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SECTION IV.

SOLUTIONS OF OBLIQUE ANGLED TRIANGLES.

ART. 143. THE sides and angles of oblique angled triangles may be calculated by the following theorems.

THEOREM I.

In any plane triangle, THE SINES OF THE ANGLES ARE AS

THEIR OPPOSITE SIDES.+

Let the angles be denoted by the letters A, B, C, and their opposite sides by a, b, c, as in Fig. 23 and 24. From one of the angles, let the line p be drawn perpendicular to the opposite side. This will fall either within or without the triangle.

1. Let it fall within as in Fig. 23. Then, in the right angled triangles ACD, and BCD, according to art. 126,

Rb sin A: p
Rasin B Ρ

Here, the two extremes are the same in both proportions. The other four terms are, therefore, reciprocally proportional: (Alg. 387.*) that is,

ab sin A: sin B.

2. Let the perpendicular p fall without the triangle, as in Fig. 24. Then, in the right angled triangles ACD and BCD;

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Sin A is here put both for the sine of DAC, and for that 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

OPPOSITE 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 have then,

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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,

CG: DB GE: EB.

THEOREM III.

Q. E. D.

145. If upon the longest side of a triangle, a perpendicular 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 center C, and with the radius BC. Through A and C, draw the diameter LD, and extend BA to Then by Euc. 35. 3.

H.

ABXAH-ALXAD

Therefore,

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

* See note F.

greater than AP.
With the radius AC, describe the arc AN.
The segment NP=AP. (Euc. 3. 3.) But BP is greater 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.)

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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 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. If BP is greater than AP, the angle BCP is greater than ACP; 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 I, (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.

* Euc. 11. 5.

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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,

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