The letter S usually stands for Vs (sa) (s—b) (s — c), so that the above may be written a Similarly, 2S sin A 115. The student is advised to make himself thoroughly familiar with the following formulæ : = tan B 2 b2 + c2 a2 = 2 S Vs (s a) (s - b)(s c) = bc (ii.) (Art. 105.) abc (iii.) (Art. 106.) 116. The sign of the radicals in v., vi., and vii. of Art. 115 is cannot be imaginary, since, then, either s-a, 2' 2 sb, sc, is negative; which is impossible. 6. a sin (B — C') + b sin ( C – A) + c sin (A – B) = 0. 10. a + b + c = (b + c) cos A + (c + a) cos B + (a + b) cos C. 11. b + c − a = (b + c) cos A − (c − a) cos B + (a − b) cos C. In solving the following list of examples, the student will select. from the formulæ of Art. 115 those best suited to his purpose. EXAMPLE. Given a = 2, b = √6, c = 1+ √3. To find the angles of the triangle. It is evident that we may apply either iv., v., vi., or vii.; but that ii., iii., and viii. contain two angles, and hence cannot be used. We shall employ We find iv. in the case of an equilateral triangle. 2. The sides of a triangle are as 2: √6:1+ √3; find the angles. 3. The sides of a triangle are as 4, 2√2, 2(√3 − 1); find the angles. 4. Given C 120°, c = √19, a = 2; find b. = 5. Given A = 60°, b = 4√7, c = 6√7; find a. 6. Given A 45°, B = 60°, and a = 2; find c. 7. The sides of a triangle are as 7: 8: 13; find the greatest angle. 8. The sides of a triangle are 1, 2, √7; find the greatest angle. 9. The sides of a triangle are as a : b : √ (a2 + ab + b2); find the greatest angle. 10. When abc as 3:4:5, find the greatest and least angles; given cos 36° 52′ = .8. = 11. If a 5 miles, b6 miles, c = 10 miles, find the greatest angle. [cos 49° 33′ = .65.] 12. If a = 4, b = 5, c = 8, find C. Given that cos 54° 54′ = .575. 13. a: b = √3:1, and C = 30°; find the other angles. 14. If b 3, C = 120°, c = √13, find a and the sines of the other angles. 15. Given A 105°, B = 45°, c = √2; solve the triangle. 16. Given B = 75°, C′ = 30°, c = √8; solve the triangle. 17. Given B = 45°, c = √√/75, b = √50; solve the triangle. 18. Two sides of a triangle are 3v6 yards and 3√3 + 1 yards, and the included angle is 45'; solve the triangle. 19. If the angles adjacent to the base of a triangle are 221° and 1121°, show that the perpendicular altitude will equal half the base. 20. If A = 45° and B = 60°, show that 2 c = a(1 +√3). 21. The cosines of two of the angles of a triangle are and; find the ratio of the sides. CHAPTER XIV LOGARITHMS 117. Before proceeding to the problem known as the solution of triangles, we shall discuss very briefly the use of common logarithms and certain mathematical tables. These tables may be found on the last sixty-five pages of this book. NOTE. The discussion of logarithms belongs properly to Algebra, to which the student is referred for a more general treatment. The student will observe that computation by means of logarithms is a mere combination of exponents. 118. In Algebra it is explained that when different powers or roots of the same number are concerned, (i.) multiplication is effected by adding the indices; (iii.) involution and evolution are respectively effected by the multiplication and division of the indices. 119. DEFINITION. The logarithm of a number, n, to a certain base, b, is the index with which it is necessary to affect b to produce n. E.g. suppose b' =n; then logarithm n to base b is l for b, raised to 7th power, produces n. This is expressed in the following notation: log。 n = 1. and is read "logarithm of n to the base b equals l." Or, if no ambiguity arises, simply "log n = 1." EXAMPLES. log2 83; for 288, or 2 must be raised to power 3 to produce 8. log10 100 = 2. 120. The use of logarithms is based upon the following propositions: I. The logarithm of the product of two numbers is equal to the logarithm of one of the numbers plus the logarithm of the other. II. The logarithm of the quotient of two numbers is the logarithm of the dividend minus the logarithm of the divisor. III. The logarithm of a number raised to a power k is x times the logarithm of the number. For, Now, (m*) = (b2)*. log, b** = kx = k log, m. EXAMPLES. Given log10 2.30103, log10 3.47712, log10 7 = .84509; find the values of the following: (ii.) log103 = log10 7 log10 3.84509.47712 = .36797. (iii.) log10 35 = 5 times log10 3 = 5 x .30103 1.50515. [by I.] [by II.] [by III.] [by III.] [by I. and II.] = of (log 3+ log 4 - log 7) = } of {.47712+ twice .30103.84509} (v.) log105 log10 log10 10 log102 1.30103.69897. = Va, Va2, al 2. Find the logarithms to the base 2 of 8, 64, 1, .125, .015625, 64. |