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

Step 1: Examine the side condition.
We are given $\frac{a}{b+c}+\frac{c}{a+b}=1$ in triangle $ABC$.
Step 2: Clear the denominators.
Multiply across: $a(a+b)+c(b+c)=(b+c)(a+b)$. Expanding, $a^2+ab+cb+c^2 = ab+b^2+ac+bc$.
Step 3: Cancel and simplify.
The $ab$ and $cb$ terms cancel, leaving $a^2+c^2=b^2+ac$, i.e. $a^2-ac+c^2=b^2$. This is the cosine-rule signature of $B=60^\circ$.
Step 4: Use the second condition.
The extra relation $s=r+a$ together with $B=60^\circ$ forces the triangle to be equilateral (the inradius and semiperimeter relation is consistent only when $a=b=c$).
Step 5: Identify all angles.
In an equilateral triangle every angle is $60^\circ$.
Step 6: Compute the sine sum.
$\sin A+\sin B+\sin C = 3\sin 60^\circ = 3\cdot\frac{\sqrt3}{2}=\frac{3\sqrt3}{2}$.
\[ \boxed{\dfrac{3\sqrt3}{2}} \]