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EXERCISES V.- PYTHAGOREAN RELATIONS

IDENTITIES

1. In exercises (a) — (i) determine the values of the remaining funcby each of the methods of Example 2, p. 16.

tions of the acute angle

V

(a) sin 0 = = 3/5.

(b) sin = 1/3.

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Prove the following relations for any acute angle 0:

2. (sin cos 0)2 = 1 + 2 sin cos 0. 4. tan 0+ ctn = sec 0 csc 0.

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3. cos

tan 0 = sin 0.

5. sin 0 sec = tan 0.

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sin2 0 = 1 2 sin2 0 = 2 cos2 0 1.

9. sec2 0 csc20

tan2 0 + ctn2 0 + 2.

10. If a and b are the sides of a right triangle, c the hypotenuse, and A the angle opposite a, show that the area of the triangle is equal to either of the expressions (ac cos A)/2, (bc sin A)/2.

K

B

N

α

11. Two straight pieces of railroad track MA and NB are to be connected by a circular track AKB with a radius of 500 ft. and center O, tangent to MA and NB. The straight por- M tions of the track produced intersect at a point V at an angle of 100°. (a) How far back from V should the track begin to turn? (b) How far from V along the bisector OV of the angle AVB is the center O? (c) Find the shortest distance from V to the curved portion.

12. If, in a figure similar to that of Ex. 11, ZAVO is any angle, and Z VOA is denoted by a, and OA=r, show that (a) AV = r tan α ; (b) KV = r exseca; (c) AB = 2 r sin α.

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(a) LADB · A/2; (b) BD = 2 c cos (A/2).

(c) From the right triangles DCB and ACB,

c sin A = a = 2 c cos (A/2) sin (4/2);

sin A

(d) Likewise, show that

=

2 sin (A/2) cos (A/2);

c cos A = b 2 c cos2 (A/2) - c;

hence cos A = 2 cos2 (A/2) — 1 = cos2 (A/2) — sin2 (A/2).

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PART II. ACCURACY OF SOLUTIONS-APPLICATIONS

11. The Question of Greater Accuracy. The degree of accuracy of the results obtained by using the values of the trigonometric functions to three places of decimals, while sufficient for many ordinary applications, is not satisfactory for some purposes; for example, in extended surveys, in astronomy, and in any work for which the data must be determined by using instruments of precision.*

More accurate values have been calculated. The values for angles at intervals of 1' are given to five decimal places in the Tables (Table II). Still more accurate values are available in separately printed tables.

12. Functions of 30°, 45°, 60°. To determine the functions of the angle 45° construct an isosceles right triangle having each of its equal sides m units in length. Each of the equal angles is 45°. The hypotenuse is m√2; hence sin 45°= m/(m√2) = 1/√2 = √2/2 = .7071+. Compute the values of the other functions of 45° in a similar manner.

=

To determine the functions of 30° and 60°, draw an equilateral triangle of side m and drop a perpendicular from one vertex to the opposite side. The acute angles of each of the right triangles thus formed are 30° and 60°. Find the hypotenuse and the sides of one of these right triangles in terms of m, and compute the values of each of the functions for each of these angles.

The values in the following table should be memorized:

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*Thus it is obviously unwise to use values to more than three decimal places in reducing any large measurements on the earth's surface unless a standard steel tape and other standard surveying instruments are available.

13. Use of the Large Tables. Five-place tables are used in precisely the same manner as the small table of p. 15.

Example 1. One angle of a right triangle is 42° 20′ and the hypotenuse is 28 ft. 6 in. long. Find the remaining sides and the other angle. Draw a diagram to illustrate the problem, indicating the given parts. Denote the unknown parts by the letters a and b, as in Fig. 13.

To find b, note that it is the side adjacent to the given angle, and that the hypotenuse is given. Hence, by (14), § 6,

a

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H

b

42

20

h

FIG. 13

Note that a is opposite the given angle; hence by (13), § 6,

a = 28.35 sin 42° 20′ = 28.5 × .67344 = 19.09;

the sine and the cosine of 42° 20' being found in a table.

To find ẞ, note that it is the complement of 42° 20′; hence ẞ = 47° 40'.

