Mathematical Drawing (Paperback)

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1906 Excerpt: ...makes the right-hand side equal to-695. 0 = 155, which makes the right-hand side equal to-701. 5. Show how to solve the equation sin" 0 (a sin 0 + b cos 0) = k, where a, b, k are given quantities, and n is an integer. Another way of solving the equation sin2 0 (a sin 0 + b cos 0) = k, suggested by Mr Alfred Lodge, is this: draw a line OB = b, and at right angles to it take OA = a; describe a circle through B, 0, and A; if any line, OR, drawn through 0 meets the circle in R and makes an angle 0 with OB, we have OR = a sin 0 + b cos 0. If then from R we drop a perpendicular Rp on OA, and from p a perpendicular pP on OR, it is clear that OP = sin2 0 (a sin 0 + b cos 0). Tracing the locus of P, we obtain an oval like OPDQ in Fig. 47; but its axis of symmetry is inclined to OA. The points of intersection of this curve with the circle of radius k described about O as centre give the values of 0. This curve is, however, not so readily traced as that in Fig. 47. 6. Show how to solve the equation sin4 0 = k sin (0--a), where k and a are given constants. Fig. 48. Take OA on the axis of y (Fig. 48) equal to a and describe a circle with A as centre; take OB = b, and describe a circle with B as centre. Let ON be any value of a; then drawing NA to meet the first circle in C, we have NG = Va? + a?+a. Similarly, drawing NB to meet the second circle in D, we have ND = Va? NC the y of the point on the curve sought is--62 + b. Hence To construct this take a unit length, NU, along NB and from U draw a parallel, Up, to DC; then NG Np y=m='.-. y=NP' on the ordinate NP drawn at N measure NP = Np, and we have the point P on the required curve. Repeating this operation for all positions of N along Ox, we have the whole curve required. There is a similar portion of the curve (n...