Question
If in a $$\vartriangle ABC,{a^2}{\cos ^2}A = {b^2} + {c^2}$$ then
A.
$$A < \frac{\pi }{4}$$
B.
$$\frac{\pi }{4} < A < \frac{\pi }{2}$$
C.
$$A > \frac{\pi }{2}$$
D.
$$A = \frac{\pi }{2}$$
Answer :
$$A > \frac{\pi }{2}$$
Solution :
$$\eqalign{
& \cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{{a^2}{{\cos }^2}A - {a^2}}}{{2bc}} = \frac{{{a^2}\left( {{{\cos }^2}A - 1} \right)}}{{2bc}} < 0. \cr
& {\text{So, }}A > \frac{\pi }{2}. \cr} $$