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As the Sine of the Angle at A, 36°. 40', Co. Ar. 0.223909

Is to the Side B C, 51.55,

So is the Angle at C, 60°.51',

To the Side A B, 75.4,

And,

1.712229 9.941187

1.877325

As the Sine of the Angle at A, 36°.40', Co. Ar. 0.223909 Is to the Side B C, 51.55,

So is the Sine of the Angle at B, 82°. 29',

To the Bafe A C, 85.6,

-

1.712229 9.996252

1.932390

The Side A B and Bafe A C may be measured by the general Rule, Prob. 19.

Prob. 28. In the oblique-angled Triangle ABC, given the Side BC 51.56, the Side A B 75.4, and the Angle at A 36°.40', to find the Bafe AC, and the Angles at B and C.

Conftruction. Draw the Bafe A C at Pleasure, and on any Point affumed, as A, by Prob. 8, make an Angle of 36° 40′; take 75.4 from the Scale, by the general Rule, Prob. 19, and fet from A to B; then take 51.56 from the

A

B

Scale, and fet one Foot of the Compaffes in B, and with the other interfect the Bafe in C; laftly, draw B C, and the Triangle is completed.

Here we have the Side BC oppofite the known Angle at A, and the Side A B oppofite the unknown Angle at C, which may be found by the following Proportion, according to the first Rule:

As the Side B C, 51.55,

Is to the Sine of the Angle at A, 36°40′,
So is the Side A B, 75-4,

To the Sine of the Angle at C, 60°. 51',

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Having found the Angle at C, the Angle at B is found to be 82°. 29', by fubtracting the Sum of the other two Angles from 180°, both which may be measured by Prob. 9; then, for the Bafe A C,

As

As the Sine of the Angle at A, 36°. 40', Co. Ar. 0.223910

Is to the Side B.C, 51.55,

So is the Sine of the Angle at B, 82°. 29',

To the Bafe A C, 85.6,

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The Bale AC may be measured by the general Rule, Prob. 19.

But here it is to be obferved, that this Cafe, where two Sides and an Angle oppofite one of them are given, is an ambiguous Cafe, and admits of two Solutions, unless, as in thefe Examples, the Side oppofite the given Angle is longer than the other given Side: This will appear by the Conftruction: For, if, in the Triangle ABC, the Base A B 45.3, the Side BC 34.6, and the Angle at A 40°.20, were given to find the Side A C, and the two Angles B and C, it admits of two Solutions.

Conftruction. Draw the Bafe AB, and fet off 45.3, by the Scale, from A to B; then, by Prob. 8, make an Angle at A of 40°. 20', and draw the Line A C to a convenient

A

B

Length; take 34.6 from the Scale, by the general Rule, Prob. 19, in the Compaffes, and fet one Foot in B, and with the other interfect the Line A C, which it will do both in C and D; whence it becomes uncertain whether the Side A D or AC is the true Solution, which can only be determined by the Circumftances of the Triangle: For, if the Angle at C was to be acute, it will be the Side A C; but, if the Angle was to be obtufe, as the Angle D in the Triangle AD B, then A D would be the Side fought.

Prob. 29. In the oblique-angled Triangle ABC, given the Side AB 75.4, the Bafe AC 85.6, and the included Angle at A, to find the other two Angles B and C, and the Side B C.

Conftruction. Draw the Bafe-Line AC at Pleasure, and from the diagonal Scale fet off 85.6 from A to C; make an Angle at A of 36°.40′, by Prob. 8, and draw A B; then take 75.4 from the Scale, and fet it from A to B; laftly, draw the Line B C, and the Triangle is completed.

