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236°. Ex. 1. Let four rays be disposed in the order OA, --POB, OC, OP, and let OP be perpendicular

B

A

B

to OA.

Denote LAOC by A, and LAOB by B. Then 234° becomes

or,

sin B cos A+ sin (A-B)sin-sin Acos B=o, sin (A-B)=sin Acos B-cos Asin B. Similarly, by writing the rays in the same order and making LBOP a, and denoting LAOB by A and BOC by B, we obtain sin (A+B)=sin A cos B+cos A sin B.

Also, by writing the rays in the order OA, OP, OB, OC, and denoting LAOP by A and BOC by B, we obtain

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cos (A+B) = cos A cos B - sin A sin B ;

which are the addition theorems for the sine and cosine.

Ex. 2. ABC is a triangle and P is any point. Let PX, PY, PZ be perpendiculars upon BC, ACA, AB, and be denoted by Pa, Pb, Pe respectively.

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Draw AQ | to BC to meet PX in Q. Then (235°)

PQ sin A+ PY sin B + PZ sin C=0.

But if AD is to BC, AD=b sin C=QX.

.'.

or

Similarly,

(PX -b sin C) sin A+ PY sin B +PZ sin C =0,

(Pasin A)=b sin A sin C.

(Pasin A)=csin B sin A

= a sin C sin B,

(Pasin A)= {abc sin2A sin2B sin C}.

Hence the function of the perpendicular

Pa sin A+ Po sin B+ P, sin C

is constant for all positions of P. This constancy is an important element in the theory of trilinear co-ordinates.

237°. A, B, C being a range of three, and P any point not

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(AB+BC+CA)(PQ2+CQ2)+BC.CA(BC - AC)

=BC. CA(BC+CA) = BC. CA. BA

=-AB. BC. CA.

EXERCISES.

1. A number of stretched threads have their lower ends fixed to points lying in line on a table, and their other ends brought together at a point above the table. What is the character of the system of shadows on the table when (a) a point of light is placed at the same height above the table as the point of concurrence of the threads? (b) when placed at a greater or less height? 2. If a line rotates uniformly about a point while the point moves uniformly along the line, the point traces and the line envelopes a circle.

3. If a radius vector rotates uniformly and at the same time lengthens uniformly, obtain an idea of the curve traced by the distal end-point.

4. Divide an angle into two parts whose sines shall be in a given ratio. (Use 231°, Cor. 3.)

5. From a given angle cut off a part whose sine shall be to that of the whole angle in a given ratio.

6. Divide a given angle into two parts such that the product of their sines may be a given quantity. Under what condition is the solution impossible?

7. Write the following in their simplest form :

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sin (2+0),

sin {-(0-)},

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8. Make a table of the variation of the tangent of an angle in magnitude and sign.

9. OM and ON are two lines making the MON=w, and PM and PN are perpendiculars upon OM and ON respectively. Then OP sin w=MN.

IO. A transversal makes angles A', B', C' with the sides BC, CA, AB of a triangle. Then

sin A sin A'+ sin B sin B' + sin C sin C'=0.

II. OA, OB, OC, OP being four rays of any length whatever, ДАОВ. ДСОР+ДВОС. ДАОР+ДСОА. ДВОР=0. 12. If r be the radius of the incircle of a triangle, and î1 be that of the excircle to side a, and if p1 be the altitude to the side a, etc.,

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13. The base AC of a triangle is trisected at M and N, then

BN2=3(3BC2+6BA2-2AC2).

SECTION II.

CENTRE OF MEAN POSITION.

238°. A, B, C, D are any points in line, and perpendiculars

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The point O is called the centre of mean position, or simply the mean centre, of the system of points A, B, C, D.

Again, if we take multiples of the perpendiculars, as a. AA', b. BB', etc., there is some point O, on the axis of the points, for which

a. AA'+b. BB'+c. CC'+d.DI)'=(a+b+c+d)ON.

Here again ON lies between AA' and DD'.

O is then called the mean centre of the system of points for the system of multiples.

Def. For a range of points with a system of multiples we define the mean centre by the equation

Σ(a. AO)=0,

where Σ(a. AO) is a contraction for

a. AO+b. BO+c. CO+...,

and the signs and magnitudes of the segments are both considered.

The notion of the mean centre or centre of mean position has been introduced into Geometry from Statics, since a system of material points having their weights denoted by a,

b, c, and placed at A, B, C, would "balance" about

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the mean centre O, if free to rotate about O under the action of gravity.

The mean centre has therefore a close relation to the "centre of gravity" or "mass centre" of Statics.

239°. Theorem.—If P is an independent point in the line of any range, and O is the mean centre,

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a. AP=a. AO+a. OP, b. BP=b. BO+b. OP, etc.

Σ(a. AP)=2(a. AO) + 2(a). OP.

But, if O is the mean centre,

(a. AO)=o, by definition,

Σ(a. AP)=2(a). OP.

Ex. The mean centre of the basal vertices of a triangle

when the multiples are proportional to the opposite sides is the foot of the bisector of the vertical angle.

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AM, BM, CM,

240°. Let A, B, C, be a system of points situated anywhere in the plane, and let AL, BL, CL, ..., denote perpendiculars from A, B, C, and M.

...

upon two lines L

Then we define the mean centre of the system of points for a system of multiples as the point of intersection of L and M

also

N

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AL. sin MON+AM. sin NOL+AN. sin LOM=0,
BL. sin MON + BM, sin NOL+ BN. sin LOM=0,

and multiplying the first by a, the second by b, etc., and adding,
Z(a.AL) sin MON+2(a.AM) sin NOL+2(a.AN) sin LOM=0.
But
Σ(a. AL)=2(a. AM)=o, by definition,
Σ(a.AN)=0.

.'.

241°. Theorem.-If O be the mean centre of a system of points for a system of multiples, and L any line whatever, (a. AL)=2(a). OL.

Proof. Let M be || to L and pass through O. Then
AL AM+ML,

adding,

BL=BM+ML,

.. a. AL=a. AM+a. ML,
.. b. BL=b. BM+6. ML,

(a.AL)=2(a. AM)+2(a). ML.

But, since M passes through O,

Σ(a. AM)=0, and ML=OL,

Σ(a. AL)=2(a). OL.

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