Let o be the circular measure of the angle between the initial line and the radius drawn from the origin to a point of intersection of these curves, then

w will be an approximate solution of the original equation; and w + 8 a more exact one.

PROBLEM 173.

To solve the quadratic equation x2 - 2Аx + B2 = 0 (Fig. 178).

Draw two lines Oa, Ob at right angles to each other, and on one of them make Ob = B.

With b as centre and A as radius describe an arc cutting Oa in a, so that Oa = √Ã2 – B2; and with centre a and radius ab

describe arcs cutting Oa in d and d1. Od and Od1 are lines representing the two values of x in the above equation. If the numerical values of the roots are required they must be measured of course on the same scale which has been used for laying off the lengths A and B.

If A is numerically less than B the roots become imaginary, and the graphic method is not applicable.

As a numerical example we may take the equation to determine the length AK in problem 169.

Here AK is one of the roots of kr2 − 2 + k|cr + c2 = 0,

7c

and B

√2

=

6

3 Suppose k= then A= 2

where c is a given length.

Make Oe in fig. 178 this given length c.

=

On Oa, Ob, take Of = Of1, then ff, represents √2 the length Of being the unit; make OF on Ob=ff1.

With centre f, and radius = 2. Of describe an arc cutting Oa in G, then OG represents √3, the length Of being the unit. Through e draw a parallel to FG meeting Ob in b.

7c
6

With centre b and radius: describe an arc cutting Oa in

a, and with centre a and radius ab describe arcs cutting Oa in d and d,. Od, Od, represent the values of r in the equation,

and the particular value of AK in Problem 169 is Od =

PROBLEM 174. To solve the quadratic equation x2+2 Ax + B2 = 0.

Ое

3

The solution is exactly the same as that of the last problem, but both roots are negative.

PROBLEM 175.

To solve the quadratic equation

x2 – 2 Ax – B2 = 0 (Fig. 179).

Draw 2 lines Oa, Ob at right angles to each other and on them make Oa = A, Ob = B:

With centre a and radius ab describe an arc cutting Oa in d

and d. 6. Od represent the re of the equation but the

The solution is idential with that of the last problem, but the greater root must be taken with negative sim

As a numerical example take the equation to determine the length AL in Problem 163

suppose

then

Make Oe = c; bisect Oe in a and make Oa, = Oa so that

Make OP on Obaa, and through e draw eb parallel to aF:

=

Ob = B.

With centre a and radius ab describe arcs cutting Oɑ in d and d. Od is the positive root of the equation, and is the length AL in curve No. 3 of Problem 169.

АО

PROBLEM 177. To solve graphically the equation

a cos 0 + b sin 0=c (Fig. 180).

Draw 2 lines at right angles to each other as AO, AB. Make

=a and AB=b on any convenient scale. Describe a circle

round OAB (its centre will of course be at the middle point of OB) and with centre O and radius OD =c describe an arc cutting it in D, the angle AOD is the required angle 0.

[In the figure a=2·5, b=1·3, c=2·65, the unit being the length L and = 47.5°.]

The second point D, in which the arc described with centre O and radius c would cut the circle gives when c is greater than a a second solution, the angle AOD, being the value of in this case.

1

When c is less than a so that D, falls between 0 and A the second solution corresponds to a cos 0-b sin 0 = c.

PROBLEM 178. A and B are two fixed points and P a variable point, the position of which is defined by the angles PAB (= 0) and PBA (=); draw the locus represented by the equation

sin 0 + sin = a,

where a is constant. [a may be either positive or negative but its numerical value cannot be greater than 2.] (Fig. 181.)

On AB make BC =a. AB, and describe a semi-circle on AB. Draw a line Ap meeting the semi-circle in p and on BA make

Bb= Bp. With centre A and radius = bC describe an arc cutting the semi-circle in 7, and draw Bq cutting Ap in P. P will be a point on the required locus. Similarly any number of points can be determined.

1

If a is greater than unity, i.e. if BC is greater than AB, the locus will meet AT, BT, drawn perpendicular to AB, in points T and T, determined by inflecting AR, BR in the semi-circle each equal to AC and drawing BR, AR, meeting AT, BT, in T and T, respectively. BT, AT, are tangents to the required locus at T and T. Lines drawn from A to points between R, and B do not intersect the locus in real points.

If a is less than unity, i.e. if BC, is less than AB, the curve passes through A and B and the tangents at those points can be drawn by inflecting BV, AV, in the semi-circle each equal to BC. AV, BV, are tangents to the required curve. In the figure the value of a for the upper curve is and for the lower. There