Properties of Triangles

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.