Problem - 2991
Let $\triangle{ABC}$ be an acute triangle and $a, b, c$ be the three sides opposite to $\angle{A}, \angle{B}, \angle{C}$ respectively. If vectors $m=(a+c,b)$ and $n=(a-c, b-a)$ satisfy $m\cdot n = 0$,
(1) Compute the measurement of $\angle{C}$.
(2) Find the range of $\sin{A} + \sin{B}$.
(i) $(a+c)(a-c) + b(b-a)=0 \implies a^2+b^2-c^2=ab$
$$\cos{C}=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{ab}{2ab}=\dfrac{1}{2} \implies C=\boxed{\dfrac{\pi}{6}}$$
(ii) $C=\dfrac{\pi}{3} \implies A + B =\dfrac{2\pi}{3}$
\begin{align*}
& \sin{A} + \sin{B} \\
=& \sin{A} + \sin{(\dfrac{2\pi}{3}-A)}\\
=& \sqrt{3}\sin(A+\dfrac{\pi}{6})
\end{align*}
$0 < A < \dfrac{2\pi}{3} \implies \dfrac{\pi}{6} < A + \dfrac{\pi}{6} < \dfrac{5\pi}{6} \implies \dfrac{1}{2} < \sin(A+\dfrac{\pi}{6}) \le 1 \implies \boxed{\dfrac{\sqrt{3}}{2} < \sin{A} + \sin{B} \le \sqrt{3}}$