4. Two sides, the one 30 and the other 35, and the included angle 20°, of a triangle, are given, to find the other two angles and the third side. Let BAC be the triangle, in which BC = 35, BA = 30, and the angle B = 20°. From A, the extremity of the into the two right-angled triangles BAD and CAD. Then, from the triangle BAD, we have 1st, or, 2d, or, sin. D sin. B :: BA : AD; D 1 sin. 20° :: 30* : AD = 30 sin. 20°. = In the table of natural sines, we find sin. 20° cos. 20° = .93969; hence, AD = 30 ×.34202 = 10.26060, and .34202, and the If, now, we add angles B and C, and take the sum from 180°, the remainder will be the angle BAC. 5. Two sides, the one 18 and the other 24, and the angle opposite the side 24 equal to 76°, are given, to find the remaining side and the other two angles. Let x denote the angle opposite the side 18. Then, or, 24 18 sin. 76° : sin. x, (Prop. 4, Trig.). 4: 3 :: sin. 76° : sin. x. sin. x = † sin. 76° = 1 × .97030 = .72772; whence the angle opposite the side 18 is 46° 41′ 45′′. Adding this to the given angle, and taking the sum from 180°, we get 57° 18′ 15′′ for the third angle. To find the remaining side, denoted by y, we have 6. The three sides of a triangle are 18, 24, and 20.815; required the angles. This problem may be solved by Prop. 6, or by Prop. 8, Trigo nometry. First. By Prop. 6. In the triangle ABC, make CB = 24, AC 20.815, and AB Then, = 18. It will be observed that Examples 5 and 6 refer to the same triangle, and that in Example 5 the angle B was 57° 18′ 15′′. This slight discrepancy in the results should be expected, on account of the small number of decimal places used in the computations. Hence, cos. A = .78800, and A (Table II, page 59) = 38° very nearly; the angle A is therefore equal to 76°, which agrees with Example 5. 7. Given, the three sides, 1425, 1338, and 493, of a triangle; required, the angle opposite the greater side, using the formula for the sine of one half an angle. Make a = 1425, b = 1338, and c = 493; then the LA is opposite the side a, and the formula is in which s denotes the half sum of the three sides. Then we have s = 1628, s- b= 290, 8-c= 1135, (s—b) = .498988 sin. 44° 56' 28.5".70638. Therefore, A 44° 56′ 28.5", and A= 89° 52' 57"; - but little less than a right angle. In these seven examples we have shown that it is possible to solve any plane triangle, in which three parts, one at least being a side, are given, without the aid of logarithms. But, when great accuracy is required, and the number of decimal places employed is large, the necessary multiplications and divisions, the raising to powers, and the extraction of roots, become very tedious. All of these operations may be performed without impairing the correctness of results, and with a great saving of labor, by means of logarithms; but, before using them, the student should be made acquainted with their nature and properties. LOGARITHMS. Logarithms are the exponents of the powers to which a fixed number, called the base, must be raised, to produce other numbers. The exponent of a number is also a number expressing how many times the first number is taken as a factor. Thus, let a denote any number; then a3 indicates that a has been used three times as a factor, a* that it has been used four times as a factor, and a" that it has been thus used n times. Now, instead of calling these numbers 3, 4, —— n, exponents, we call them the logarithms of the powers a3, a1, an. To multiply a' by a3, we have simply to write a, giving it an exponent equal to 2 + 5; thus, a2 x a5 = a". α. Hence, the sum of the logarithms of any number of factors is equal to the logarithm of the product. 12 = To divide a" by ao, we have only to write a, giving it an exponent equal to 12-9; thus, a12÷a a3; and, generally, the quotient arising from the division of am by a", is equal to am-n. Hence, the logarithm of a quotient is the logarithm of the dividend diminished by the logarithm of the divisor. If it is required to raise a number denoted by a3, to the fifth power, we write a, giving it an exponent equal to 3×5; thus, (a) a", and, generally, (a")" anm。 5 = Hence, the logarithm of the power of a number is equal to the logarithm of the number multiplied by the exponent of the power. number a3, we write a, ; thus, Vaa3, and, To extract the 5th root of the giving it an exponent equal to generally, to extract any root of a number, we divide the exponent of the number by the index of the root, and the quotient will be the exponent of the required root. Hence, the logarithm of a root of a number is equal to the quotient obtained by dividing the logarithm of the number by the index of the root. Now, understanding that by means of a table of logarithms we may find the numbers answering to given logarithms, with as much facility as we can find the logarithms of given numbers, we see from what precedes that multiplications, divisions, raising to powers, and the extraction of roots, may be performed by logarithms; and the utility of logarithms, in trigonometrical computations, mainly consists in the simplification and abridgment of these operations by their use. |