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To find the hypothenuse AC.

As radius 90° 10.00000
Is to the base 690 2.83.885
So is secant [C 54° 51: 10.23979
13.0786,
10,00000

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Is to radius 90° 10.00000
So is the base 690 2.83.885
12.83.885

2.99.123

To tangent [A 35° 9 9.84762

To find the hypothenuse AC.

As radius 90° 10.00000
Is to the perpendicular 980 2.99123
So is secant [A 35° 9 10.08743
13.07866
10.00000

To the hypothenuse 1198 3.07866

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The extent from 690 to 980, on the line of numbers, will reach from radius 45°, to 54° 51, on the line of tangents.

The extent from 33°9, to radius 90°, on the line of sines will reach from 690, the base, to 1198, the hypothenuse.

The work to the preceding cases being all set down; and as a good knowledge of Trigonometry is the basis of the Mathematics, an example in each case, with their answers, is annexed, for the benefit of the learner.

1. Given the hypothenuse, 250 perches or chains; the angle opposite the base, 54° 30', to find the base and perpendicular.

Answer; the base 203.5; perpendicular 145.2.

2. Given the angle opposite the perpendicular 33° 15': the base 325, to find the hypothenuse and perpendicular.

Answer; the hypothenuse 388.6; perpendicular 213.1.

3. Given the perpendicular 91; the hypothenuse 170, to find the angles and base.

Answer: the angle opposite the perpendicular 32° 22'; consequently, the other angle 57 38, (by position 2, page 10) and the base 143.6.

4. Given the base 787 ; the perpendicular 890, to find the angles and hypothenuse. f

Answer; the angle opposite the base, 41° 29'; consequently, the other angle (by position 2, page 10) 48° 31', and the hypothenuse 1188.

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IN all plane triangles, the sides are in direct proportion to the sines of their opposite angles, and the contrary.

When the angles, and one side are given, to find the other sides, or either of them.

RULE 1.

As the sine of the angle opposite the given side,
Is to the given side;
So is the sine of the angle opposite the side required,
To the side required.

When two sides, and an angle opposite to one of them, are given, to find the other angles and side.

RULE 2.

As the side opposite to the given angle,
Is to the sine of the given angle;
So is the other given side,
To the sine of its opposite angle.

Then find the side (if required) by Rule 1.

CASE 1. Two Jingles, and one Side given, to find the other two Sides. The angle BDC 101° 25, and CBD 44° 42, and the side BC 76 perches given, to find the sides CD and BD; consequently, (by position 3, page 11) the other angle is 33° 53'.

Tor the construction, see problem 14, in Geometry.

To find DC.

As sine supplement of angle D 101° 25' 9.99132
Is to the side BC 76 1.88084
So is sine [B 44° 42 9.84.720

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To find BD.

As supplement of sine [D 101° 25' 9.99132

Is to the side BC 76 1.88081

So is sine [C 33° 58' 9.74.625

11.6.27.06

9.991.32

To the side BD 43.23 1.63574
BY GUNTER.

The extent from the supplement of [D = 78° 35', to [B 44° 42, on the line of sines, will reach from the side BC 76, to the side DC 54.53, on the line of numbers.

The extent from 78° 35', to [C 33° 53, on the line of

sines, will reach from the given side 76, to the side : BD 43.23, on the line of numbers.

CASE 2 & 3.

Two Sides, and an Angle opposite one of them given, to jind the other Jingles, and third Side.

The side BC 106, and BD 65, and the angle BCD 31°49. given to find the other angles, and side CD.

For the construction, see problem 15, in Geometry.

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