B

C

Here

Here we have given the two Sides A C and AB, with the Angle included between them; and, therefore, this Cafe cannot be done by the first general Rule, as neither of the Sides is oppofite the given Angle; but must be folved by the fecond Rule, Page 224. Now, as the three Angles of every Triangle are equal to two right Angles, or 180°, the Angle at A 36°. 40', being fubtracted from 180°, leaves 143°. 20', the Sum of the two unknown Angles B and C, Half of which is 71°. 40; and Half their Difference may be found by the following Proportion, according to the faid fecond Rule:

As the Sum of the two Sides, 161,
Is to their Difference, 10.2,

Co. Ar. 7-793174

1.008600

So is the Tangent of the Half-Sum of the } 10.479695

unknown Angles, 71°.40′,

To the Tangent of Half their Difference, 10°.1', 9.281469

If to the Half-Sum of the unknown Angles B and C Half their Difference is added, the Sum is the greater Angle; and, if from the Half-Sum the Half-Difference is fubtracted, the Remainder is the leffer of the two unknown Angles;

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Now, becaufe the greatest Side is always oppofite to the greatest Angle, the Angle at B must be greater than the Angle at C: And, having found the two Angles B and C, the Side BC may be found by the following Proportion, as at Prob. 27 :

As the Sine of the Angle at B, 82°. 29',
Is to its oppofite Side A C, 85.6,
So is the Sine of the Angle at A, 36°.40',

To its oppofite Side B C, 51.56,

Co. Ar. 0.003748

1.932474 9.941187

1.877409

Which may be measured by the general Rule, Prob. 19; and the Angles B and C may be measured by Prob. 9.

Prob.

Prob. 30. In the oblique-angled Triangle ABC, given the Side A B 75.4, the Side BC 51.56, and the Bafe AC 85.6, to find the Angles.

B

Conftruction. Draw the Bafe A C, and fet off 85.6 from A to C; take 75.4 in the Compaffes, and fet one Foot in A, and with the other draw an Arch; then take 51.56 from the fame Scale, and fet one Foot of the Compaffes in C, and with the other interfect the former Arch in B; from B draw Lines to A and C, and the Triangle is completed.

A

E.

D

C

Here being no Angle given, this Problem must be folved by the third Rule, in the following Manner: Place one Foot of the Compaffes in B, and extend the other fo as to take in the fhorteft Side B C, and draw the Arch CE; from B let fall a Perpendicular on the Bafe A C, which will divide it into two Segments, CD the lefs, and AD the greater, whofe Difference is AE; then by the following Proportion, according to the faid third Rule,

As the Bafe A C, 85.6,

Is to the Sum of the two Sides, 126.96,
So is the Difference of the Sides, 23.84,

Co. Ar. 8.067526

2.103667

1.377306

To the Difference of the Segments of the Base, 1.548499 or A E, 35-359,

Half the Difference of the Segments, being added to Half the Bafe, gives the greater Segment A D; and Half the Difference, being fubtracted from Half the Bafe, gives the leffer Segment CD:

42.8
Half the Bafe.
17.679 Half the Difference.

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And as ED is equal to DC, therefore the Half of 50.24, or 25.12, is D C, the leffer Segment: And as ED is the

fame as DC, that is,

If to that we add AD, or

The Sum is AD, the greater Segment,

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Thus is the oblique-angled Triangle ABC divided into two right-angled Triangles, ABD and CBD, both right-angled at D, in each of which are given the Base and Hypothenufe to find the other Parts, which may be done in the following Manner Firft, for the Angle at B in the right-angled Triange A BD,

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To the Sine of the Angle at B, 53°.20′,

Co. Ar. 8.122629

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The Angle at A is 36°.40', being the Complement of

the Angle at B to 90°.

Then for the Angle at B in the right-angled Triangle CBD,

As the Hypothenufe B C, 51.56,

Is to the Radius,

So is the Bafe D C, 25.12,

To the Sine of the Angle at B, 29°. 9′,

Co. Ar. 8.28777°

10.

1.400020

9.687790

Whence the Angle at C is 60°. 51', the Complement of 29°.9′ to 90°; and the Angle at B in one Triangle, being added to the Angle at B in the other, is 82°. 29: The Angle at B in the oblique-angled Triangle ABC and the Angles at A and C being found before, the Solution of the Problem is finifhed.

To prove the Truth of the Calculation, the Angles may be measured by Prob. 9.

Having, in this plain and eafy Manner, explained to the Reader how to conftruct the Triangles from the Parts given, and to find the required Parts by Calculation, fhall now fhew the Ufe and Application of Trigonometry, and first in finding the Multipliers for computing the Areas of regular Polygons, as was promised in Art. 60.

Prob.

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