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find p, the Pole of the oblique Circle; then, thro' the three Points R, p, P, draw, by Prob. 17, the oblique Circle Rp P, which will be perpendicular to QTS, and divide the Triangle RST into two right-angled Triangles, a RT and a RS, both right-angled at a: In the Triangle R Sa, given the Side RS, and the Angle at S, to find the Perpendicular a R; which, being an Extreme disjunct, and the Side a R the middle Part, may be found by the third Rule, and its Note:

As the Radius

Is to the Sine of the Hypothenufe RS, 52°. 28',
So is the Sine of a ST, 45°. 36',

To the Sine of the middle Part a R, 34°. 31',

10.

9.899273 9853986

9-753259

In the fame Triangle, the fame Parts being given, the Side S may be found by the fecond Rule, and its Note, fince it is an Extreme conjunct, and the required Side a S is one of the Extremes; therefore,

As the Co-tangent of R S, 52°. 28', Co. Ar. 0.114496

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10.

Is to the Radius,
So is the Co-fine of the middle Part aSR, 45°. 36′, 9.844889

To the Tangent of a S, 42°. 19',

9.959385

In the fame Triangle, given the Side RS, oppofite the right Angle at a, and the Side a S, found by the laft Proportion, oppofite the Angle a RS, the next Part to be found; this may be obtained by the general Axiom; thus,

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This being taken from the given Angle TRS 73°.42", there remains 15°. 36', for the Angle TRa; therefore, in the right-angled Triangle a RT, we have the Angle TR 15°36', and the Side a R 34.31, to find the Side a T; which lying together, with nothing betwixt them but the

right Angle, it is an Extreme conjunct, and a T is one of the Extremes; therefore, by the fecond Rule, and its Note,

As the Co-tangent of T R a, 15o. 36′,

Is to the Radius,

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So is the Sine of a R, 34°. 31', the middle Part,

To the Tangent of a T, 8°.59',

10.554077

10.

9.753312

9.199235

Which being added to the Side Sa 42°. 19', before found, makes 51°. 18, the Side TS, required.

Prob. 89. In the oblique-angled spherical Triangle DEF, given the Angle at D 63°.45', the Angle at E 71°. 49′, and the Side DE 86°. 25', to find the third Angle at F.

The fame Parts being given here as in the laft Problem, the Conftruction is the fame. The Perpendicular Eb B, being let fall from the Angle at E, on AFD, as in the laft Problem, A

will interfect the Bafe FD, produced in b, making the two right-angled Triangles bED and bEF: In the Triangle ED, given the Side ED, and the Angle at D, to find the Perpendicular bE; which is an

E

F

D

P

B

Extreme disjunct, and the Side bE is the middle Part; there

fore, by the third Rule, and its Note,

As the Radius

"

10.

Is to the Sine of the Hypothenufe E D, 86°.25, 9.999150 So is the Sine of b D E, 63°. 45',

To the Sine of the middle Part b E, 63.31',

29.952731

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9.951881

And, in the fame Triangle, from the fame Parts given, to find the Angle bED, is an Extreme conjunt, and the required Angle is one of the Extremes; therefore, by the fecond Rule, and its Note,

PP

As

1

As the Co-tangent of b D E, 63°.45,

Co. Ar. 0.307025

10.

Is to the Radius,
So is the Co-fine of the middle Part ED, 86°.25, 8.795881

To the Co-tangent of b E D, 82°.47',

9.102906

From which fubtract the given Angle FED 63°.45, and the Remainder is the Angle EF 19°.2. Then, in the Triangle EF, right-angled at b, given the Side 6 E, and the Angle bEF, to find the Angle bFE, the Supplement of the required Angle DFE to 180°; this is an Extreme disjunct, and the Angle bFE is the middle Part; therefore, by the third Rule, and its Note,

As the Radius

Is to the Co-fine of b E, 63°. 31′,
So is the Sine of b EF, 19°.2",

10.

9.649781 9.543375

To the Co-fine of the middle Part bFE, 81°. 21, 9.163156

This being fubtracted from 180°, the Remainder is the required Angle DFE 98°. 39; which, being made by the Interfection of two oblique Circles at F, may be measured by by Prob. 56, Cafe 2.

Prob. 86. In the oblique-angled spherical Triangle ABD, given the Side AB 74°. 20, the Side BD 58°. 46', and the Side A D 47° 38', to find an Angle, as the Angle at A.

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Conftruction. With the Chord of 60°, draw the Circle ABEF; draw, at Pleasure, the right Circle E CA; lay 74°. 20 from A to B, and draw BF; by Prob. 61, draw a Circle, a Db, at 47°. 38' diftant from E A; alfo, by the fame Problem, draw dDc, at 58°.46′ diftant from B, which will interfect the former in D; laftly, thro' the Points A, D, and E, by Prob. 17, draw the oblique

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Circle EDA; and, thro' the Points B, D, and F, by the

fame

fame Problem, draw the oblique Circle BDF, which will complete the Triangle A B D.

Here being no given Angle, the Perpendicular must be let fall oppofite to the required Angle at A; therefore, thro' D," and C, the Pole of the primitive Circle, draw the right Circle CDe, which will be perpendicular to the primitive Circle: Make ef equal to A e, then will A B be the Base, and ƒ B the alternate Bafe; to find which, add the two Sides A D and BD together, and take the Half of their Sum; then fubtract the leffer Side A D from the greater Side B D, and take Half the Difference; alfo take Half the Bafe AB:

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Having the Half of the Bafe A B, and Half the Sum and Difference of the Sides A D and D B, the alternate Base Bƒ is thus found:

As the Tang. of Half the Bafe AB, 37°. 10, Co. Ar. 0.120259 Is to the Tangent of Half the Sum of the Sides, 10.126043 So is the Tangent of Half the Difference of the 8.988842

53°. 10,

Sides, 5° 34',

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To the Tangent of Half Bf, the alternate Base, } 9.235144

9°.45',

Which, being added to Half the Bafe A B 37°. 10', is the Side Be 46° 55', the Bafe of the right-angled Triangle BDe; and, being fubtracted from the Half-Bafe, the Remainder is A e 27°.25', the Bafe of the right-angled Triangle A De; in which there is given the Side A D 47°.38′, and the Side A e 27°. 25', to find the Angle at A; which is an Extreme conjunt, and the Angle at A is the middle Part; therefore,

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As the Radius

Is to the Co-tangent of A D, 47°.38',
So is the Tangent of A e, 27°.25',

10.

9.960023 9.714933

To the Co-fine of the Angle at A, 61°.46′, 9.674956

This Problem may be folved another Way, without drawing a Perpendicular: Thus, Add the three Sides together, and take Half the Sum; then fubtract each of the two Sides containing the Angle from the Half-Sum:

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90°. 22' Half-Sum.

74

16

. 22 Half-Sum.

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20 A B fubtract.

2 Remainder.

90°. 22'
47 38 AD.

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Then add the Sines of thefe Remainders, viz. 16°.2', the Difference between Half the Sum of the Sides and A B, and 42°.44, the Difference between Half the Sum of the Sides and the Side AD, to the arithmetical Complements of the Sines of the two Sides containing the required Angle; Half that Sum is the Sine of Half the required Angle, which, being doubled, gives the required Angle at one Operation :

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The Angle at A, being made at the Periphery, may be measured by Prob. 54: The other Angles may now be found by the general Axiom, Page 277; by which the Angle B DA will be found to be 67°. 28', and the Angle DBA 49° 35', which the Learner may try for his Improvement.

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Prob.

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