in triangle abc find the value of r1/bc + r2/ca +r3/ab
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Answer:
+
ca
r
2
+
ab
r
3
=
bc
stan
2
A
+
ac
stan
2
B
+
ab
stan
2
C
=
abc
s
[atan
2
A
+btan
2
B
+ctan
2
C
]
Now using a=2RsinA=2R.2sin
2
A
.cos
2
A
=4Rsin
2
A
.cos
2
A
, and similarly using the formula for b and c, we get
bc
r
1
+
ca
r
2
+
ab
r
3
=
abc
s
.4R[sin
2
2
A
+sin
2
2
B
+sin
2
2
C
]
Using r=
s
△
and R=
4△
abc
, we get
abc
s
.4R=
r
1
.
.
⇒
bc
r
1
+
ca
r
2
+
ab
r
3
=
r
1
[
2
1−cosA
+
2
1−cosB
+
2
1−cosC
]
.
=
r
1
[
2
3
−
2
cosA+cosB+cosC
]
Now using cosA+cosB+cosC=1+
R
r
, we get
bc
r
1
+
ca
r
2
+
ab
r
3
=
r
1
[1−
2R
r
]
.
⇒
bc
r
1
+
ca
r
2
+
ab
r
3
=
r
1
−
2R
1
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