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211. To express r and r, in terms of R.
a sin } B sin 10

sin } A sin } B sin {C
cosA sin A

4R sin } A sin } B sin }C,
a cos } B cosic 2a
ri=

sin } A cos B cos C
cos } A sin A

= 4R sin } A cos ? B cos C.

212. The distances of the orthocentre from the sides and vertices of the triangle.

See fig., Art. 191.
AO= AM sec OAM = AB cos A cosec C = cos A = 2R cos A.

sin C

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If 4 is obtuse, 40 is arithmetically – 2R cos 4, $ being on

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the side of A remote from BC.
OL = BL tan LBO = AB cos B cot C = 2R cos B cos C.

If B or C is obtuse, OL is arithmetically – 2R cos B cos C.

COR. The distances of O from the sides are inversely as those of s.

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a

213. The distances of the centre of gravity from the sides and vertices of the triangle.

See fig., Art. 192.
Since GD=LAD, .. the distance of G from BC

s = }AL = }b sin C= . = R sin B sin C ......(1). Also, producing AD to A', so that A'D= AD, ABA'C is a parallelogram, and the angle ABA' is the supplement of A; .. 4AD2 = A A"? = 12 + co + 2bc cos A ........

..(2), 262 + 2co - a?.....

..(3). :: GA= { AD= 1/(262 + 2c - a) ............ (4). COR. The distances of G from the sides are inversely as the

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214. Expressions for the radius (p) of the Cosine-Circle ; for the distances of its centre from the sides, and for its intercepts on the sides.

See fig., Art. 193.
Draw KX, KY, KZ perpendiculars on the sides.

Then the antiparallels through K, which are diameters of the cosine-circle, cut BC at an inclination A, .. the intercepts on the sides are, respectively,

2p cos A, 2p cos B, 2p cos C,
and are :: proportional to the cosines of the angles.

[Hence the name of the circle.]
And the perpendiculars on the sides are, respectively,

p sin A, p sin B, p sin C,
.. KX : KY : KZ=sin A : sin B : sin C

... (1). If now x, y, z be the perpendiculars on the sides from any point within the triangle, it follows, as in Art. 208, that

xa + yb + zc = 2 . (area of triangle) = 28.

KX KY KZ 28 ..

(2). 6

a + b + c

=

= a :

6

: C.......

a

с

This gives the distances of K from the sides.

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a? + b2 + ca 2bc cos A + 2ca cos B + 2ab cos C
Now
4S

48
- cot A + cot B + cot C.
za 16 c R

cot A + cot B+ cot C... ... ... (5). KX KY KZ

р

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с

=

sin w

с

sin w

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Let CA, AN, BQ subtend equal angles w at A, B, C respectively. Then clearly the angles ANB, BNC, CNA must be the supplements of A, B, C. AS2

ΑΩ
Thus

and
sin A

sin (C w) sin c' sin (C – w)

sin c sino C sin C sin (A + B) 6 sin A sin A sin B sin A sin B .:. cot w=

cot A +cot B + cot C ........ (1). If again BO', CO, AN' subtend equal angles wat A, B, C respectively, the same equation (1) will give w'. .. w = w.......

..(2). If 2X, NY, OZ be perpendiculars on the sides, since circles go round A YZA, BZX, CXY, it follows that XYZ is similar to CAB, and has 12 and w for Brocard point and Brocard angle respectively. And since 2X :NC = sin w, .. sin w is the ratio of similitude of XYZ to CAB.

b sino w Now

ΩΧ = ΩC sin ω =

sin C

c sino w If N'X' be perpendicular to BC, N'X':

sin B .:. ΩΧ. Ω'Χ' = ΩΥ. ΩΥ" = ΩΖ. Ω' Ζ = 4R2 sin4 ω. .: (IN'N2)= (radius of circle XX' YY' ZZ')2 – 4R- sin* w =Rsin’ w (1 – 4 sinow)=Rsin’ w (cos” w – 3 sino w).

. .. N'N=R sin 2w /(1-3 tanow) ..(3).

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216. Since BO' and CA are inclined to BC at equal angles w, therefore if their point of intersection D' is joined to D (the middle point of BC), we have

16 C DD

EE FF"

cot w.

р

tan w.

=

1

Comparing this with Art. 215 (1), and Art. 214 (5), we have

KX= DD', KY = EE', KZ = FF", and

KX KY KZ

La 16 C R
Thus
р

R tan w.
COR. (Lemoine-Radius)*=(1 circumradius)2 + (1 cosine-radius).
.. Lemoine-Radius IR sec w.

:. the sides of A,B,C, and of A,B,C, (where the Lemoine circle cuts ABC) are to those of ABC in the ratio į sec w to 1.

217. The distance between the Circumcentre and Incentre.

By Prop. IX. in the last chapter, the rectangle of the segments of any chord of the circumcircle drawn through the incentre= 2 Rr.

Draw through I the diameter DSIE of the circumcircle. Then

DI.IE + SI = DS, ..

SI2= DSP DI.IE = Ro 2 Rr.

=

=

Similarly

SIP = R + 2 Rri:
218. The distance between the Circumcentre and Orthocentre.
SOP = SA+ OA’ – 2SA.QA cos SAO,

= Ro + 4Ro coso A – 4R* cos A cos (C B)
=R2 4R2 cos A {cos (C + B) + cos (C B)}

=R2 (1 - 8 cos A cos B cos C).
Cor. Since the Nine-Points Centre T bisects SO,

.. ST2 = pl (1 -8 cos A cos B cos C), where t= nine-points radius = {R.

219. The distance between the nine-points centre and incentre.

Drop perpendiculars from T on BC, and from I on this perpendicular. Let AI cut the circumcircle in U.

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Then, by Prop. IX. of last Chapter, UI. IA = 2 Rr = 4tr; and by Prop. XI, Cor. II, - DPL= 2. ZUIX = 20 say. :: T'I'=(cos 20 - r) + (7 sin 20 - DX)"

= p + go– 27r cos 20 DX (27 sin 20 – DX)
=p? + 2tr cos 20 - DX, XL
= p+ go? - 277 cos 20 UI. IA sino 6

= p? +92 – 271;
::. TI =1-1.

.

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