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To find the Angle T Sa: In the fame Triangle T Sa, we have given the fame Parts as before, to find the Angle TSa; which is an Extreme conjunct, and the Angle at S is one of the Extremes; therefore, by the fecond Rule, and its Note, As the Co-tangent of S T a, 59°. 10',

Is to the Radius,

Co. Ar. 0.224092

So is the Sine of the middle Part TS, 69°. 18',

To the Co-tangent of T Sa, 59°. 22′,

IO.

9.548359

9.772451

Having found the Angle TSa, we are next to find the Angle R Sa, the other Part of the required Angle R ST; to find which, we have, in the Triangle R S a, right-angled at a, given the Side Sa, before found, and the given Angle TRS, or, which is the fame, the Angle a RS; which is an Extreme disjunct, and, the Angle at R being the middle Part, the Angle T Sa, which is to be found, is one of the Extremes; therefore, by the fourth Rule, and its Note,

As the Co-fine of S a, 53°. 26',

Is to the Radius,

So is the Co-fine of a R S, 82°. 19′,

To the Sine of R Sa, 12°. 58',

Co. Ar. 0.224930

IO.

9.126125

9.351055

To this adding the Angle T Sa, found before, it makes the Angle TSR 72°. 20', which, being made by the Interfection of two oblique Circles, may be measured by Prob. 56, Cafe 2.

Prob. 85. In the oblique-angled fpherical Triangle A D B, given the Side AD 79°. 25, the Side A B 72°. 36", and the Angle at B 54°. 29', to find the third Side DB.

The fame Parts being given here as in Prob. 81, the Conftruction will be the fame. And, to let fall a Perpendicular from the Angle at A, on the Side B D, find p, the Pole of the Circle B DE; then, thro' the three Points F, p, A, draw the oblique Circle FpA, which will be perpendicular to the oblique Circle B D E, and divide the Triangle B A D into two right-angled

P

B

E

A

Triangles, A a B and A a D, both right-angled at a: In the

002

Triangle

Triangle A a B, there is given the Side A B, and the Angl at B, to find the Perpendicular A a; which being separated. from the two other Parts, it is an Extreme disjunct, and the Part required is the middle Part; therefore, by the third Rule, and its Note,

As the Radius

Is to the Sine of the Hypothenufe A B, 72°. 36′,
So is the Sine of a B A, 54°. 29,

To the Sine of the middle Part a A, 50°. 58′,

10.

9.979658

9.910596

9.890254

To find Ba: With the fame Parts given, the Part of the Bafe à B may be found, this being an Extreme conjunct, and the required Side is one of the Extremes; therefore, by the fecond Rule, and its Note,

As the Co-tangent of A B, 72°. 16′,

Is to the Radius,

Co. Ar. 0.495146

IO.

So is the Co-fine of the middle Part a BA, 54°. 29', 9.764131

To the Tangent of a B, 61°. 10',

10.259277

In the right-angled Triangle A a D, we have given the Hypothenufe A D, and the Perpendicular a A, found before, to find a D, the other Part of the required Side B D, which is an Extreme disjunct, and the required Side is one of the Extremes; therefore, by the fourth Rule, and its Note,

As the Co-fine of A a, 50. 58′,

Is to the Radius,

So is the Co-fine of AD, 79°.25',

To the Co-fine of a D, 73°. 2′,

Co. Ar. 0.200816

10.

9.264027

9.464843

This, being added to a B 61°. 10, before found, is 134. 12, the Side B D, fought, which may be measured by Prob. 46.

Prob. 86. In the oblique-angled Spherical Triangle DEF, given the Side DE 79°. 56', the Side E F 66°. 29', and the Angle at D 62°. 40′, to find the Angle at E.

Conftruction. With the Chord of 60°, defcribe the primitive Circle AEDB; draw a right Circle, A D, at Pleasure;

from

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E

D

S

P

B

from D lay 79°. 56' to E, and draw the right Circle E B, which cross at right Angles with the right Circle st; by Prob. 48, draw the oblique Circle AF D, to make an A Angle of 62°. 40' at the Point D; by Prob. 61, draw theparallel Circle b Fc 23°. 31', the Complement of EF, diftant from st, interfecting A F D in F; laftly, thro' the three Points E, F, B, by Prob. 17, draw the oblique Circle EF B, which forms the required Triangle DEF. Draw an oblique Circle from the Angle at E, thro' p, the Pole of AF D, to B, which will be perpendicular to AF D, and fall without the Triangle on the Bafe D F, produced at the Point a; by which is found the right-angled Triangle a E D, right-angled at a; in which is given the Side D E, and the Angle at D, to find the Perpendicular a E: This is an Extreme disjunct, and a E is the middle Part; therefore, by the third Rule, and its Note,

As the Radius

IO.

