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(1) The functions of one angle are equal to the conamed functions of the complementary angle, hence the term cofunction. Example:

sin 65o=cos (90—65)=cos 25°
cot 20°=tan (90—20)=tan 70°

(2) Therefore, any function of an angle between 45° and 90° may be found by taking the conamed function of the complementary angle which is between 0° and 45°. Thus we need never have a direct table of functions beyond 45°.

6. Exercises.-Express the following functions as functions of angles less than 45o.

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sin 44°,

67. Use of tables.-2. The functions of angles, other than those discussed previously, are not so easily computed. More advanced methods are employed in securing these functions for other angles. Table II of appendix II has been computed by these methods, which will be used. The table gives the values of the functions to four decimal places for every degree from 0° to 90°. All such values are only approximate as the fourth decimal place has been rounded off to the nearest integer. As previously explained, cos 45o=sin 45°, cos 46°=

etc. Hence the column of sines from 0° to 45° is the same as the column of cosines from 45° to 90°. Thus, in finding the functions of angles from 0° to 45°, read from the top down; in finding the functions of angles from 45° to 90°, read from the bottom up.

b. Exerciser.-(1) From table II (app. II) find the values of the following: sin 5° cot 5° =11.4301

cot 45° sin 14° .2419 cot 82° =

cot 82°

. 1405 sin 35° tan 75°= 3. 7321

tan 17°
1219
tan 19°

sin 90° 1. 0000 sin 0° . 0000

cos 90°=-
. 9925
cos 15°

sin 3° .0523 sin 75°= tan 42°= .9004

tan 46°

=

=

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cos 83° cos 720 cos 7°

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(2) Given the following values from the table find the corresponding angles: sin A=.2588 LA=

sin A= .6428 LA=40° sin B=.9205 ZB=67°

tan B= 4245 ZB=

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68. Methods of solving right triangles.-a. Given any two sides or one side and one angle, other than the 90° angle, all other parts of a right triangle, including the area, can be found. In working with problems where one angle and a side are given there is a choice of two functions to use. In one of these, multiplication is all that is necessary; but in the other, division by a 4- or 5-place decimal becomes necessary. (1) Example:

Given: Figure 114, where ZA=20° and c=50 units. Find side a.

B

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From this comparison it is evident that the proper choice of functions will save a great deal of unnecessary work. (The easier method is to start with the unknown in the numerator.) When any problem of this type is encountered there is always one function that can be chosen to assure the multiplication rather than the division operation. (2) Example:

Given: A=40°, b=20 units. Find side a.

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43---8

113

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Solution: In this case side 6 is known and side a is to be deter

mined, so the fraction should be s

By definition: tan A, or 6 tàn A=a

Therefore a=20 tan 40o=(20) (.8391)=16.782

b. Another type problem is one where two or three sides are known and the angle is unknown.

In this kind a division is necessary.
Example:
Given: a=5.5 and c= =11. Find angle A.

B

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Looking this up in the table it is found that sin 30°=.5000, therefore
LA=30°.
c. Exercises.
ZA=18°, c=18. Find a
ZA=47°, b=29. Find c

42.57 Answer. LA=32°, a=42. Find 6 ZA=75°, b=37. Find a

138.09 Answer. ZA=29°, c=72. Find b ZA=55°, a=49. Find c

59.82 Answer. a=18, b=45. Find ZA to nearest degree a=12, c=27. Find ZA to nearest degree

26° Answer. An airplane at an elevation of 17,000 feet sights an airfield. Measurements to the nearest boundary show the angle of depression to be 7o. How far in a straight line is the airplane from the field and what is the horizontal distance from a point directly above the boundary?

Seen from a point on the ground, the angle of elevation of an airplane is 64o. If the airplane is 10,000 feet above the ground, how far is it in a straight line from the observer?

11,300 feet. Answer. 69. Interpolation.-a. Interpolation is a method of estimating the value of functions of angles which are not given in the tables or estimating the angle, given the function which is not listed in the tables. Briefly, it is a process that assumes a straight line difference between two values such that sin of 27.5° has a value halfway between 27° and 28°, and may be found by adding one-half their difference to the function of 27°.

(1) Example: Find the sin of 47.5°

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Therefore

The desired function is .5 of the way from .7314 to .7431.

.5 X.0117=.0059 and

sin 47.5o=.7314+.0059=.7373 (2) Example: Find the sin of 18°20'.

Answer.

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Therefore: cot 34.2°=1.4826 .0109=1.4717

Answer. (4) It will be noted that in the third example the process varies slightly. It is necessary to subtract the difference from the value of the smaller angle. This is true 'in the case of all cofunctions because their values decrease as the angle increases. varies slightly when an angle is desired from a given function.

(5) Example: Find the angle whose sin is .4295

The process

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Difference between sin 25° and sin 26°=.0158
Difference between sin 25° and sin X =.0069

.0069 From this the desired angle is

.0158

of the way from 25° to 26°.

Changing this to a decimal, it becomes .44 or rounding off to tenths, 0.4.

Therefore the desired angle is 25.4°.

b. Exercises.—(1) Find the required functions of the following angles: (a) sin 18.4°

(d) cot 53°20' .7446 Answer. (6) cos 23.7° .9156 Answer. (e) sin 27°49' (c) tan 72.9°

(f) tan 44°50' .9943 Answer. (2) Find the angles whose functions are: (a) sin A=.7590

(d) cot R=1.0990 (6) cos B=.8028

ZR=42.3°

Answer.
ZB=36.6° Answer. (e) tan A=1.0490
(c) tan D=.0700

() sin B=.3810
ZB=22.4

Answer. (3) Find the unknown sides and angles in the following right triangles by trigonometry:

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