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And it is evident that these comprehend all the combina. tions of three that can possibly arise, the case of the three angles being excluded, as insufficient to determine any triangle (11).

Let ABC be any plane triangle; and let the angles be

4.A.

BA

denoted by the capital letters A, B, C at their vertices; the sides opposite to them by the small letters a, b, c.

From either vertex, as C, draw the perpendicular CD to the opposite side. Then CD will obviously be the sine of A to the radius b; and consequently its length will be 6 times the trigonometrical sine of A; that is, CD=b sin A.

In like manner the sine of B to the radius a is the same line CD, whose length is therefore a times, the trigonometrical sine of B; that is, CD a sin B. We have thus two expressions for CD, which therefore furnish the equation

a sin Bb sin A;

or, which amounts to the same thing, the proportion

ab sin A: sin B.

And this expressed in words, gives the following rule for the first case, viz.

CASE I. A side and angle opposite to it given.

RULE.

As given side,

: sine of its opp. angle,

:: any other side,

: sine of opp. angle.

NOTE. As remarked at (11), there may, under certain circumstances, be ambiguity in this case; that is, the same data may belong equally to two different triangles. Thus the annexed diagram shows,

B

that with two given sides AC, CB, and the acute angle A opposite to one of them, we may always construct two triangles, ABC, AB'C; where the angle B, opposite to the A other given side in the one triangle, will be the supplement of the corresponding angle B', in the other, provided CB is less than CA. With this proviso, therefore, the sine determined by the rule will belong not only to the acute angle immediately furnished by the tables, but equally to the obtuse angle which is the supplement of it; thus furnishing, in combination with the data, two distinct triangles in accordance with the geometrical figure. But if the given angle A be obtuse, or, if being acute, CB is not less than CA, there can be but one triangle, and consequently no ambiguity. Hence the unambiguous circumstances are these two:

1. When the given angle is obtuse; for then the sought angle must be acute, since a triangle cannot have two obtuse angles.

2. When the given angle is acute, and moreover the side opposite to it greater than the side opposite to the sought angle; for then the sought angle must be acute, since the greater angle is opposite to the greater side.

These precepts will enable us to distinguish the ambiguous examples from the others.

EXAMPLES.

1. In the triangle ABC are given b=153, c=137, and B=78° 13′, to find the remaining parts.

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There is no ambiguity here, because the side opposite to the given acute angle is greater than the side opposite to the sought angle C.

II. To find A.

(B+C) = 180°

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The angle A being equal to 180°

139° 26′ 47′′ is equal to 40° 33′ 13′′, therefore

As sin B 78° 13' arith. comp. 0.0092498

: b 153

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2.1846914

9.8130198

2.0069610

2. Given A=74° 14′, B=49° 23′ and c=408 yards, to find the sides a and b.

a=471.5 yards, b=371-9 yards.

3. Given A=58° 7′, B=22° 37′ and c=408 yards, to find a and b.

a=351·03 yards, b=158.98 yards.

4. Given A=90°, C=33° 30′ and b=45.54, to find

a and c, by logarithms.

a=54.612, c=30.142.

5. Given A22° 37′, a=117 and b=216, to find

B and c.

B=45° 13′ 55′′, or 134° 46′ 5′′, c=281-785.

6. Given A=32° 40′, a

195, c=318, to find C.

C

61° 40′ 3′′, or 118° 19' 57".

(20.) We shall now proceed to investigate a rule for the solution of the second case of plane triangles, or that in which the data are two sides, and the included angle. Let ABC be any plane triangle, in which the sides AC, BC, as also the included angle C, are given: we are required from these to determine the third side AB, as also the angles A and B.

From CA, the longer of the A4 two given sides, cut off CD equal to CB; draw BD, and perpen

F

E

dicular to it draw CE, which produce to F, and then draw EG parallel to FA. It is evident that CE bisects the given angle C, and the line BD, from the 26th of the First Book, because the right angled triangles CEB, CED have the angles at B and D, as well as the right angles equal, and have also the side CE common to both.

As DB is thus bisected by GE, a parallel to AB, it follows (2 of vI.) that AD is also bisected in G. Hence, CD is the half of BC and CD together; and DG is the half of the remaining part DA, that is,

1. CG is half the sum of the given sides.

Moreover, AD is by construction the difference of those sides; and, as just proved, AG is half that difference: therefore,

2. AG is half the difference of the given sides.

Again, the two unknown angles A and B are made up of these three, viz. A, ABD, CBD. But (32 of 1), the first

two of these are together equal to CDB, and therefore to CBD; so that the sum of the unknown angles is equal to twice CBD; therefore

3. CBD is half the sum of the unknown angles.

Moreover the angle B is equal to the two angles CDB, ABD; of which the first, as just remarked, is equal to the two, A, ABD; so that B is equal to A and twice ABD; therefore,

4. ABD is half the difference of the unknown angles.* Now, by right-angled triangles

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which, expressed in words, gives this rule for the second

case.

CASE II. When two sides and the included angle are given.

* This might have been inferred at once from the preceding conclusion, since the greater of any two quantities diminished by the half sum gives the half difference. And the conclusion 2 might have been inferred from 1, upon the same principle.

Y

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