In what follows, only the steps in the solutions will be indicated. The examples that are worked may be saved, so that the amount of labor required by the method of solution shown here can be compared with the amount required by another method which will be described later.

There are four cases in the solution of oblique triangles; these cases correspond to the four problems of construction stated above. CASE I. Given two angles and a side opposite to one of them.

A A

α

с

B

D

In ABC let A, B, a be known. Angle C and sides b, c are re quired. From C draw CD at right angles to AB or AB produced. In triangle CBD, angle CBD and side CB are known. .. BD and DC can be found.

Then, in triangle ACD, side DC and angle A are known. .. AC and AD can be found.

=

Side AB AD – BD when B is obtuse, and AB= when B is acute.

Another method of solution is given in Art. 55.

AD + DB

1. Ex. 1, Art. 55.

3. Ex. 2, Art. 60.

EXAMPLES.

2. Ex. 2, Art. 55.

4. Other Exs. in Arts. 55, 60.

CASE II. Given two sides and an angle opposite to one of them.

N.B. The first part of the text in Art. 56 should be read at this time.

Let (Fig. 30) AC, BC, angle A be known. [In a certain case, as shown in Art. 56, two triangles can be drawn which satisfy the given conditions.] From C draw CD at right angles to AB or AB produced.

34.]

OBLIQUE TRIANGLES.

59

In ACD, AC and A are known. .. AD, DC, angle ACD, can be found.

Then, in BCD, BC and CD are known. .. BD, angle DBC, can be found.

In one figure, AB = · AD – BD, angle ABC = 180° — CBD.

In other figure, AB = AD + DB. In both figures,

angle ACB = 180° — (CAB + ABC). ·

Another method of solution is given in Art. 56.

EXAMPLES.

1. Ex. 1, Art. 56.

3. Ex. 1, Art. 60.

2. Ex. 2, Art. 56.

4. Other Exs. in Arts. 56, 60.

CASE III. Given two sides and their included angle.

In ABC let b, c, A be known. Side a, B, C are required. From C draw CD at right angles to AB or AB produced.

In ACD, AC and angle A are known. .. CD and AD can be found.

Then, in triangle CDB, CD is now known, and BD=AD-AB or AB - AD. .. Angle CBD can be found. Angle ABC= 180° - CBD in figure on the left. Angle ACB = 180° — (A + B). Another method of solution is given in Art. 57.

In ABC let a, b, c be known. The angles A, B, C are required.

From any vertex C draw CD at right angles to AB or AB pro

duced,

also,

CD2 = b2 — AD2;

---

CD2 = a2 — DB2 = a2 — (c — AD)3.

.. b2 — AD2 = a2 - (c — AD)2

b2 + c2 - a2

(1)

(2)

Also, DB = AD - c (one figure), or c AD (other figure).

Hence, in ACD, AC, AD are known.
Also, in CDB, CB, DB are known.

.. A can be found.

.. B can be found.

Another method of solution is given in Art. 58.

1. Ex. 1, Art. 58.

3. Ex. 1, Art. 62.

EXAMPLES.

2. Ex. 2, Art. 58.

4. Other Exs. in Arts. 58, 62.

5. Solve some of the problems in Art. 63 by means of right-angled triangles.

34 a. The area of a triangle in terms of the sides. (See Fig. 30.) From (1), (2), Case IV., Art. 34,

Let

then

(a + b + c)(− a + b + c) (a − b + c)(a + b −c).

4 c2

a+b+c=2s;

2 (s− a) = a+b+c−2 a = − a+b+c.

Similarly, 2 (sb) = a − b + c; 2 (s—c) = a + b — c.

.. Area ABC = 1⁄2 AB • CD = √ s(s — a)(s — b) (s — c).*

Ex. Find the areas of the triangles in Exs. Case IV., Art. 34. Check the results by finding the areas by the method of Art. 31.

* This is sometimes known as Hero's Formula for the area of a triangle. It was discovered by Hero (or Heron) of Alexandria, who lived about 125 B.C., and placed engineering and land surveying on a scientific basis.

34 b, 34 c.]

DISTANCE OF VISIBLE HORIZON.

34 6. Distance and dip of the visible horizon.

Let C be the centre of the earth, and let the radius be denoted by r.
Let P be a point above the earth's surface, and let

its height PL be denoted by h.

Join P, C; draw PB from P to any point in the visible horizon; draw the horizontal line PH in the same plane with PC, PB. Then angle HPB is called the dip of the horizon. By geometry,

Take r = 3960 mi., and let h be measured in feet. Then of P in miles.

61

P

H

height

5280

Hence, the distance of the horizon in miles is approximately equal to the square root of one and one-half times the height in feet.

EXAMPLES.

1. A man whose eye is 6 ft. from the ground is standing on the seashore. How far distant is his horizon?

Distance = √ × 6 mi. = 3 mi.

2. Find the greatest distance at which the lamp of a lighthouse can be seen, the light being 80 ft. above the sea level.

3. Find the height of the lamp of a lighthouse above the sea level when it begins to be seen at a distance of 12 mi.

4. From the top of a cliff, 40 ft. above the sea level, the top of a steamer's funnel which is known to be 30 ft. above the water is just visible. What is the distance of the steamer?

5. Find the distance and dip of the horizon at the top of a mountain 3000 ft. high?

6. Find the distance and dip of the horizon at the top of a mountain 21 mi. high.

34 c. Examples in the measurement of land. In order to find the area of a piece of ground, a surveyor measures distances and angles sufficient to provide data for the computation. An account

of his method of doing this, and of his arrangement of the data and the results in a simple, clear, and convenient form, belongs to special works on surveying. This article merely gives some examples which can be solved without any knowledge of professional details. The various rules for finding the area, of a triangle and a trapezoid, are supposed to be known. In solving these problems, the student should make the plotting or mapping an important feature of his work.

The Gunter's chain is generally used in measuring land. It is 4 rods or 66 feet in length, and is divided into 100 links.

An acre 10 square chains=4 roods=160 square rods or poles. The points of the compass have been explained in Art. 30.

EXAMPLES.

1. A surveyor starting from a point A runs S. 70° E. 20 chains, thence N. 10° W. 20 chains, thence N. 70° W. 10 chains, thence S. 20° W. 17.32 chains

D

IN

D

Ꮯ

C1

H

to the place of beginning. What is the area of the field which he has gone around?

Make a plot or map of the field, namely, ABCD. Here, AB represents 20 chains, and the bearing of B from A is S. 70° E. BC represents 20 chains, and the bearing C from В is N. 10° W., and so on. Through the most westerly point of the field draw a north-and-south line. This line is called the meridian. In the case of each line measured, find the distance that one end of the line is east or west from the other end. This easting or westing is called the departure of the line. Also find the distance that one B end of the line is north or south of the other end. This northing or southing is called the latitude of the line. For example, in Fig. 30 b, the departures of AB, BC, CD, DA, are B1B, BL, CH, DD1, respectively; the latitudes of the boundary lines are AB1, B1С1, C1D1, D¡A, respectively. It should be observed (Art. 36) that the algebraic sum of the departures of the boundary lines is zero, and so also is the algebraic sum of their latitudes. The following formulas are easily deduced:

B IS

FIG. 30b.

L

Departure of a line = length of line × sine of the bearing;

Latitude of a line = length of line x cosine of the bearing.

By means of the departures, the meridian distance of a point (i.e. its distance from the north-and-south line) can be found. Thus the meridian