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Mathematics
Mathematics
Problem
In triangle
A
B
C
ABC
A
B
C
, if
∑
(
b
+
c
)
cos
A
4
cos
A
2
⋅
cos
B
2
⋅
cos
C
2
=
8
\frac{\sum (b+c) \cos A}{4 \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}} = 8
4
c
o
s
2
A
⋅
c
o
s
2
B
⋅
c
o
s
2
C
∑
(
b
+
c
)
c
o
s
A
=
8
, then the circum radius of triangle
A
B
C
ABC
A
B
C
is equal to 1 4
1
2
\frac{1}{2}
2
1
2
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