angles ADB and CDB are equal and of an observed magnitude 0; prove that the error in the calculated length of DB consequent on a small error d in the observed magnitude of 6, is 2ab (a+b)2 sin 0 (a2 + b2 – 2ab cos 20) approximately, where AB-a and BC=b. δ 7. In measuring the three sides of a triangle small errors x and y are made in two of them, a and b; prove that the error in the angle C will be --1/ cot A b X a cot B, and find the errors in the other angles. 8. In a triangle ABC we have given that approximately a=36 feet, 3 b=50 feet, and C-tan-1 4' find what error in the given value of a will cause an error in the calculated value of c equal to that caused by an error of 5" in the measurement of C. 9. A triangle is solved from the parts C=15°, a=√6, and b=2; prove that an error of 10" in the value of C would cause an error of about 13.66" in the calculated value of B. 10. Two sides b and c and the included angle A of a given triangle are supposed to be known; if there be a small error in the value of the angle A, prove that (1) the consequent error in the calculated value of B is. - 0 sin B cos C cosec A radians, (2) the consequent error in the calculated value of a is c sin B. 0, and (3) the consequent error in the calculated area of the triangle is O cot A times that area. 11. There are errors in the sides a, b, and c of a triangle equal to x, y, and z respectively; prove that the consequent error in the calculated value of the circum-radius is 12. The area of a triangle is found by measuring the lengths of the sides and the limit of error possible, either in excess or defect, in measuring any length is n times that length, where n is small. Prove that in the case of the triangle whose sides are measured as 110, 81, and 59 yards, the limit to the error in the deduced area of the triangle is about 3:1433n times that area. 13. The three sides of a triangle are measured and found to be nearly equal. If the measurements can be wrong one per cent. in excess or defect, prove that the greatest error that can arise in calculating one of the angles is 80' nearly. 14. It is observed that the elevation of the summit of a mountain at each corner of a plane horizontal equilateral triangle is a; prove that the height of the mountain is where a is the side of the triangle. If there be a small error n” in the elevation at C, shew that the true height is CHAPTER XXXII. MISCELLANEOUS PROPOSITIONS. Solution of a Cubic Equation. 396. The standard form of a cubic equation is Put y=x-a, and this equation becomes x3 − 3 (α2 − b) x + (2a3 − 3ab+c) = 0, i.e. it becomes of the form Hence any cubic equation can be reduced to the form (1), which has no term containing a2. 397. To solve the equation x3-3px+q=0. Now, by Art. 107, we always have cos 30 = 4 cos3 0 – 3 cos 0, Now (2) and (3) are the same equation if The equation (4) can always be solved (by means of the tables if necessary) if [The student who is acquainted with the Theory of Equations will notice that is the case which cannot be solved by Cardan's Method. It is the case when the roots of the original cubic are all real. ] If be the smallest angle satisfying equation (4), then the values 2π 477 Ө+ and 0+ 3 also satisfy it, so that the roots of the equation x3-3px+q=0 1 COS Ө+ 1 cos,cos (+), and cos (+) 3 i.e. 2 √p cos 0, 2 √p cos (0+ 3 398. Ex. Solve the equation x3+6x2+9x+3=0. Put xy-2, and the equation becomes y3 − 3y+1=0. .. y=2 cos 40°, or 2 cos 160°, or 2 cos 280°. .. x=y-2=-2+2 cos 40°, or -2-2 cos 20°, or -2+2 cos 80°. On referring to the tables we then have the values of x. Maximum and Minimum Values. 399. In Art. 133 we have given one example of the maximum value of a trigonometrical expression. We add another example. If x and y be two positive angles whose sum is a constant angle a(π), find when sin x sin y is a maximum, and extend the theorem to more than two angles. |