Question
In a $$\vartriangle ABC,$$ the angles $$A$$ and $$B$$ are two values of $$\theta $$ satisfying $$\sqrt 3 \cos \theta + \sin \theta = k,\left| k \right| < 2.$$ The triangle
A.
is acute angled
B.
is right angled
C.
is obtuse angled
D.
has one angle $$ = \frac{\pi }{3}$$
Answer :
is obtuse angled
Solution :
$$\eqalign{
& \sqrt 3 \cos A + \sin A = \sqrt 3 \cos B + \sin B \cr
& {\text{or, }}\frac{{\sin A - \sin B}}{{\cos B - \cos A}} = \sqrt 3 \cr
& \therefore \,\,\sqrt 3 = \frac{{2\cos \frac{{A + B}}{2} \cdot \sin \frac{{A - B}}{2}}}{{2\sin \frac{{A + B}}{2} \cdot \sin \frac{{A - B}}{2}}} = \cot \frac{{A + B}}{2} \cr
& {\text{or, }}\frac{{A + B}}{2} = \frac{\pi }{6}\,\,\,{\text{or, }}A + B = \frac{\pi }{3} \cr
& \therefore \,\,C = \pi - \frac{\pi }{3}. \cr} $$