Don't Find Those Nasty roots!

$\begin{aligned} \sqrt 3 \cos A + \sin A & = \frac 12 \\ 2 \left(\frac {\sqrt 3}2 \cos A + \frac 12 \sin A \right) & = \frac 12 \\ \sin \left(A+60^\circ \right) & = \frac 14 \\ A + 60^\circ & = \color{#3D99F6}\sin^{-1} \frac 14 & \small \color{#3D99F6} \text{Note that }\sin^{-1} \frac 14 < 30^\circ \\ & > \color{#3D99F6} 180^\circ - 30^\circ & \small \color{#3D99F6} \text{For positive }A \\ \implies A & > 180^\circ - 30^\circ - 60^\circ \\ & > 90^\circ \ \boxed{\text{Obtuse}} \end{aligned}$

$A$ definitely cannot be in the first quadrant:

If $A < 30^\circ$ , $\sqrt{3}\cos A+\sin A > \frac{3}{2}+\sin A > \frac{1}{2}$
If $30^\circ \leq A \leq 90^\circ$ , $\sqrt{3}\cos A + \sin A > \sqrt{3}\cos A + \frac{1}{2} > \frac{1}{2}$

Hence $A$ must fall on the second quadrant, and form an obtuse triangle.

Divide both sides by 2

Now note that lhs is sine of $A+\frac{\pi}{3}$

For lhs to be 0.25, A has to be obtuse as A is positive and arcsin of 0.25 is lesser than $\frac{\pi}{3}$

.

$2*\left(\sqrt3/2*CosA+1/2*SinA \right )=1/2.\\ \implies\ Cos(A-60^o)=1/4<1/2=Cos60^o.\\ Cos\ is\ a\ decreasing\ function. So \ Cos^{-1}1/4>60^o.\\ \therefore\ Cos(A-60^o)>Cos60^o.\\ \implies\ A>120^o.\\ \ \$

Let cos A = c and sin A = s.

Squaring the equation,
3c^2 + s^2 + 2sc * sqrt(3) = 1/4
= (s^2 + c^2) + 2c^2 + 2sc * sqrt(3)
= 1 + 2c * (c + s * sqrt(3))
2c * (c + s * sqrt(3)) = 1/4 - 1 = -3/4
s = sin A > 0 for any A in a triangle, so RHS could only happen iff c < 0 AND c + s * sqrt(3) > 0
c = cos A < 0 for 90 < A << 180.
==> Triangle ABC has an obtuse angle A.