Question
In a $$\vartriangle ABC,\cos A + \cos B + \cos C > 1$$ only if the triangle is
A.
acute angled
B.
obtuse angled
C.
right angled
D.
the nature of the triangle cannot be determined
Answer :
the nature of the triangle cannot be determined
Solution :
$$\eqalign{
& \cos A + \cos B + \cos C > 1 \cr
& \Rightarrow \,\,1 + 4\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} > 1 \cr
& \therefore \,\,\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} > 0 \cr
& \Rightarrow \,\,\frac{A}{2},\frac{B}{2},\frac{C}{2}\,\,{\text{are acute}}{\text{.}} \cr} $$