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Prob. 74. In the right-angled spherical Triangle A B C, right-angled at B, given the Perpendicular BC 64°.10', and the Angle at A 680.30', to find the Hypothenufe A C.

See the Figure of the laft Problem, the Construction being the fame.

The Bafe lying between the two given Parts, BC and the Angle B A C, this is an Extreme disjunct, and the Hypothenufe AC, the Part to be found, is one of the Extremes; therefore, by the fourth Rule, and its Note,

As the Sine of the Extreme BAC, 68°. 30', Co, Ar, 0.031322 Is to the Radius,

So is the Sine of the middle Part B C, 64°, 10',

To the Sine of the Hypothenuse A C, 75°. 20 ́,

10.

9.954274

9.985596

The Hypothenuse A C may be measured by Prob. 46.

Prob. 75. In the right-angled fpherical Triangle ABC, right-angled at B, given the Perpendicular BC 64°. 10, and the Angle at A 68°. 30', to find the Angle at C.

The fame Parts being given as in the two laft Problems, the Conftruction is the fame. See the Figure at Prob. 73.

The Hypothenufe A C lying between the two Angles BAC and BCA, this is an Extreme disjunct, and the required Angle at C is one of the Extremes; therefore, by the third Rule, and its Note,

As the Co-fine of the Extreme BC, 64°. 10, Co. Ar. 0.360758
Is to the Radius,
So is the Co-fine of the middle Part BAC, 68°. 30, 9.564075

To the Sine of the Angle at C, 57°. 16′,

10.

9.924833

The Angle at C, made by the oblique Circle CAE interfecting the primitive Circle at the Point C, may be measured by Prob. 54.

Prob. 76. In the right-angled spherical Triangle ABC, right-angled at B, given the Perpendicular BC 64°. 10', and the Bafe AB 54°.28', to find the Hypothenufe A C.

Conftruction. With the Chord of 60°, draw the primitive Circle BCDE; thro' the Pole or Center draw the right

N n

Circle

Circle BD; take 64°. 10' from
the Line of Chords, and lay it
from B to C; draw the right
Circle CO E, and crofs it at B
right Angles with the right Circle
FOG; by Prob. 42, Cafe 2, lay G
on the right Circle BD 54°. 28,
from B to A; then, by Prob. 17,
draw an oblique Circle thro' the
three Points C, A, E, and the Tri-
angle ABC is completed.

[blocks in formation]

The Angles lying between the two given Parts and the Part required, it is an Extreme disjunct, and the middle Part A C is to be found; therefore, by the third Rule, and its Note,

As the Radius

10.

9.764308

Is to the Co-fine of the Extreme A B, 54°. 28',
So is the Co-fine of the other Extreme BC, 64°.10, 9.639242

To the Co-fine of the middle Part A C, 75°. 20, 9.403550

The Hypothenufe may be measured by Prob. 46.

Prob. 77. In the right-angled spherical_Triangle ABC, right-angled at B, given the Perpendicular BC 64°. 10′, and the Bafe A B 54°.28′, to find one of the Angles, as C.

See the laft Figure, the Construction being the fame as in the laft Problem.

No Part lying between the two given Sides and the Angle required, it is an Extreme conjunct, and the Angle at C, one of the Extremes, is to be found; whence, by the second Rule, and its Note,

As the Tangent of the Extreme A B, 54°.28′,
Is to the Radius,

So is the Sine of the middle Part BC, 64°. 10',

To the Co-tangent of BCA, 57°. 16′,

10.146198

10.

9.954274

9.808076

The Angle at C, made by the oblique Circle CA E interfecting the primitive Circle at the Point C, may be measured by Prob. 54.

If the fame Things are given, and the Angle at A required, it is ftill an Extreme conjunct, and the Bafe A B is now the middle Part; to find which, by the firft Rule, and its Note the Proportion is as follows:

As the Tangent of BC, 64°. to',

Is to the Radius,

So is the Sine of AB, 54°. 28',

10.315032

IO.

9.910506

19.910506

10.315032

9.595474

To the Co-tangent of B AC, 68°.30',

The Angle at A, made by the right Circle BAD and oblique Circle CAE interfecting each other at the Point A, may be measured by Prob.-55.

Prob. 78. In the right-angled spherical Triangle ABC, right-angled at B, given the Angle at A 68°. 30', and the Angle at C 57°. 16', to find the Hypothenufe A C.

