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1950 Moscow Mathematical Olympiad
178
MMO 178 Moscow MO 1950 sinx =(sinBsinC)/(1 - cosB cosCcosA)
MMO 178 Moscow MO 1950 sinx =(sinBsinC)/(1 - cosB cosCcosA)
Source:
August 6, 2019
trigonometry
angles
Problem Statement
Let
A
A
A
be an arbitrary angle,let
B
B
B
and
C
C
C
be acute angles. Is there an angle
x
x
x
such that
sin
x
=
sin
B
⋅
sin
C
1
−
cos
B
⋅
cos
C
⋅
cos
A
?
\sin x =\frac{\sin B \cdot \sin C}{1 - \cos B \cdot \cos C \cdot \cos A} ?
sin
x
=
1
−
cos
B
⋅
cos
C
⋅
cos
A
sin
B
⋅
sin
C
?
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