« ForrigeFortsett »
At any point p of the line we have, if pAq = 0, pBq=4,
.. (a) may be written k (Aq + Dq) = 1 (Dq − Bq),
i.e. k. AD-l. BD,
which by construction it does.]
To find points on the locus represented by (1). With the points A and B as centres describe two circles S and T of radii
AB and -. AB respectively. Draw any common ordinate NLQ,
meeting S in L and Tin N; then the lines AL and BN intersect in a point, P, on the required locus; for
AQ+QB = AB,
AL cos 0 + BN cos &= AB, if BAL is 0 and ABN is 4,
which is the given equation.
If the line DC meet the curve in R and R, the angles RAB, RAB are the required values of 0, and the angles RBA, R,BA those of p.
There is a precisely similar loop on the other side of AB.
In the particular case in which ab the locus is the Magnetic Curve. (Prob. 168.)
PROBLEM 182. To find 0 and from the equations
SLCL G sin
Take two points A and B such that AB = a+b; make 40 = a, 07-5 and draw OD perpendicular to AB; with A as centre and
radius describe a circle, and draw any radius AC meeting Pin 4; inflect BJ LC (J being on OD); then P, the point of mersection of AC and BJ is a point on the locus represented by (1), the angles @ and 4 being ALO and BJO respectively.
There is a precisely similar loop on the other side of AB.
Again the equation cos 0= k cos & gives sin PAB = k. sin PBA og PBk. PA, i. e. P is the vertex of a triangle on a given base 1% and with sides in a given ratio (Problem 17), i. e. the locus reprosouted by the second equation is a circle whose diameter V
the line joining the points which divide AB internally and externally in the ratio 1: k; ie.
AQ: QB :: 1: k :: AQ1 : Q1B.
The values of @ and which satisfy both equations are those the points of intersection of this circle and the
[Trace the loci y = sin x (harmonic curve) and y
line through the origin): the values of x corresponding to their points of intersection are solutions.]
2. Solve the equation sin x = ax + b.
[The intersections of the harmonic curve y = sin x and of the straight line y = ax + b where a and b are constants.]
3. Solve the equation 2o = 5 sin 0.
[The intersections of the equiangular spiral r = 2o and of the
where a, b, c and n are given constants.
[The 2nd equation represents a locus identical with (1) of Problem 181, attention being paid to the usual conventions as to sign.
The 1st equation represents a right line perpendicular to AB (fig. 185), the base of this locus, and meeting it in D so that
where l, m, n, a and a are constants.
[Draw two lines AB, AC including an angle a, and mal AB= = a and AC = m. With centres B and C and radii = n and
respectively describe arcs intersecting in D on the same side of BC as AB. Let CD meet AB in E. BED is the required value of and DBE that of p.]
6. Determine 0 from the equation
a cos λ. cos (λ + 20) = c. cos (a + 0)
where a, c, λ and a are given constants.
[The locus represented by the right-hand side of the above equation is a circle of radius c, the origin (0) being the extremity of a diameter, and the initial line making an angle a therewith.
To draw the locus represented by the left-hand side :—draw a line through the origin O making an angle λ with the initial line, and on it measure a length OL = a. Draw LN perpendicular to the initial line meeting it in N so that ON = a cos λ. With centre and radius ON describe a circle. From any point Q on this circle draw QM perpendicular to OL meeting it in M. Draw OP bisecting the angle NOQ and make OP = OM. P will be a point on the second locus, and any additional number of points may be similarly determined. Let the two loci intersect in X, and the angle between OX and the initial line is the required angle 0.]
This equation defines the position of equilibrium of a uniform rectangular board resting in a vertical plane against two equally rough pegs in a horizontal line.
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