difference, the greater angle is found by adding the half difference to the half sum, and the less angle is found by subtracting the half difference from the half sum. solution is completed as in Case I.

Then the

c+ b = 990; cb90; } (C+B) = (180° 80°) = 50°.

—

Applying logarithms to Formula (14), we have,

log sin (C B)

log (c—b) + log tan (C + B) +(a. c.) log (c + b) - 10;

log tan (C + B) (50°) 10.076187

(a. c.) log (c+b)· (990) 7.004365

log tan (C B) 9.034795 .. (C—B) = 6° 11';

C = 50° + 6° 11′ = 56° 11′ ; B = 50° — 6° 11′ = 43° 49′.

From Formula (13), we have,

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

2. Given c = 1686 yds., b = 960 yds., and A 128° 04',

6. Given a 3754, b 3277.628, and C 57° 53' 17",

to find A, B,
A, B, and C.

=

Ans. A = 68° 02′25′′, .B = 54° 04′ 18′′, c = 3428.512.

CASE IV.

Given the three sides of a triangle, to find the remaining

parts.*

46. Let ABC represent any plane triangle, of which BC is the longest side. Draw AD perpendicular to the base, dividing it into two segments BD and BD.

A

D

The angles may be found by Formula (A) or (B), Lemma. Pages

109, and 110, Mensuration.

From the right-angled triangles BAD and CAD, We

(DC + BD) (DC – BD).

(AC + AB) (AC — AB) Converting this equation into a proportion (B. II., we have,

P. II.),

DC - BD;

DC + BD : AC + AB :: AC-AB : DC

that is, if in any plane triangle, a line be drawn from the vertex of the vertical angle perpendicular to the base, divid ing it into two segments; then,

The sum of the two segments, or the whole base, is to the sum of the two other sides, as the difference of these ides is to the difference of the segments.

The half difference added to the half sum, gives the greater, and the half difference subtracted from the half sum gives the less segment. We shall then have two rightangled triangles, in each of which we know the hypothenuse and the base; hence, the angles of these triangles may be found, and consequently, those of the given triangle.

Applying logarithms to Formula (15), we have,

log (ss') = log (b+c) + log (bc) + (a. c.) log (s+s')-10;

Ans. A = 82° 49′ 09′′, B = 55° 46′ 16′′, °C = 41° 24′ 35′′.

3. Given a = 71.2 yds., b = 64.8 yds., and yds., to find A, B, and C.

c = 37.4

Ans. A = 83° 44′ 32′′, B = 64° 46′ 56′′, °C = 31° 28′ 30′′.

2. At what horizontal distance from a column, 200 feet high, will it subtend an angle of 31° 17′ 12′′ ?

and the angles of elevation at A and B being respect. ively 15° 36′ and 27° 29'.

Ans. DC 587.61 yds.

4. The distances AC and BC are found by measurement to be, respectively, 588 feet and 672 feet, and their included angle 55° 40'. ed the distance AB.

Requir

Ans. 592.967 ft.

A

B

5. Being on a horizontal plane, and wanting to ascertain. the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring in a direct line 180 feet farther from the hill, the