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(4.) Sin. (60°+A)- sin. (60°-A)=2 cos. 60° sin. A (60)= sin. A.

For by (Art. 84) cos. 60°, hence 2 cos. 60°=1. Again, cos. (30°-A)-cos. (30+ A)=2sin. 30° sin. A. (62)= sin. A.

For by (Art. 84) sin. 30°, hence 2 sin. 30°=1.

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(6.) Article 36 gives ==

hence

(Alg. 108) and

b

sin. A
sin. B'

hence

sin. A+ sin. B

=

a+b sin. Asin. B

=

a-b sin. A--sin. B' tan. (AB)

sin. Asin. B tan. (A-B)

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-b tan. (A-B)'

(59.)

and since a and b are the sides

opposite to the angles A and B, the expression proves, that the sum of the sides is to their difference, as the tangent of half the sum of the opposite angles is to the tangent of half their difference, which is the rule.

(7.) Let (AD-DC)=d then,

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c―a: d=p cosec. A-p cosec. C: p cot. A-p cot. C (÷p). =cosec. A- -cosec. C: cot. A- -cot. C(× sin.A sin. C). sin. C-sin. A: sin. C cos. A-cos. C sin. A. sin. C-sin. A: sin. (C-A)

(53.

=2sin.(C-A)cos. (C+A):2sin. (C-A)cos.§(C−A) (60 and 69).) =cos.(C+A): cos.(C-A). (÷2 sin.C-A). Again,

c+a: d=p cosec. A+p cosec. C:p cot. A-p cot. C. =cosec. A+cosec. C: cot. A-cot. C (÷p).

sin. C+sin. A: sin. C cos. A-cos. C sin. A. 2 sin. (C+A) cos. § (C-A): 2 sin. § (C—A) cos. & (C—A). sin. (CA): sin. (C-A)

(÷2 cos. C-A). Q. E. D.

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(2.)

206.148

Art. 87.

160: the height above the eye.

=

10.000000

=10.099395

2.204120

2.303515

height of the eye.

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R: tan. 47° 30' 100: the height above the eye.

(3.)

Log. R

Log. tan. 47° 30'

Log. 100

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10.000000

10'034948

tower.

2.000000

2.037948

R: tan. 50° 43′=130: height above the eye.

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(1.) Sin. 131° 26', or cos. 41° 26': sin. 38° 40′ 266 : the ne dist and Sin. 131° 26': sin. 92° 46′ 266: the other distance.

Ar. co. Log. sin. 131° 26'-Log, cosec. 131°26′ =10.125097

Log. sin. 38-40
Log. 266

Log. 221-672

= 9.795733 -2-424882

= 2.345712

Ar. co. Log. sin. 131° 26'-Log. cosec. 131°26′ =10-125097
Log. sin. 92° 46'=Log. cos. 2° 46′
Log. 266

Log. 354-382

9.999493 2.424882

= 2.549472

(2.) NE. N. is 3 points to the east of the north, and NW. W. is 44 points to the west of the north, the sum of these, or 7 points, is the angle formed by the ship's course, and the bearing of the headland. Again, the course the ship had when at the second station was SE. E., which is just 4 points from the bearing of the headland: hence the angles at the extremity of the base were 73 points and 4 points, the sum of which is 11 points, and since the base is given 20 miles, we have

C

Sin. 44 pts.: sin. 4

Sin. 4 pts.: sin. 7

pts.=20: Dist. from first station.
pts.=20: Dist. from second station.

Ar. co. Log. sin. 44 pts. Log. cosec. 44 pts.

Log. sin. 4 pts.

Log. 20

Log. 19-0864, or 19.09 nearly

10.130210 = 9.849485 1.301030

ar. co. Log. sin. 44 pts.=Log. cosec. 44 pts. 10-130210

Log. sin. 7 pts.

Log. 20

Log. 26.96

= 12-80725

= 9.999477

= 1.301030

= 1.430717

(3.) Sin. 60°: sin. 80°-260 : side op. 80°.
R: sin. 40°-side op. 80°: : perp. breadth.
Compounding these proportions, we have
R sin. 60°: sin. 80° sin. 40°—260: perp.
Hence, making R=1, and solving the proportion, we

obtain perp. breadth=

260xsin. 80° sin. 40°

sin. 60°

breadth.

=260 × sin. 80° sin. 40% cosec. 60°; hence

Log.cosec. 60°

Log.' sin. 40°

Log. sin. 80°

Log. 260

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Log. 190-046

=2.278860

Art. 89.

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180°-174° 13′=5° 47′ the angle directly under the top of Arthur's Seat subtended by the distance of the stations, on the same horizontal plane as the base.

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3.259 correction for curvature.

821.174 height above the medium level of the tide.

4.067220

2

8.134440

Constant 8.378641

=0.513081

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Constant 8.378641

2.808 correction for curvature.

821-019 height above the medium level of the tide.

= 448417

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(1.) In ABCD there are given LBCD, 37°; LBDC= LADB+LADC; :. [BDC=96°; hence the third angle CBD 47°, and the side CD=300 yards, to find ČB and DB.

=

}

By

Art. 36.

10.135873 = 9-779463

Sin. CBD: sin. BDC-CD: CB
Sin. CBD sin. BCD-CD: BD
Log.cosec. 47° 10-135873 Log.cosec. 47°
Log. sin. 96° = 9.997614 Log. sin. 37°
Log. CD, 300,= 2-477121 Log. CD, 300
Log. CB 407-952-610608 Log. BD 246.864 2.392457
Again, in the AACD there are given LADC, 41°;

= 2.477121

LACD LACB+LBCD; .. LACD=93°; hence the third angle CAD=46°, and the side CD=300 yards, to find AC and AD.

Sin. CAD: sin. ACD=CD: AD

Sin. CAD: sin. ADC=CD : AC ƒ By Art. 36.

10-143066

= 9.816943

Log.cosec. 46° 10-143066 Log.cosec. 46° Log. sin. 93° = 9.999404 Log. sin. 41° Log. CD 300 = 2.477121 Log. CD, 300, = 2.477121 Log. AD416-477= 2.619591 Log.AC, 273-609= 2·437130 Next, in the AACB there are given the side AC= 273 609, and CB=407·95, and the contained [ACB=56o, to find AB (180°—56°)—26°, which is therefore half the sum of the angles at A and B. Hence by Art. 40,

CB=407.95

AC=273.609

Ar. co. Log. 681-559, sum of sides,

17-166496

Log. 134-341, dif. of sides, = 2.128208

Log. tan. 62°

=10.274326

Log. tan. 20° 20' 24"

= 9.569030

LCAB 82° 20′ 24′′

LCBA 41° 39' 36"

Lastly, to find the side AB from ABC.

Log. cosec. CAB=82° 20′ 24′′-10-003893

Log. sin. 56°

Log. CB=407.95

Log. AB=341-252

=9-918574

= 2.610608

=2.533075

The same by the method of rectangular co-ordinates.

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