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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.
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.)
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
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.
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?
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.
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
DC (2) In triangle BDC, cos C= and DC=a cos C=15.62 cos
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=
BD=C sin A=73.34 sin 32.4°
BD =39.29 (3) In right triangle ADB,
CD=39.29 cot 60.6°
Add log cot 60.6°= 9.7509-10