Is to the Sine of the Hypothenufe ED, 79°. 56', 9.993262 So is the Sine of a DE, 62°.40',

To the Sine of the middle Part a E, 61°,

9.948584

9.941846

The fame Parts being given, in the fame right-angled Triangle Ea D, to find the Angle a ED, will be an Extreme conjunct, and the required Angle at E is one of the Extremes; therefore, by the fecond Rule, and its Note,

Co. Ar. 0.286614

As the Co-tangent of a DE, 62°. 40',
Is to the Radius,
So is the Co-fine of the middle Part E D, 79°. 56', 9.242526

To the Co-tangent of a E D, 71°. 19',

10.

9.529130

Having found the Angle a ED, to find the Angle a E F, in the Triangle Ea F, right-angled at a, given the Hypothenufe E F 66°. 29', and the Perpendicular a E 61°,

before

before found; whence this is an Extreme conjunƐ, and the middle Part is required; therefore, by the first Rule, and

its Note,

As the Radius

Is to the Co-tangent of E F, 66°. 29',
So is the Tangent of a E, 61°,

To the Co-fine of a EF, 38°. 17',

10.

9.638647 10.256248

9.894895

This, being fubtracted from the Angle a E D, before found, leaves the required Angle FED 33°. 2′; which, being made at the Periphery of the primitive Circle by the oblique Circle EFB, may be measured by Prob. 54.

Prob. 82. In the oblique-angled Spherical Triangle BDE, given the Side B D 35°. 27, the Side BE 82°. 10 ́, and the Angle at D 79°.44', to find the third Side E D.

The fame Parts being given as in Prob. 81, the Conftruction of the Triangle EBD is the fame. And, to let fall a Perpendicular from the Angle at B, on the Side DE, find p, the Pole of the oblique Circle BEA; A then, thro' the three Points A, P, B, draw, by Prob. 17, the oblique Circle Ap FB, which will be perpendicular to D E C, and divide the Triangle B DE

C

P

E

D

F

B

into two right-angled Triangles, EBF and BDF, both right-angled at F: In the Triangle D B F, is given the Side BD, and the Angle at D, to find the Perpendicular B F ; which, being an Extreme disjunt, and the required Part B F being the middle Part, may be found by the third Rule, and its Note,

As the Radius

10.

Is to the Sine of the Hypothenufe BD, 35°. 27', 9.763422 So is the Sine of BDF, 79°. 44',

9.992990

To the Sine of the middle Part, BF, 34°.48',

9.756412

And,

And, in the fame Triangle, from the fame Parts given, to find the Side D F, will be an Extreme conjunct, and D F, the Part required, is one of the Extremes; therefore, by the fecond Rule, and its Note,

As the Co-tangent of B D, 35°.27',

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10.147534

IO.

9.250980

9.103446

Now, in the Triangle B F E, right-angled at F, having found BF, and the Side B E being given, the Side E F may be found by the fourth Rule, and its Note, fince it is an Extreme disjunct, and E F is one of the Extremes; therefore,

As the Co-fine of BF, 34°. 48',
Is to the Radius,

So is the Co-fine of BE, 82°. 10',

To the Co-fine of E F, 80°. 26',

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Co. Ar. 0.085578

10.

9.134470

9.220048

Then D F and F E, being added together, make 87°.40' for the required Side ED; which, being a Part of an oblique Circle, may be measured by Prob. 46.

Prob. 88. In the oblique-angled Spherical Triangle RST, given the Angle at R 73°.42, the Angle at S 45°. 36', and the Side RS 52°. 28′, to find an oppofite Side, as S T.

Conftruction. With the Chord of 60°, draw the primitive Circle PQRS; from any Point, as S, fet off 52°. 28′ from S to R; then, by Prob. 48, draw an oblique Circle, ŔT P, to make an Angle at R of 73°. 42'; and, by the fame Problem, draw another oblique Circle, QTS, to make an Angle at S of 45°.36', interfecting the former oblique

P

R

Circle in T, and thereby forming the Triangle TRS. To let fall a Perpendicular from the Angle at R, on the Side ST,

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