFG;, but it is plain, that EC is the sum of the sides, BA and AC, also that CF is their difference; and since EG and BC are parallel, EC :CF as EB : BG, that is, the sum of the fides is to their difference: as the tangent of half the sum of the angles at the base, is to the tante gent of half their difference.

THEO.

THEOREM III.

In a plane triangle, the base is to the sum of the sides, as the

difference of the fides, is to the difference of the segments of the base, made by the perpendicular upon it from the vertex. Fig. 55. plate 4.

LET ABC be a plane triangle, if from B the vertex a perpendicular BD be dropped on the base, AC: AB+BC::BC-AB: DC-AD. Upon B as centre with BC, the greater fide for a radius, describe a circle meeting BA and CA, produced in F and E. It is manifeft, that AF is the difference of the sides, and that EA is the difference of the segments of the base, for ED and DC are equal, and AG is the sum of AB and BC; but, because FG and EC cut each other within a circle in the point A, the rectangle contained by the segments of the one, is e qual to the rectangle contained by the segments of the other, that is, FAXAG=EAXAC, and by Euclid vi. 16. AC: AG::FA:EA.

Wherefore, in any plane triangle, the base &c.

is to,

Note, The fum and difference of two magnitudes being given,

to find each of them.

Rule, To half the sum, add half the difference, the sum will be the greater, and from half the fum, subtract half the difference, the remainder will be the less.

In plane triangles may be given,

The three angles and one side.
'Two sides and the angle opposite to one of them. to find the
Two fides and the angle contained between them.
The three sides

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other parts.

PROBLEM I.

The angles and one fide given, to find the two remaining fides.

Plate 3. fig. 47

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Ang. C 52°157 Req. AC and BC
Ex. I. Given Ang. A 59
AB 276.5

5180-52°15+59°=68°43'=an.B.

To find AC

To find BC.

As fine ang. C52° 15' 9.89801 | As fine ang. C 52° 15' 9.89801 is to A B 276.5 2.44170

is to AB 276.5 2.44170 So is fine an. B68° 45' 9.96942 | So is fine ang. A 59° 9.93307

12.41112
2.51311 | To BC 299.8

12.37477
2.47676

TO AC 325.9

EXAMPLE 2. plate 4. fig. 56.

AB=2600

Req. AC BC.
Given Ang. A 47°30'

(Ang.C 41°15'J180--47°30+41°15'=21°15'ang.B.

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The sine, tangent, secant, &c. of any arch, is the fine, tangent, secant, &c. of its fupplement. Hence the fine of 91° 15' may be obtained thus, 180°-91° 15'=88° 45 the fupplement of g1° 15.

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To find BC.
As fine ang. C 41° 15' 9.81911
is to AB 2600

3.41497
So is sine ang. A 47° 30' 9.86763

13.28260 To BC 2907

3-46349

PROBLEM II. I wo sides and the angle opposite to one of them being given, to find the other angles and the third fide. Fig. 48. plate 3.

26 Ex. Given AC

39 42 Req. ang. A; ang. C and BC. ang. B 91°

91° 15'
180°-ang. C+ang. B.=ang. A 47° 30'.
To find angle C.

To find BC. 1.59572 | As fine ang. C41° 15' 9.81911 is to sine ang. B 91°15' 9.99990

is to AB 26

1.41497 So is AB 26 1.41497 Sois sine ang. A 47°30' 9.86763 11.41487

11.28260 To fine C 41° 15" 9.81915 | To BC 29.07 1.46349

PROBLEM III. Tavo fides and the angle contained being given, to find the remaining angles, and the third side. Fig. 49. plate 3.

АС 60
Ex. Given BC
Req. ang. A, ang. B. and AB.

Ta

AB

As AC 39.42

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