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82. Application to right triangles.-a. Logarithm table.-(1) Table IV of appendix II gives common logarithms of circular or trigometric functions and will be used to solve trigonometric problems by logarithms. It is to be understood that a (-10) must be supplied whenever a characteristic of 8 or 9 is given in a log column. Interpolation is applied as in common logarithms.

(2) Corresponding to 18°, one finds log sin 18°=9.4980—10. For 62°, it is found that log tan 62°=.2743. Again, by interpolation, log cos 7.7°=9.9961–10.

b. Solving right triangles.The following process gives the unknown parts of a right triangle: By use of the definition of the sine, cosine, tangent, and cotangent functions, write an equation which involves just one missing part; then solve this equation for the missing part, and perform the indicated arithmetic by logarithms.

Example 1: Given A=62°10', a=78; find B, b, and c.

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b=41.19

a

(3)

sin A

a

78
sin A Sin 62010

62°10'
log 78=11.8921-107

Subtract
log sin 62°10'= 9.9466-10

tract

log c= 1.9455

C=88.2 Example 2: Given a=40, c=59.3; find A, B, and b. (Since the pythagorean relation' is not well adapted to logarithms, find the angles first, and then find b by a trigonometric relation.)

40
(1) sin A

59.3
log 40=11.6021-101

Subtr
log 59.3= 1.7731

a

с

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log b=11.6414-10

b=43.79 83. Evaluating formulas with logarithms.-a. There has been given a number of formulas for calculating areas and volumes of various objects. By solving the equation of the formula for a missing part, logarithms can often be used to compute the missing part when the others are given. Example 1: Find the volume of a sphere whose radius is 6 feet.

Use a=3.1416, or log r=.4971

4 V=

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1 The relation of the hypotenuse to the legs of a right triangle, namely that the square of the hypotenuse is equal to the sum of the squares of the legs.

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log V=2.9567

1-905.2 cubic feet Erample 2: A right circular cone has an altitude of 18 inches and a volume of 364 cubic inches. What is the radius of its base?

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log denom.=1. 7524 b. Exercises.-Using the conventional triangle, solve for the missing parts of the following right triangles, finding sides to four significant digits and angles to tenths of degrees or minutes.

(1) A=23.5°, c=627 (2) B=76°15', c=93.4 A=13°45', a=22.20, b=90.7 Answer. (3) A=60°, b=4 (4) B=68°a=73

A=22°, b=180.7, c=194.9 Answer. (5) B=3.5, b=2

(6) a=21.9, c=91.9

A=13°47'1" B=76°13' b=89.25 Answer. (7) b=18.3, c=30.75

(8) A tin can has base diameter 4.5 inches and height 5 inches. What is its cubic capacity?

79.52 cubic in. Answer. (9) What is the radius of a sphere whose volume is 700 cubic feet? (10) What is the volume of a sphere whose radius is twice that of the sphere in the preceding problem?

5,600 Answer. (11) Find the angle at the vertex of the cone discussed in example 2, a above.

84. Oblique triangles.-a. It is often possible to find the missing parts of oblique triangles by constructing auxiliary lines, generally altitudes, which will allow the use of the theory developed for right triangles.

Example 1: Find side b of the triangle shown in figure 127.

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Construct the altitude, BD, to the side AC, and two right triangles are formed: ABD and BDC.

AD (1) In triangle ABD, cos A and AD=C COS A=20.95 cos 20.5°. log 20.95= 1.3212

ladd
log cos 20.5°= 9.9716-10
log AD =11.2928-10
AD =19.62

DC (2) In triangle BDC, cos C= and DC=a cos C=15.62 cos

a

27.5o.
log 15.62= 1.1937

Add
log cos 27.5o= 9.9479-10
log DC =11.1416-10

DC =13.85 (3) b=AD+DC=19.62+13.85, b=33,47 Example 2: Find sides a and b of the triangle shown in figure 128.

Drop a perpendicular from B to the line AC, extended, meeting it at D.

AD (1) In the right triangle ABD, cos A= and AD=C cos A=

с

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BD=C sin A=73.34 sin 32.4°
log 73.34= 1.8653

>Add
log sin 32.4° 9.7290–10
log BD = 11.5943-10

BD =39.29 (3) In right triangle ADB,

ZDBA=90°--32.4°=57.6°
In right triangle BCD

ZDBC=57.6°- 28.2°=29.4°
ZBCD=90°—24.4°=60.6°

CD
cot BCD

BD
CD=BD cot ZBCD

CD=39.29 cot 60.6°
log 39.29= 1.5943

Add log cot 60.6°= 9.7509-10

log CD=11.3452-10

CD=22.14

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