16. The Use of Logarithms. Logarithms may be used to shorten computations involving multiplications, divisions, raising to powers or extracting roots, but not involving additions or subtractions. In much of the numerical work which follows, the use of logarithms is very advantageous in saving time and labor, but the student should bear in mind that logarithms are not necessary. They are merely convenient, and they belong. no more to trigonometry than to arithmetic. One of the questions which a computer has to decide is whether or not it will be advantageous to use logarithms in a given problem. Tables usually contain (1) tables of logarithms; (2) tables of the trigonometric functions; (3) tables of logarithms of the trigonometric functions. The notation log tan 61° means the logarithm of the tangent of 61°; i.e. the tangent of 61° is a number, 1.804+, and the logarithm of this number 1.804+ is .25625. A formula which has been arranged so as to involve only products and quotients of powers and roots of quantities either known or easily computed is said to be adapted to logarithmic computation. Thus the formula h = √a2 + b2, which gives the hypotenuse h of a right triangle in terms of the sides a and b, is not adapted to logarithmic computation. On the other hand, the formula b = √h2 — a2 = √(h + a)(h − a) which gives one side in terms of the hypotenuse and the other side is adapted to logarithmic computation because (h + a) and (h − a) are easily obtained from h and a. Thus, if the hypotenuse is 17.34 and one side is 12.27, the other side is Again, such an expression as 12.5 sin 42° 37', which might occur in solving a triangle, would be computed as follows: 17. Products with Negative Factors. To find by use of logarithms the product of several factors some of which are negative, the product of the same factors, all taken positively, is first obtained, and the sign is then determined in the usual way by counting the number of negative signs. Thus, to obtain the product of the four factors 115, 23.41, .6422, .1123, we write x = :(115)(23.41)(.6422)(.1123); then Here the fact that the first factor and the last two are negative is indicated by writing an (n) in parenthesis to the right of the corresponding logarithms. The product of the given factors all taken positively is 194.15; since the number of negative factors is odd the product is really — 194.15. For other processes, see the Explanation of the Tables. EXERCISES VIII. — RIGHT TRIANGLES — MISCELLANEOUS 1. Solve by means of logarithms the following right triangles, where h denotes the hypotenuse, other small letters the sides, and the corresponding capital letters the angles opposite those sides. 2. A tree stands on the opposite side of a small lake from an observer. At the edge of the lake the angle of elevation of the top of the tree is found to be 30° 58'. The observer then measures 100 ft. directly away from the tree and finds the angle of elevation to be 18° 26'. Find the height of the tree and the width of the lake. 3. From a point 250 ft. from the base of a tower and on a level with the base the angle of elevation of the top is 62° 32'. Find the height. 4. To determine the height of a tower, its shadow is measured and found to be 97.4 ft. long. A ten-foot pole is then held in vertical position and its shadow is found to be 5.5 ft. Find the height of the tower and the angle of elevation, of the sun. 5. Find the length of a ladder required to reach the top of a building 50 ft. high from a point 20 ft. in front of the building. What angle would the ladder in this position make with the ground? 6. The width of the gable of a house is 34 ft.; the height of the house above the eaves is 15 ft. Find the length of the rafters and the angle Find the pitch of the roof. (Ex. 10, p. 11.) of inclination of the roof. 7. A kite string is 250 ft. long and makes an angle of 40° with the level ground. Find approximately the height of the kite above the ground, neglecting the sag in the string. 8. One bank of a river is a bluff rising 75 ft. vertically above the water. The angle of elevation of the top of the bluff from the water's edge on the opposite bank is 20° 27'; find the width of the river. 9. A taut rope 100 ft. long is attached to the top of a building. The free end reaches the ground 24 ft. 7 in. away from the base of the building. Find the height of the building and the angle which the rope makes with the ground. 10. Find the angles which the diagonal of a rectangle 12 ft. wide and 17 ft. long makes with the sides. 11. A chord of a circle is 100 ft. long and subtends an angle of 40° 42' at the center. Find the radius of the circle. 12. A hill rises 8 ft. vertically in a horizontal distance of 40 ft. Find the angle of inclination of the hill with the horizontal. What is the difference in elevation of two points that are 500 ft. apart measured up the hill ? 13. Find the length of a side of an equilateral triangle circumscribed about a circle of radius 15 inches. 14. Devise a formula for solving an isosceles triangle when the base and the base angles are given; when the base and one of the equal sides 18. From the same figure show that sin = 2 sin (0/2) cos (0/2). [HINT. Find the area of ▲ ABC, using first AB, then BC, as base.] CHAPTER III SOLUTION OF OBLIQUE TRIANGLES PART I. FUNCTIONS OF OBTUSE ANGLES 18. Obtuse Angles. The solution of oblique triangles involves obtuse * as well as acute angles. Let an obtuse angle a be placed on the coördinate axes with the vertex at the origin and one side along the x-axis to the right; then the other side will fall in the second quadrant. The ratios sin a, cos a, etc., are defined in terms of x, y, and r=√x2+ y2 precisely as for acute angles. (See § 6.) It should be noticed, however, that since x is negative while y and rare positive, every ratio which involves x is negative for an obtuse angle; thus x/r: r = cos a, y/x= tan a, and their reciprocals, sec a and ctn α, are all negative for obtuse angles. FIG. 19 We have seen how it is possible to find the ratios of angles greater than 45° from a table extending no farther than 45°, by means of the relations sin (90° — α) = cos a, etc., which were proved on p. 13. By means of similar relations it is possible to find the ratios of obtuse angles from the same table. FIG. 20 (ab) Let a be placed on coördinate axes as described above, and let the supplement of a be denoted by B (which is an acute angle). Lay off B from Ox so that the other side of ẞ falls in the first quadrant. From a point P in the side of a (in second quadrant) and a point P' in the side of B (in first quadrant) at the same distance r from the origin, draw the * An obtuse angle is an angle which is greater than 90° and less than 180°. 25 perpendiculars PM, P'M', as in Fig. 20. The value of x for the point P will be negative since P is in the second quadrant. Let its coördinates be (a, b); then, since the triangles OPM, OP'M' are symmetric, the coördinates of P' are (a, b). As in § 6, we have It follows that if a is an obtuse angle we find its sine by looking for the sine of its supplement, which is an acute angle, and similarly for the other functions, always having regard for the proper sign. The relations just found, together with those of § 9, enable us to find the values of the functions for any angle which can occur in a triangle, from a table which gives them from 0° to 45°. EXERCISES IX.-FUNCTIONS OF OBTUSE ANGLES 1. From the accompanying figure prove the following relations: 3. Find the values of the remaining functions of the angles of Ex. 2. 4. Express the following as functions of an angle less than 45°, and look up their values in a table. |