(19) If on one side BC of a triangle ABC a triangle A'BC' is described without it such that the angles BA'C, CBA', A'CB are equal to aẞy respectively, then

AA2 sin a sin ẞ sin y (a2 cot a+b2 cot B+c2 coty+4A).

=

(20) If a triangle be cut out in paper and doubled over so that the crease passes through the centre of the circumscribing circle and one of the angles A, the area of the doubled portion is 62 sin2 C cos Ccosec (2C – B) sec (C-A), C being greater than B.

(21) If 10=IN, prove that one angle of the triangle ABC is 60o.

(22) If two of the angular points and the radius of the circumscribing circle of a triangle are given, the loci of the centre of the nine-point circle and of the ortho-centre are circles.

(23) Prove that a triangle can be constructed whose sides are a cos A, b cos B, c cos C and that its area is

2A.cos A. cos B. cos C.

(24) If R1, R2, R3 are the radii of the circumscribed circles of BIC, CIA, AIB, prove that R12. R2. R ̧2=R3. AI.BI.CI.

(25) If the two straight lines which bisect the angles A and C of a triangle ABC, meet the circumference of the circumscribing circle in R and S, then RS is divided by CB, BA into three parts which are in the ratio

sin24 2sin4. sin B. sin C: sin3C.

(26) If a point be taken in an equilateral triangle such that its distances from the angular points are proportional to the sides of a triangle ABC, the angles between these distances will be π+A, }π+В, π+ C.

CHAPTER XI.

ON THE USE OF SUBSIDIARY ANGLES TO FACILITATE
NUMERICAL CALCULATION.

124. In the Elementary Trigonometry, Art. 185, we have shewn how the Tables may be made use of in the solution of Simple Trigonometrical Equations.

It is usual to shew how the Tables may be made use of to facilitate the calculation of the roots of quadratic and cubic equations.

The solution of such equations is however rarely required in practical work, so that the method is not of much practical importance.

125. To obtain the numerical values of the roots of a quadratic equation.

I. Let the equation be x 2px +q=0, where p and 7 are positive.

Solving, we obtain

x=p±√(p2 - 9)

−p{1 ± √(1–2)}.

First, let զ be less than p2; then we can find from the

Tables an angle a such that sin3a =

Whence we obtain

x = p {1 + cos a}.

Secondly, let q be greater than p2; then we can find

II. Let the equation be x2+2px+q=0. Then the roots of this equation are equal to those in Case I. with the signs changed.

III. Let the equation be x-2px-q=0.

We can find from the Tables an angle a such that

and then

IV.

The roots of x2 + 2px − q = 0 are equal to those of Case III. with the sign changed.

EXAMPLE. Calculate the value of the roots of the equation

126. The student will observe that this method is the

same as that of adapting the expression p {1 ± √(1 + 2})

to Logarithmic calculation by means of the Trigonometrical Tables.

127. To obtain the numerical value of the roots of a

cubic equation.

Let the equation be

x3 + 3px3 + 3qx + r = 0.

Write y-p for x, and the equation becomes

y3 − 3 (p2 − q) y + (2p3 − 3pq + r) = 0.

Therefore any cubic equation can be transformed into another in which the second term is wanting.

But, if a be any angle, we have (E. 167)

then cosa is one of the roots of the equation.

Also since cos (2nπ ± 3a) = cos 3a, the other two values of y are cos (3π + a) and cos (3π − a).

But x= ny.

Therefore the required roots are

2(a) cosa, 2(a) cos (+α), 2(a)* cos(π — a).

cos(π

3a can be found provided ba is less than unity, i.e. provided 6 is less than 4a".