of BAC. For, as one of these angles is the supplement of the other, they have the same sine. (Art. 90.)

The sines which are mentioned here, and which are used in calculation, are tabular sines. But the proportion will be the same, if the sines be adapted to any other radius. (Art.119.)

THEOREM II.

144. In a plane triangle,

As the sum of any two of the sides,

To their difference;

So is the tangent of half the sum of the oppo

site angles,

To the tangent of half their difference.

Thus the sum of AB and AC (Fig. 25.) is to their difference; as the tangent of half the sum of the angles ACB and ABC, to the tangent of half their difference.

Demonstration.

Extend CA to G, making AG equal to AB; then CG is the sum of the two sides AB and AC. On AB, set off AD equal to AC; then BD is the difference of the sides AB and AC.

The sum of the two angles ACB and ABC, is equal to the sum of ACD and ADC; because each of these sums is the supplement of CAD. (Art. 79.) But, as AC=AD by construction, the angle ADC=ACD. (Euc. 5. 1.) Therefore ACD is half the sum of ACB and ABC. As AB=AG, the angle AGB=ABG or DBE. Also GCE or ACD=ADC =BDE. (Euc. 15. 1.) Therefore, in the triangles GCE and DBE, the two remaining angles DEB and CEG are equal; (Art. 79.) So that CE is perpendicular to BG. (Euc. Def. 10. 1.) If then CE is made radius, GE is the tangent of GCE, (Art. 84.) that is, the tangent of half the sum of the angles opposite to AB and AC.

If from the greater of the two angles ACB and ABC, there be taken ACD their half sum; the remaining angle ECB will be their half difference. (Alg. 341.) The tangent of this angle, CE being radius, is EB, that is, the tangent of half the difference of the angles opposite to AB and AC. We haye then,

CG=the sum of the sides AB and AC;

DB=their difference;

GE=the tangent of half the sum of the opposite angles;

EB=the tangent of half their difference.

But, by similar triangles,

145. If upon the longest side of a triangle, a perpendicu

lar be drawn from the opposite angle;

As the longest side,

To the sum of the two others;

So is the difference of the latter,

To the difference of the segments made by the

perpendicular.

In the triangle ABC (Fig. 26.) if a perpendicular be drawn from C upon AB;

AB:CB+CA :: CB-CA: BP-PA*

Demonstration.

Describe a circle, on the centre C, and with the radius BC. Through A and C, draw the diameter LD, and extend BA to H. Then by Euc. 35, 3,

ABX AH=ALxAD

And converting the equation into a proportion,

AB: AD :: AL: AH

But AD=CD+CA=CB+CA

And AL=CL-CA=CB-CA

And AH=HP-PA=BP-PA (Euc. 3. 3.)

If then, for the three last terms in the proportion, we substitute their equals, we have,

AB: CB+CA:: CB-CA: PB-PA

146. It is to be observed, that the greater segment is next the greater side. If BC is greater than AC, (Fig. 26.) PB is geater than AP. With the radius AC, describe the arc AN. The segment NP=AP. (Euc. 3. 3.) But BP is great

er than NP.

147. The two segments are to each other, as the tangents of the opposite angles, or the cotangents of the adjacent angles. For, in the right angled triangles ACP and BCP, (Fig. 26.) if CP be made radius, (Art. 126.)

R:PC:: Tan ACP: AP
R:PC:: Tan BCP: BP

Therefore, by equality of ratios, (Alg. 384.*)

Tan ACP: AP:: Tan BCP: BP

That is, the segments are as the tangents of the opposite angles. And the tangents of these angles are the cotangents of the adjacent angles A and B. (Art. 89.)

Cor. The greater segment is opposite to the greater angle. And of the angles at the base, the less is next the greater side. I BP is greater than AP, the angle BCP is greater than ACl'; and B is less than A. (Art. 77.)

148. To enable us to find the sides and angles of an oblique angled triangle, three of them must be given. (Art. 114.)

These may be, either

1. Two angles and a side, or

2. Two sides and an angle opposite one of them, or

3. Two sides and the included angle, or

4. The three sides.

The two first of these cases are solved by theorem 1, (Art. 143.) the third by theorem II, (Art. 144.) and the fourth by theorem III, (Art. 145.)

149. In making the calculations, it must be kept in mind, that the greater side is always opposite to the greater angle, (Euc. 18, 19. 1.) that there can be only one obtuse angle in a triangle, (Art. 76.) and therefore, that the angles opposite to the two least sides must be acute.

The third angle is found, by merely subtracting the sum of

the two which are given from 180°. (Art. 79.)

The sides are found, by stating, according to theorem I, the following proportion;

As the sine of the angle opposite the given side,

To the length of the given side;

So is the sine of the angle opposite the required side,

To the length of the required side.

As a side is to be found, it is necessary to begin with a tabular number.

Ex. 1. In the triangle ABC (Fig. 27.) the side b is given 32 rods, the angle A 56° 20', and the angle C 49° 10', to find the angle B, and the sides a and c.

The sum of the two given angles 56°20′+49°10′=105°30'; which subtracted from 180°, leaves 74° 30' the angle B. Then,

Sin B: 6::

{

Sin A: a
Sin C: c

The arithmetical complement used in the first term here, may be found, in the usual way, or by taking out the cosecant of the given angle, and rejecting 10 from the index. (Art.113.)

Ex. 2. Given the side 671, the angle A 107° 6', and the angle C 27° 40'; to find the angle B, and the sides a and c. The angle B is 45° 14'. Then

When one of the given angles is obtuse, as in this example, the sine of its supplement is to be taken from the tables. (Art. 99.)

151. Given

Two sides, and
An opposite angle,

CASE II.

to find

The remaining side, and
The other two angles.

One of the required angles is found, by beginning with a

side, and, according to theorem I, stating the proportion,

As the side opposite the given angle,

To the sine of that angle;

So is the side opposite the required angle,

To the sine of that angle.

The third angle is found, by subtracting the sum of the other two from 180°; and the remaining side is found, by the proportion in the preceding article.

152. In this second case, if the side opposite to the given angle be shorter than the other given side, the solution will be ambiguous. Two different triangles may be formed, each of which will satisfy the conditions of the problem.

Let the side b, (Fig. 28.) the angle A, and the length of the side opposite this angle be given. With the latter for radius, (if it be shorter than 6,) describe an arc, cutting the line AH in the points B and B'. The lines BC andB'C will be equal. So that, with the same data, there may be formed two different triangles, ABC and AB'C.

There will be the same ambiguity in the numerical calculation. The answer found by the proportion will be the sine of an angle. But this may be the sine, either of the acute angle AB'C, or of the obtuse angle ABC. For, BC being equal to B'C, the angle CB'B is equal to CBB'. Therefore ABC, which is the supplement of CBB' is also the supplement of CB'B. But the sine of an angle is the same, as the sine of its supplement. (Art. 90.) The result of the