Conftruction. With the Chord of

60°, draw the Circle E CFD;
and, thro' the Pole or Center,
draw the right Circle CD, at B
Pleasure, and crofs it at right
Angles with the right Circle E F; E
by Prob. 48, draw an oblique
Circle, CAD, to make an Angle
of 57°. 16', with the primitive
Circle, at the Point C; by Prob.
40, find p, the Pole of the oblique
Circle CAD; a Ruler, being

q D

F

G

laid from C to p, will cut the primitive Circle in s; on the primitive Circle, from s to q, fet 68°. 30', the given Angle at A; then draw the right Circle qr, perpendicular to which draw the right Circle BG, interfecting the oblique Circle CAD in A, making an Angle, CA B, of 68°. 30', and ABC is the Triangle required.

[blocks in formation]

The three Parts lying together, this is an Extreme conjun&, and the Hypothenufe AC is the middle Part; therefore, by the firft Rule, and its Note,

As the Radius

Is to the Co-tangent of B AC, 68°. 30 ́,
So is the Co-tangent of B C A, 57°. 16',

10.

9.595398 9.808083

To the Co-Sine of the middle Part AC, 75°. 20', 9.403481 The Hypothenufe A C may be measured by Prob. 46.

Prob. 79. In the right-angled spherical Triangle ABC, right-angled at B, given the Angle at A 68°. 30, and the Angle at C 57°. 16', to find one of the Legs, as the Base A B.

The Conftruction is the fame as in the laft Problem. See the Figure there.

The Hypothenufe A C lying between the given Angles, it is an Extreme disjunct, and the Bafe A B is one of the Extremes; therefore, by the fourth Rule, and its Note,

As the Sine of the Extreme BAC, 68°. 30', Co. Ar. 0.031322
Is to the Radius,
So is the Co-fine of the middle Part BCA, 57°. 16′, 9.732980

To the Co-fine of the Bafe A B, 54°.28′,

10.

9-764302

The Base A B may be measured by Prob. 43, Cafe 2.

If the Perpendicular B C had been required, it would still have been an Extreme disjunct; but the Angle at A would then have been the middle Part, and the Perpendicular might be found by the fourth Rule, and its Note; thus,

As the Sine of the Extreme BCA, 57°. 16', Co. Ar. 0.075102
Is to the Radius,
So is the Co-fine of the middle Part, BAC, 68°. 30, 9.564075

10.

To the Co-fine of the Perpendicular BC, 64°. 10', 9.639177

Of

Of Oblique Spherical Trigonometry.

The two firft Cafes of oblique-angled fpherical Triangles are folved by this Axiom: The Sines of the Sides of all spherical Triangles are in Proportion to the Sines of their oppofite Angles.

Prob. 80. In the oblique-angled Spherical Triangle ABD, given the Angle BDA 64°. 15, the Angle BAD 59° 47', and the Side B D 47°. 21', oppofite to one of the given Angles, to find the Side B A, oppofite the other given Angle.

Conftruction. With the Chord of 60°, defcribe the primitive Circle AEFD; at any Point, as D, of the Periphery, by Prob. 48, draw the oblique Circle DE, fo as to make an Angle with the primitive Circle, at D, of 64°. 15, and lay 47°21′ on this oblique Circle, by Prob. F 45, from D to B; thro' B, by Prob. 48, draw an oblique Circle, ACF, to make an Angle, DAB, of 59°.47′ with the primitive

Circle, and form the oblique Triangle A B D.

D

E

B

Here being given the Angle at A, oppofite the given Side BD, and the Angle at C, oppofite the required Side B A, this may be found by the following Proportion, founded on the above Axiom.:

As the Sine of BAD, 59.47',

Co. Ar. 0.063422

Is to the Sine of the oppofite Side B D, 47°. 21,

So is the Sine of BDA, 64°. 15',

To the Sine of the oppofite Side A B, 50°, 3',

9.866586

9.954579

9.884587

The Side A B, being Part of the oblique Circle A B F, may be measured by Prob. 46.

Prob. 81. In the oblique-angled Spherical Triangle ABD, given the Side AD 74°. 20', the Side A B 65° 34', and the Angle at D 59°. 50′, to find the Angle at B.

Conftruction. With the Chord of 60°, defcribe the primitive Circle HD AI; on which lay 74°. 20 from any Point

affumed

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