Question

If in a $△ABC,a^{2}co{s}^{2}A=b^2+c^2$ then

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{a^{2}co{s}^{2}A-a^2}{2bc}$
=$\frac{a^2(co{s}^{2}A-1)}{2bc}<=0so,A>=\frac{\pi }{2}$ but A$=\frac{\pi }{2}$ not possible form the hypothesis