rods, and the angle opposite the second side equal to 96' to solve the triangle.

4. Given in a triangle ABC, the side a equal to 32.1098 rods, the side b equal to 125.701 rods, and the angle A equal to 14° 48'; to solve the triangle.

Ans. Angle B, 90°; angle C, 75° 12'; side c, 121.531 rods.

5. Given two sides of a triangle equal to 1540.37 feet and 760.9 feet, and the angle opposite the second equal to 30° 22' 8"; to find the other side and angles. Ans. Impossible.

CASE III.

128. Given two sides and the included angle. Let there be given in the triangle ABC the sides a and b and the included angle C, to solve the triangle.

To find A and B, we have

A + B = 180° C,

and

A

B

C

≥ (A + B) = 90°C complement of C;

whence,

=

tan (A+B) = cot C.

(131)

Then, the half difference of A and B is found by means of (95),

log tan (4-B)=log (a—b) — log (a+b)+ log tan † (A+B) =log (a - b) — log (a+b)+ log cot

That is,

C.

(133)

The logarithmic tangent of half the difference of the two required angles is equal to the logarithm of the difference of the

given sides, minus the logarithm of their sum, plus the logarithmic tangent of half the sum of the required angles, or plus the logarithmic cotangent of half the given angle.

Since

have

That is,

(A+B) is known, when (AB) is found, we

The GREATER of the two required angles is equal to half their sum, plus half their difference; and the SMALLER angle is equal to half their sum, minus half their difference.

1. Given of any triangle ABC, the side a equal to 9459.31 feet, the side b equal to 8032.28 feet, and the included angle C equal to 55° 30' 26"; to find the side c and the angles A and B. Solution. A + B = 180° C

= 124° 29′ 34′′, and

Then, by (133),

ar. co. log 5.757170

log 3.154433

log tan 10.278844

log tan 9.190447

ar. co. log sin 0.024176

Ans. Side c, 8242.64 ft.; angle A, 71° 3'34"; angle B, 53° 26'.

2. Given two sides of a triangle equal to 142.6 feet and 110 feet, and the included angle equal to 33° 55'; to solve the triangle.

3. Given the two sides of a triangle equal to 153 rods and 137 rods, and the included angle equal to 40° 33′ 12′′; to find the other parts.

Ans. Side, 101.615 feet; angles, 78° 13′ 1′′ and 61° 13′ 47′′.

CASE IV.

129. Given the three sides.

Let there be given (Fig. Art. 128) the three sides a, b, and c; to solve the triangle.

To jina A, B, and C. By (102), (103), and (104), we have

log sin A=

log (~ — b) + log (s — c) — log blog c, (134)

2

log (sa)+log (sc) log a log c

log (sa) +log (s—b)

2

The logarithmic sine of half of any angle of a triangle is equal to the logarithm of the difference between half the sum of the sides and one of the adjacent sides, plus the logarithm of the difference between half the sum and the other adjacent side, minus the logarithms of those two sides, divided by 2.

130. A, B, and C can also be determined by formulæ (106), (107), and (108) for the cosine of half an angle, and by formulæ (109), (110), and (111) for the tangent of half an angle.

When the half angle is less than 45°, the table will determine it from its sine with greater precision than from the cosine, and vice versa when the half angle is greater than 45°.

The method by the tangent of half the angle is precise, and requires the use of but four logarithms.

NOTE. This case may also be solved by drawing a perpendicular from the vertex to the base of the triangle, thus dividing it into two right-angled triangles, of which the hypothenuses are known, and the sum of whose bases is the base of the original triangle. Let s and s' represent CD and DA (Fig. Art. 113), then (Geom., Prop. XI. Bk. IV.),

Knowing 88 and s-s', s and s' can at once be found, and thence the angles A, C, and B, by Art. 122.

EXAMPLES.

1. Given of any triangle ABC, the side a equal to 216 yards, the side b equal to 217 yards, and the side c equal to 235 yards; to find the angles A, B, and C.

Solution. By (134), (135), and (136) we have

= 32° 52′ 8′′.6. C65° 44' 17".2.

A 28° 27' 47"; } B= 28° 40′ 4′′.4; † C Ans. A 56° 55′ 34′′; B 57° 20' 8".8 ;

2. Given the three sides of a triangle equal to 432, 543, and C54; to solve the triangle by means of the cosine.

3. Given the three sides of a triangle equal to 95.12, 162.34, and 98; to solve the triangle by means of the tangent.

Ans. The angles, 32° 14′ 53′′; 114° 24′ 9′′; 33° 20′ 58′′.

BOOK IV.

PRACTICAL APPLICATIONS.

DETERMINATION OF HEIGHTS AND DISTANCES.

131. A HORIZONTAL PLANE is one which is parallel to the horizon.

A VERTICAL PLANE is one which is perpendicular to a horizontal plane.

A HORIZONTAL LINE is one which is parallel to the hori

zon.

A VERTICAL LINE is one which is perpendicular to a horizontal plane.

132. A HORIZONTAL ANGLE is one the plane of whose sides is horizontal.

A VERTICAL ANGLE is one the plane of whose sides is vertical.

An ANGLE OF ELEVATION is a verti- D cal angle having one side horizontal and the inclined side above it; as the angle СА В.

An ANGLE OF DEPRESSION is a vertical angle having one side horizontal and

the inclined side under it; as the angle A

DBA.

B

C

133. To determine the height of a vertical object standing on a horizontal plane.}

Let B be the top of the object, and let it be required to find its height BC.