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PART II.

OBLIQUE TRIGONOMETRY.

The solution of the first two cases of Oblique Trigonometry depends on the following PROPOSITION.

IN ALL PLANE TRIANGLES, THE SIDES ARE IN PROPORTION TO EACH OTHER, AS THE SINES OF THEIR OPPOSITE ANGLES. That is, AS THE SINE OF ONE ANGLE IS TO ITS OPPOSITE SIDE, SO IS THE SINE OF ANOTHER ANGLE TO ITS OPPOSITE

Or, AS ONE SIDE IS TO THE SINE OF ANGLE, SO IS ANOTHER SIDE TO THE SINE OF ITS OPPOSITE

SIDE.

ITS OPPOSITE

ANGLE.

Note. When an angle exceeds 90°, make use of its sup

plement, which is what it wants of 1800. Note, Def. 25. GEOM.

CASE I.

Fig. 46.
C

The angles and one side given, to find the other sides. Fig. 46.

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In the triangle ABC, given the angle at B 48°, the angle at C 72°, consequently the angle at A 60°, and the side A B 200, to find the sides A C and B C.

To find the side A C.

To find the side B C. As sine A CB, 72° 9.978206 As sine ACB, 72 - 9.978206 : side AB, 200 2.3010301: side A B, 200 . 2.301030 :: sine A B C, 480 - 9.871073 : : sine B AC, 600 » 9.937531

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BY NATURAL SINES.

AS THE NAT. SINE OF THE ANGLE OPPOSITE THE GIVEN SIDE IS TO THE GIVEN SIDE, SO IS THE NAT. SINE OF THE ANGLE OPPOSITE EITHER OF THE REQUIRED SIDES TO THAT REQUIRED SIDE.

Given side 200; nat. sine of 720, its opposite angle, 0.95115, nat. sine of A B C 48°, 0.74314; nat. sine of BAC 600, 0.86617. Thus,

0.95115 : 200 : : 0.74314 : 156.
0.95115 : 200 : : 0.86617 : 182.

CASE II.

Fig. 47.

C

200

Two sides, and an angle opposite to one of them given, to find the other angles and side. Fig. 47.

A

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In the triangle ABC, given the side A B 240, the side BC 200, and the angle at A 46° 30'; to find the other angles and the side A C.

To find the .angle A C B. As side B C, 200 2.301030 Angle at A 460 30' : sine B AC, 460 30' 9.860562

с 60 30 : : side A B, 240 2.380211

107.00 12.240773 2.301030 Sum of the three angles 1800 Sum of two

1070 :sineACB, 60°30' nearly9.939743

Angle at B

73 The side A C will be found by Case I. to be nearest 253. Note. If the given angle be obtuse, the angle sought

will be acute; but if the given angle be acute, and opposite a given lesser side, then the angle found by the operation may be either obtuse or acute.

It ought therefore to be mentioned which it is, by the conditions

BY NATURAL SINES.

AS THE SIDE OPPOSITE THE GIVEN ANGLE IS TO THE NAT. SINE OF THAT ANGLE, SO IS THE OTHER GIVEN SIDE TO THE NAT. SINE OF ITS OPPOSITE ANGLE.

One given side 200, nat. sine of 46° 30', its opposite angle, 0.72537, the other given side 240.

As 200 : 0.72537 :: 240 : 0.87044—6030'.

CASE III.

Fig. 48.

180

AL

36°40'

Two sides and their contained angle given, to find the other angles and side. Fig. 48.

240

B The solution of this case depends on the following PROPOSITION.

IN EVERY PLANE TRIANGLE, AS THE SUM OF ANY TWO SIDES IS TO THEIR DIFFERENCE, SO IS THE TANGENT OF HALF THE SUM OF THE TWO OPPOSITE ANGLES TO THE TANGENT OF HALF THE DIFFERENCE BETWEEN THEM.

ADD THIS HALF DIFFERENCE TO HALF THE SUM OF THE ANGLES, AND YOU WILL HAVE THE GREATER ANGLE, AND SUBSTRACT THE HALF DIFFERENCE FROM THE HALF SUM, AND YOU WILL HAVE THE LESSER ANGLE.

In the triangle A B C, given the side A B 240, the side AC 180, and the angle at A 360 40', to find the other angles and side. Side A B

240
A B

240 АС

180
AC

180

Sum of the two sides 420

Difference

60

The given angle BAC 36° 40', substracted from 180°, leaves 143° 20', the sum of the other two angles, the half of which

As the sum of two sides, 420 : their difference 60 :: tangent half unknown ang. 71° 40'

2.623249 1.778151 10.479695

12.257846 2.623249

: tangent half difference, 23° 20' nearly

9.634597

The half sum of the two unknown angles,
The half difference between them,

71° 40' 23 20

Add, gives the greater angle A CB

95 00

Substract, gives the lesser angle ABC

48 23

The side B C may be found by CASE I. or Il.

CASE IV.

Fig. 49.

85

The three sides given, to find the angles. Fig. 49.

75 30

A 105 D The solution of this case depends on the following PROP

OSITION.

IN EVERY PLANE TRIANGLE, AS THE LONGEST SIDE IS TO THE SUM OF THE OTHER TWO SIDES, SO IS THE DIFFERENCE BETWEEN THOSE TWO SIDES TO THE DIFFERENCE BETWEEN THE SEGMENTS OF THE LONGEST SIDE, MADE BY A PERPENDICULAR LET FALL FROM THE ANGLE OPPOSITE THAT SIDE

Half the difference between these segments, added to half the sum of the segments, that is, to half the length of the longest side, will give the greatest segment; and this half difference substracted from the half sum will be the lesser segment. The triangle being thus divided, becomes two right angled triangles, in which the hypothenuse and one leg

In the triangle A B C, given the side A B 105, the side AC 85, and the side B C 50, to find the angles. Side A C

85
AC

85 BC

В С

50

50

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Substract, gives the lesser segment BD

30.0 Thus the triangle is divided into two right angled triangles, A DC and BDC; in each of which the hypothenuse and one leg are given to find the angles.

To find the angle DCA. I To find the angle DC B. As hyp. A C, 85 1.929419 As hyp. B C, 50 1.698970 : radius 10.000000 : radius

10.000000 : : seg. A D, 75 1.875061: : seg. B D, 30 1.477121

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11.875061
1.929419

11.477121 1.698970

: sine DCA, 61° 56' 9.945642: sine DCB, 36° 52' 9.778151

The angle D C A 61° 56', substracted from 90°, leaves the angle CAD 28° 4'.

The angle DCB 36° 52', substracted from 90°, leaves the angle CBD 53° 16'.

The angle DC A 61° 56', added to the angle D C B 36° 52', gives the angle A CB 98° 48'.

This case may also be solved according to the following PROPOSITION.

IN EVERY PLANE TRIANGLE, AS THE PRODUCT OF ANY TWO

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