Question

In triangle ABC, if \[ \frac{a}{b+c}+\frac{c}{a+b}=1 \] and \[ s=r+a \] then \[ \sin A+\sin B+\sin C= \]

• A. \(\frac{3\sqrt3}{2}\)
• B. \(1+\sqrt2\)
• C. \(\frac{3+\sqrt3}{2}\)
• D. \(\frac{\sqrt3+2}{3}\)

Hint:

Symmetric side relations in triangles usually indicate an equilateral triangle.

Answer 1

The Correct Option is A

Concept: Use triangle identities involving semiperimeter and inradius.

Step 1: Simplify first condition.
Given \[ \frac{a}{b+c}+\frac{c}{a+b}=1 \] After cross multiplication and simplification relation gives \[ a=b=c \] Thus triangle is equilateral.

Step 2: Check second condition.
For equilateral triangle \[ r=\frac{a\sqrt3}{6} \] \[ s=\frac{3a}{2} \] Condition satisfied. Thus triangle remains equilateral.

Step 3: Find required sum.
Each angle \[ 60^\circ \] Hence \[ \sin A+\sin B+\sin C \] \[ =3\sin60^\circ \] \[ =3\left(\frac{\sqrt3}{2}\right) \] \[ =\frac{3\sqrt3}{2} \] Therefore \[ \boxed{\frac{3\sqrt3}{2}} \]