Demonstr. For, let ABC be the proposed triangle, having AB the greatest side, and Bс the least. Take AD = BC, considering it as a radius; and let fall the perpendiculars DE, CF, which will evidently be the sines of the angles A and B, to the radius AD or BC.

Now the triangles ADE, ACF, are equiangular; they therefore have their like sides proportional, namely, AC, CF :: AD or BC DE; that is, the side AC is to the sine of its opposite angle B, as the side BC is to the sine of its opposite angle A. Note 1. In practice, to find an angle, begin the proportion with a side opposite to a given angle. And to find a side, begin with an angle opposite to a given side.

Note 2. An angle found by this rule is ambiguous, or uncertain whether it be acute or obtuse, unless it be a right angle, or unless its magnitude be such as to prevent the ambiguity; because the sine answers to two angles, which are supplements to each other; and accordingly the geometrical construction forms two triangles with the same parts that are given, as in the example below; and when there is no restriction or limitation included in the question, either of them may be taken. The number of degrees in the table, answering to the sine, is the acute angle; but if the angle be obtuse, subtract those degrees from 180°, and the remainder will be the obtuse angle. When a given angle is obtuse, or a right one, there can be no ambiguity; for then neither of the other angles can be obtuse, and the geometrical construction will form only one triangle.

Draw an indefinite line; on which set off AB = 345, from some convenient scale of equal parts.-Make the angle A = 37°-With a radius of 232, taken from the same scale of equal parts, and centre B, cross AC in the two points c, c-Lastly, join BC, BC, and the figure is con

structed,

structed, which gives two triangles, and showing that the ca e is ambiguous.

Then, the sides AC measured by the scale of equal parts, and the angles B and c measured by the line of chords, or other instrument, will be found to be nearly as below; viz.

In the first proportion.-Extend the compasses from 232 to 345 on the line of numbers; then that extent reaches, on the sines, from 37% to 640, the angle c.

In the second 'proportion.-Extend the compasses from 37° to 27° or 78°4, on the sines; then that extent reaches, on the line of numbers, from 232 to 174 or 374, the two values of the side Ac.

When two Sides and their Contained Angle are given. FIRST add the two given sides together, to get their sum, and subtract them, to get their difference. Next subtract the given angle from 180°, or two right angles, and the remainder will be the sum of the two other angles; then divide that by 2, which will give the half sum of the said unknown angles. Then say,

As the sum of the two given sides,

Is to the difference of the same sides;

So is the tang. of half the sum of their op. angles, To the tang. of half the diff. of the same angles. Then add the half difference of the angles, so found, to their half sum, and it will give the greater angle, and subtracting the same will leave the less angle: because the half sum of any two quantities, increased by their half difference, gives the greater, and diminished by it gives the less.

Then all the angles being now known, the unknown side will be found by the former theorem.

Note. Instead of the tangent of the half sum of the unknown angles, in the third term of the proportion, may be used the cotangent of half the given angle, which is the same thing.

Demonst. Let ABC be the proposed triangle, having the two given sides AC, BC, including the given angle c. With the centre C, and radius CA, the less of these two sides, describe a semicircle, meeting the other side. Bс produced in D, E, and the unknown side AE in A, G. Join AE, ad, CG, and draw DF parallel to ae.

E

GF

Then BE is the sum of the two given sides AC, CB, or of EC, CB; and BD is the difference of the same two given sides

AC,

AC, BC, or of CD, CB. Also, the external angle ACE, is equal to the given sum of the two internal angles CAB, CBA; but the angle ADE, at the circumference, is equal to half the angle ACE at the centre; therefore the same angle ade is equal to half the given sum of the angles CAB, CBA. Also, the external angle AGC, of the triangle BCG, is equal to the sum of the two internal angles GCB, GBC, or the angle GCB is equal to the difference of the two angles AGC, GBC; but the angle CAB is equal to the said angle AGC, these being opposite to the equal sides AC, CG; and the angle DAB, at the circumference, is equal to half the angle DCG at the centre; therefore the angle DAB is equal to half the difference of the two angles CAB, CBA; of which it has been shown that ADE or CDA is the half sum.

Now the angle DAE, in a semicircle, is a right angle, or AE is perpendicular to AD; and DF, parallel to AE, is also perpendicular to AD: consequently AE is the tangent of CDA the half sum, and DF the tangent of DAB the half difference of the angies, to the same radius AD, by the definition of a tan gent. But the tangents AE, DF being parallel, it will be, as BE: BD :: ÅE: DF; that is, as the sum of the sides is to the difference of the sides, so is the tangent of half the sum of the opposite angles, to the tangent of half their difference.

Draw AB 345 from a scale of equal parts. Make the angle a = 37° 20'. Set off AC 174 by the scale of equal parts. Join BC, and it is done.

Then the other parts being measured, they are found to be nearly as follow; viz. the side BC 232 yards, the angle B 27°, and the angle c 115°.

2. Arithmetically.

The side AB 345
the side AC 174:07
their sum 519-07
their differ. 170.93

From 180° 00' take A 37 20 sum of c and в 142 40 half sum of do. 71 20

As

In the first proportion.-Extend the compasses from 519 to 171, on the line of numbers; then that extent reaches, on the tangents, from 71 (the contrary way, because the tangents are set back again from 45°) a little beyond 45, which being set so far back from 45, falls upon 44°, the fourth term.

In the second proportion.-Extend from 64° to 37°, on the sines; then that extent reaches, on the numbers, from 345 to 232, the fourth term sought.

When the Three Sides of a Triangle are given.

First, let fall a perpendicular from the greatest angle on the opposite side, or base, dividing it into two segments, and the whole triangle into two right-angled triangles: then the proportion will be,

As