SCHOLIUM.

In each of the foregoing examples where two sides, and an angle opposite to one, are given, we have found it necessary to find the angle opposite to the other given side before we could apply the rule to the determination of the third side; so that the determination of this third side requires two proportions, and there is no logarithmic method which will lead us to it by a shorter process. It may, however, be deduced directly from the formula at (17), viz.

c=AB=b cos. A ± √a2 — b2 sin.2 A ;

which expression is, however, not adapted to logarithmic computation. 3. In the plane triangle ABC are given

A=44° 13′ 24′′, B=79° 46′ 38′′, AB=368,

to find the rest.

1. To find the side AC.

The angle C is equal to 180° — (A + B)=55°59′58′′, therefore,

4. In the plane triangle ABC are given

AB=408 yards, A=74° 14', B=49° 23′;

6. In the plane triangle ABC are given

AB=318, BC=195, A= 32° 40';

to find the angle C.

C = 61° 40′ 3′′, or 118° 19′ 57′′.

CASE II.

(24.) When two sides and the included angle are given.

RULE, (ART. 19.)

As the sum of the two given sides,

: their difference,

:: tangent of half the sum of the opposite angles

: tangent of half their difference.

Having thus found the half difference of the unknown angles, we obtain the angles themselves, by first adding and then subtracting this half difference from the half sum. The angles being thus known, as well as two sides, the third side is found by the first case.

The student will find a more compendious method of solution for this case in Prob. I., Part IV.; but the rule here given will be more easily remembered.

EXAMPLES.

1. In the triangle ABC are given

AB=137, AC=153, A=40° 33′ 12′′;

to find the other parts.

1. To find the other two Angles.

The sum of the other two angles is (B+ C)= 180° — A = 139°

2. In the triangle ABC are given

AC 378, BC= 526, C=32° 18′ 26′′;

to find the other parts.

1. To find the Angles.

The sum of the angles A, B is (A + B) = 180°

32° 18′ 26′′=

If we wish to obtain the third side of the triangle immediately, without first finding the angle, we may do so by means of the formula at (17), adverted to in the scholium to last case; but as the computation will not be adapted to logarithms, it will in general be the shortest method to proceed as above, by two proportions.

3. In the triangle ABC are given

AB 1637, AC2065, A = 132° 7′ 12′′;

=

to determine the remaining parts.

B=26° 52′ 42′′, C=21° 0′59′′, BC=3387·974.

4. In the triangle ABC are given

AB=1686, BC=960, B=128° 4′;

to find the rest.

A 18° 21′ 20′′, C= 33° 34′40′′, AC=2400.364.

CB, describe a circle, and produce the sides AB, BC, to meet the circumference; then it is plain that

GB AC+ CB, BF = AC - CB, BE AD
=

Now (Geom. Prop. 24, book 6.)

- DB.

GB.BF = ABBE ... AB· (AD — DB) = (AC + CB) (AC — CB) ·.·. AB : AC + CB :: AC — CB : AD — DB ;

hence the following rule.

RULE I.

Consider the longest side of the triangle as the base, and demit upon it a perpendicular from the opposite vertex, dividing the base into two segments; then say

As the base,

: the sum of the other two sides,

:: the difference of those sides

: the differénce of the segments of the base.

Having thus the sum and difference of the segments, each segment becomes known, and, therefore, in each of the two right-angled triangles into which the proposed is divided, there will be known the base and hypotenuse, and this is enough to determine all the other parts.

Both these rules are adapted to logarithmic computation, and this last is much the shortest; when, however, the three sides are small numbers, it will be best to operate without logarithms, by means of the