We take the smaller value of b, for the reason that A, being greater than B, requires that α should be greater

than b.

Applying logarithms to Proportion (12), Art. 83, we have,

(a. c.) log cos (4−B)+log cos 1⁄2 (A+B) +log tan § (a+b)—10 = log tan c;

we have,

and,

† (A + B) = 87° 59′ 05′′, (a + b) = 57° 13′ 58′′,

··· + c = 3° 09′ 09", and c = 6° 18′ 18′′.

Applying logarithms to the proportion,

sin a : sin c :: sin A

: sin C,

we have,

(a. c.) log sin a+ log sin c + log sin A 10 = log sin C;

The smaller value of C is taken, for the same reason

as before.

2. Given A = 50° 12′, B = 58° 08′, and to find b, c, and C.

a = 62°42′

CASE III.

Given two sides and their included angle.

87. The remaining angles are found by means of Napier's Analogies, and the remaining side, as in the preceding cases.

Applying logarithms to Proportions (10)

Proportions (10) and (11),

Art. 83, we have,

(a. c.) log cos(a + b) + log cos(a - b) + log cot C10

UNI OF

The greater angle is equal to the half sum plus the half difference, and the less is equal to the half sum minus the half difference. Hence, we have,

A 27° 31' 44", and B 5° 17' 58".

Applying logarithms to the Proportion (13), Art. 83, we have,

(a. c.) log sin (A− B) + log sin † (A+B) + log tan † (a−b) −10 = log tanc;

and

B.

C = 39° 23′ 23′′, to find c, A,

Ans. A = 120° 59′ 47′′, B = 33° 45′ 03′′, c = 43° 37′ 38′′

Ans. A 130° 05′ 22′′, B 32° 26' 06".

CASE IV.

Given two angles and their included side.

88. The solution of this case is entirely analogous to Case III.

Applying logarithms to Proportions (12) and (13), Art. 83, and to Proportion (11), Art. 83, we have,

MIA OL

(a. c.) log cos † (A + B) + log cos † (A — B) + log tan c log tan (a + b);

=

tanc 10

(a. c.) log sin (A + B) + log sin † (A − B) + log tan c log tan (ab);

=

(a. c.) log sin (a - b) + log sin (a + b) + log tan † (A — B)—10 = log cot C.

The application of these formulas are

solution of all cases.

sufficient for the

Ans. C = 64° 46′ 24′′, a = 70° 04′ 17", b = 63° 21′ 27′′.

Ans. 121° 36′ 12′′, a = 40° 0′ 10′′, b = 50° 10′ 30′′.

CASE V.

Given the three sides, to find the remaining parts.

89. The angles may be found by means of Formula (3), Art. 81; or, one angle being found by that formula, the other two may be found by means of Napier's Analogies.

EXAMPLES.

1. Given a 74° 23', b = 35° 46' 14", and c 100° 39', to find A, B, and C.

Applying logarithms to Formula (3), Art. 81, we have,

log cos A 10+ [log sin s + log sin (8 — a)

+(a. c.) log sin b + (a. c.) log sin c - 20];

Using the same formula as before, and substituting B for

Using the same formula, substituting C for A,

for d

and α for c, recollecting that s c = 4° 45′ 07′′, have,

we