The work can be more compactly arranged, as follows: - 10 log tan B = 10.12494 - 10 ... B= 53° 7' 48" c + b = 126 с b = 14 log (c+b)= 2.10037 log (cb) 1.14613 ... log a2 = 3.24650 NOTE 1. The latter form is preferable when all the parts of a triangle are required. - log sin A in the second NOTE 2. If there is difficulty in calculating log a form, write log sin A on the edge of a piece of paper and place it immediately beneath log a. NOTE 3. The formula tan B b = α can be used instead of (2). A check then is A + B = 90°. Instead of (3), one of the following formulas can be used, viz. There is often a choice of formulas that can be used in a solution. NOTE 4. In every example it is advisable to make a complete skeleton scheme of the solution, before using the tables and proceeding with the actual computation. In the last exercise, for instance, such a skeleton scheme can be seen on erasing all the numerical quantities in the equations that follow the logarithmic formulas. NOTE 5. Time will be saved if all the logarithms that can be found at one place in the tables, be written at one time. Thus, for example, in the preceding exercise find log sin A immediately after A has been found. NOTE 6. The logarithmic formulas can be written on a glance at the formulas such as (1), (2), (3). The writing of the logarithmic formulas may be dispensed with when the student has become familiar with calculation by logarithms. A glance at the original formulas will show how the logarithms are to be combined in the computation. 2. In a triangle ABC right angled at C, c = 60 ft., b = 50 ft.; find side a and the acute angles. Logarithmic formulas: log cos A = log blog c. a = b tan A. NOTE. There is a slight difference between the results obtained by the two methods. This is due to the fact that the calculations have been made with a four-place table in one case, and with a five-place table in the other. A four-place table will give an angle correctly to within one minute; a fiveplace table will give it correctly to within six seconds, and sometimes, to within a second. Ex. Make the computation I. with a five-place table. 3. In a triangle right angled at C, the hypotenuse is 250 ft., and angle A is 67° 30'. Solve the triangle. I. Computation without logarithms. B = 90° - A = 90° — 67° 30′ = 22° 30'. a = c sin A = 250 x sin 67° 30' = 250 × .9239 = 230.98. bc cos A = 250 x cos 67° 30′ 250 x .3827 = 95.68. C-250 Ft. B a Logarithmic formulas: log a = log clog sin A. 4. In a triangle ABC right angled at C, b = 300 ft. and A= 37° 20' Solve the triangle. B N.B. Check all results in the following examples. The given elements belong to a triangle ABC which is right angled at C. Solve Exs. 17-24 by two methods, viz.: (1) with logarithms; 16. c 50.13, a = 24.62. N.B. Questions and exercises suitable for practice and review on the subject-matter of Chapter III. will be found at page 183. CHAPTER IV. APPLICATIONS INVOLVING THE SOLUTION OF Some practical applications of trigonometry will now be given. It is not necessary that all the problems be solved, or all the articles be considered, before Chapter V. is taken up. 28. Projection of a straight line upon another straight line. If from a point P a perpendicular PO be drawn to the straight line ST, then O is called the projection of the point P upon the line ST. If perpendiculars be drawn from two points A, B, to a line LR, and intersect LR in M, N, respectively, then MN is called the projection of AB upon LR. Let be the length of AB, and let a be the angle at which the two lines AB, LR are inclined to each other. Through A draw AD parallel to LR. Then Projection = MN=AD = AB cos DAB = l cos a. That is, the projection of one straight line upon another straight line is equal to the product of the length of the first line and the cosine of the angle of inclination of the two lines. NOTE. The projection discussed here, is orthogonal (i.e. perpendicular) projection. If a pair of parallel lines AM1, BN1, not perpendicular to LR, be drawn through A, B, then M1N1 is an oblique projection of AB on LR. |