Example 2. The perpendicular sides of a right triangle are 22 ft. 6 in. and 54 ft., respectively. Find the hypotenuse and the angles. Draw a diagram, indicating the given parts and lettering the parts to be found, as in Fig. 14. To find α, note that the given parts are the sides opposite and adjacent to it; hence by the definition of tangent, we may write = 22.5 ÷ 54 =.41667.

A

54 ft.

tan α =

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Another method is the following: By the Pythagorean theorem of plane geometry, using a table of squares and square roots or by direct calculation,

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FIG. 14

Hence h = 58.5.
table of squares.

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But this method is arduous without a See Table VI.

Having found α = 22° 37' we might find h as follows. By (15), § 6, h = 54/(cos 22° 37′) = 54/.92310 = 58.498. This method is no shorter than the one used above, and is open to the objection that any error made in computing a vitiates the resulting value found for h. In general, compute each unknown part from the given parts; i.e. do not use computed parts as data if it can be avoided.

As in these Examples, observe the procedure suggested on

EXERCISES VI.-RIGHT TRIANGLES

LARGE TABLES

1. Solve the following right triangles. The hypotenuse is denoted by h, other sides by other small letters, and any angle by the capital letter corresponding to the small letter that denotes the side opposite it.

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2. The base of an isosceles triangle is 324 ft., the angle at the vertex is 64° 40'. Find the equal sides and the altitude.

3. A chord of a circle is 21.5 ft., the angle which it subtends at the center is 41°. Find the radius of the circle.

4. To determine the width BA of a river, a line BC 100 rods long is laid off at right angles to a line from B to some object A on the opposite bank visible from B. The angle BCA is found to be 43° 35'. Find AB. 5. The shadow of a tower 200 ft. high is 252.5 ft. long. angle of elevation of the sun?

What is the

6. Two ships in a vertical plane with a lighthouse are observed from its top, which is 200 ft. above sea level. The angles of depression of the two ships are 15° 17' and 11°22'. Find the distance between the ships.

14. Projections. The projection of a line segment AB upon a line is defined to be the portion MN of the line 7 between perpendiculars drawn to it from A and B, respectively. The

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or, the projection of a segment upon a given line is equal to the product of the length of the segment and the cosine of the angle the segment makes with the given line.

The projections of a segment upon the coördinate axes are frequently used. If the segment makes an angle a with the horizontal, the projections on the x and y axes are, respectively,

(1) Proj, AB=AB cos a, Proj, AB=AB sin α,

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FIG. 16

where Proj, AB and Proj, AB denote the projections of AB on the x-axis and the y-axis, respectively.

15. Applications of Projections. In mechanics and related subjects, forces and velocities are represented graphically by line segments. A force, say of 10 lb., is represented by a segment 10 units in length in the direction of the force. A velocity of 20 ft. per sec. is represented by a segment 20 units in length in the direction of motion.

The projection upon a given line 7, of a segment representing a force, represents the effective force in the direction 7; this is called the component of the given force in the direction 7.

Example 1. A weight of 50 lb. is placed upon a smooth plane inclined at an angle of 27° with the horizontal. What force acting directly up the incline will be required to keep the weight at rest?

N

27°

R

50 sin27

M

127

Draw to some convenient scale a segment 50 units in length directly downward to represent the force exerted by the weight. Project this segment upon a line inclined at an angle of 27° with the horizontal. The length of this projection is 50 sin 27° = 22.7 nearly. This represents the component of the force down the plane. Therefore, a force of 22.7 lb. acting up the plane will be required to keep the weight at rest.

EXERCISES VII. - PROJECTIONS

FIG. 17

1. Find the horizontal and vertical projections of the segments:
(a) length 42, making an angle of 37° with the horizontal.
(b) length 5.5, making an angle of 50° with the vertical.

2. A straight railroad crosses two north and south roadways a mile apart. The length of track between the roadways is 14 mi. A train travels this distance in 2 min. Find the components of the velocity of the train parallel to the roadways and perpendicular to them. Find the angle between the track and either roadway.

3. The eastward velocity of a certain train is 24 mi. per hour. The northward velocity is 32 mi. per hour. Find its actual velocity along the track and the angle the track makes with the east and west direction.

4. A car is drawn by means of a cable. If a force of 5000 lb. exerted along the track is required to pull the car, what force will be required when the cable makes an angle of 15° with the track?

5. Find the horizontal and vertical components of a force represented by a segment 30 units long at an angle of 40° with the horizontal.

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