Imágenes de páginas
PDF
EPUB
[ocr errors]

Similar formulas hold for B and C. They can be deduced in the same manner as those for A; or, they can be written immediately, from the symmetry apparent in the formulas (3)–(5). The student is advised to derive the similar formulas for † B, C, viz.:

[ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small]

Formula (5) can be given a more symmetrical form; for, on multiplying and dividing its second member by (s — a),

[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]

When all the sides are known, the angles can be found by means of formulas (3)–(5′) or, by (7)-(8'). When all the angles are required, the tangent formulas are better, since fewer logarithms are required than in (3), (4), (3′), (4'). It will be shown in Art. 69 that r is the radius of the circle inscribed in the triangle.

EXAMPLES.

[merged small][merged small][merged small][ocr errors][merged small]

1. In triangle ABC, a = 25.17, b = 34.06, c = 22.17. Find

A, B, C.

[ocr errors]

(s − a) (s — b) (s

[ocr errors][ocr errors]

b = 34.06

s-a

== ; tan C =
S b

r

=

[ocr errors]

.. logr = log tan

[ocr errors]

A C-22.17 B

25.17

n

[log (s − a) + log (s − b) + log (s − c) — log s].

[blocks in formation]
[blocks in formation]

FIG. 60.

A = logr — log (s — a); log tan B = log r — log (s — b);

[blocks in formation]

2. Solve ABC, given a = 260, b =

3. Solve ABC when a = 26.19,

4. Solve PQR, given p = 650, q

b

[ocr errors]
[blocks in formation]

736, r = 914.

= 2592.

5. Solve RST, given r = 1152, s = 2016, t 6. Solve Exs. 1, 4, Art. 58, using formulas (3)–(8′), without logarithms.

63. Problems in heights and distances. Some problems in heights and distances have been solved in Art. 29 by the aid of right-angled triangles. Additional problems of the same kind will now be given, in the solution of which oblique-angled triangles may be used. It is advisable to draw the figures neatly and accurately. The graphical method should also be employed.

[blocks in formation]

In the triangle ABP (Fig. 23), AB = 100 ft., BAP = 30°, PBA = 180° - 45° = 135°. Hence the triangle can be solved, and BP can be found. When BP shall have been found, then in the triangle CBP, BP is known and BP = 45°; hence CP can be found. The computation is left to the student.

2. Another solution of Ex. 3, Art. 29. In the triangle CBP (Fig. 24), BP 30 ft., BCP 40° 20′ — 38° 20′ = 2°, PBC = 90° + LCB = 128° 20'. Hence CBP can be solved and the length of CB can be found. When CB shall have been found, then, in the triangle LCB, angle C = 38° 20', CB is known, and hence LB can be found. The computation is left to the student.

3. Find the distance between two objects that are invisible from each other on account of a wood, their distances from a station at which they are visible being 441 and 504 yd., and the angle at the station subtended by the distance of the objects being 55° 40'.

4. The distance of a station from two objects situated at opposite sides of a hill are 1128 and 936 yd., and the angle subtended at the station by their distance, is 64° 28'. What is their distance ?

5. Find the distance between a tree and a house on opposite sides of a river, a base of 330 yd. being measured from the tree to another station, and the angles at the tree and the station formed by the base line and lines in the direction of the house being 73° 15′ and 68° 2′, respectively. Also find the distance between the station and the house.

6. Find the height of a tower on the opposite side of a river, when a horizontal line in the same level with the base and in the same vertical plane with the top is measured and found to be 170 ft., and the angles of elevation of the top of the tower at the extremities of the line are 32° and 58°, the height of the observer's eye being 5 ft.

7. Find the height of a tower on top of a hill, when a horizontal base line on a level with the foot of the hill and in the same vertical plane with the top of the tower is measured and found to be 460 ft.; and at the end of the line nearer the hill the angles of elevation of the top and foot of the tower are 36° 24', 24° 36', and at the other end the angle of elevation of the top of the tower is 16° 40'.

8. A church is at the top of a straight street having an inclination of 14° 10' to the horizon; a straight line 100 ft. in length is measured along the street in the direction of the church; at the extremities of this line the angles of elevation of the top of the steeple are 40° 30', 58° 20'. Find the height of the steeple.

9. The distance between the houses C, D, on the right bank of a river and invisible from each other, is required. A straight line AB, 300 yd. long, is measured on the left bank of the river, and angular measurements are taken as follows: ABC = 53° 30', CBD=45° 15', CAD=37°, DAB=58° 20'. What is the length CD?

10. A tower CD, C being the base, stands in a horizontal plane; a horizontal line AB on the same level with the base is measured and found to be 468 ft.; the horizontal angles BAC, ABC, are equal to 125° 40', 12° 35', respectively, and the vertical angles CAD, CBD, are equal to 30° 20'′, 11° 50′, respectively. Find the height of the tower and its distances from A and B.

11. A base line AB 850 ft. long is measured along the straight bank of a river; C is an object on the opposite bank; the angles BAC, ABC, are observed to be 63° 40′, 37° 15', respectively. Find the breadth of the river.

12. A tower subtends an angle a at a point on the same level as the foot of the tower and, at a second point, h feet above the first, the depression of the foot of the tower is B. Show that the height of the tower is h tan a cot ß.

13. The elevation of a steeple at a place due south of it is 45°, and at another place due west of it the elevation is 15°. If the distance between the two places be a, prove that the height of the steeple is a(√3-1)÷2√3.

14. The elevation of the summit of a hill from a station A is a; after walking c feet toward the summit up a slope inclined at an angle ẞ to the horizon the elevation is 7. Show that the height of the hill above A is c sin a sin (y - ẞ) cosec (y — a) ft.

64. Summary. The preceding discussions on the solution of triangles have shown that a triangle may be solved in the following ways:

I. By the graphical method. [Arts. 10, 14, 21–24.]

II. If the triangle is right-angled, it can be solved, either with or without logarithms, by the methods shown in Arts. 25–27.

III. If the triangle is oblique, it can be divided into rightangled triangles, each of which can be solved by either of the methods II. [Art. 34.]

IV. The triangle, whether right-angled or oblique, can be solved without using logarithms, by means of formulas (1), (3), Art. 54; (1) or (2), Art. 61; (3)-(8), Art. 62.

V. The triangle, whether right-angled or oblique, can be solved with the use of logarithms, by means of formulas (1), Art. 54; (1) or (2), Art. 61; (3)-(8), Art. 62.

Checks: Any formula not employed in the computation can be employed as a check; that is, as a test for the correctness of the result.

Two things are necessary on the part of one who wishes to do well in the solution of triangles:

(1) The formulas referred to above should be clearly understood and readily derived.

(2) The arithmetical work required should be done accurately.

N.B. Questions and exercises suitable for practice and review on the subject-matter of this chapter will be found at pages 189–193,

CHAPTER VIII.

SIDE AND AREA OF A TRIANGLE.

CIRCLES CON

NECTED WITH A TRIANGLE.

65. Length of a side of a triangle in terms of the adjacent sides and the adjacent angles. In this proof, regard is paid to the conventions about signs, described in Arts. 36, 37. Let ABC be any triangle. From A draw AD perpendicular to BC, or BC pro

duced. The positive direction of BC is in the direction of V. [At the first reading, only Fig. 61 a may be regarded.]

[blocks in formation]

Therefore, in any triangle each side is equal to the sum of the products of each of the other sides by the cosine of the angle which it makes with the first side.

When C is a right angle, (1) reduces to a =

EXAMPLE.

formulas.

c cos B.

Write the corresponding formulas for

and c.

Derive these

116

« AnteriorContinuar »