O is the center of a circle. AB, CD and EF are three parallel chords of this circle having respective lengths 2, 3 and 4.
It is known that ∠AOB = m, ∠COD = n and, ∠EOF = m+n, where m+n < 180o
Determine cos m
Let the radius of the circle be R.
Applying the law of cosines to triangle AOB
we get
cos(m) = 1 - 2^2/(2*R^2) (1)
So all we need is R^2.
Let G be the point on the smaller arc EF
such that |GF| = |AB| = 2.
|GF| = 2 ==> /GOF = /AOB = m
==> /EOG = /EOF - /GOF
==> /EOG = (m+n) - m = n
==> |EG| = 3
Therefore, R is the circumradius of triangle
EFG whose side lengths are 2, 3, and 4.
The circumradius squared of a triangle whose
side lengths are a, b, and c is
(a*b*c)^2
R^2 = ------------------------
16*s*(s-a)*(s-b)*(s-c)
,where s is its semiperimeter.
Plugging in 2, 3, and 4 for a, b, and c we get
R^2 = 64/15
Plugging this into equation (1) gives
cos(m) = 17/32 = 0.53125
QED
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Posted by Bractals
on 2015-12-06 15:46:04 |
Solution
