Given data indicates: ‌r1+r2=3R ‌r2+r3=2R Let's analyze these conditions: From r1+r2=3R : We start by substituting in the formula: ‌
∆
s−a
+‌
∆
s−b
=‌
3abc
4∆
Simplifying, we have: ∆2(‌
1
s−a
+‌
1
s−b
)=‌
3abc
4
Using the identity: s(s−b)(s−c)+s(s−a)(s−c)=‌
3abc
4
Combining terms: s(s−c)(s−b+s−a)=‌
3abc
4
Further simplification: s(s−c)c=‌
3abc
4
Resulting relationship:
s(s−c)=‌
3ab
4
Therefore: √‌
s(s−c)
ab
=‌
√3
2
Which implies: cos(‌
C
2
)=‌
√3
2
Hence, ‌
C
2
=30∘‌ or ‌C=60∘ From r2+r3=2R : Use: s(s−a)=‌
2bc
4
Gives: √‌
s(s−a)
bc
=‌
1
√2
Therefore: cos(‌
A
2
)=‌
1
√2
Thus, ‌
A
2
=45∘‌‌‌ or ‌‌‌A=90∘Final Angles of the Triangle: ‌∠A=90∘ ‌∠B=60∘ ‌∠C=30∘ This implies the triangle does not fit typical ratios like a:b:c=2:1:√3 ; instead, the lengths of sides must be inconsistent, i.e., a≠b≠c , which corresponds to a non-equilateral, non-isosceles right triangle.