Question

In a triangle $ABC$, if $a^{2}co{s}^{2}A=b^2+c^2$, then:

• A. $0<A<\frac{\pi }{4}$
• B. $\frac{\pi }{4}<A<\frac{\pi }{2}$
• C. $\frac{\pi }{2}<A<\pi$
• D. $A=\frac{\pi }{2}$

Answer 1

The correct option is C $\frac{\pi }{2}<A<\pi$

$a^2\cos {}^{2}A=b^2+c^2{a}^2\cos {}^{2}A=2bc\cos A+a^2{a}^2\cos {}^{2}A-2bc\cos A-a^2=0\cos A=\frac{2bc\pm \sqrt{4b^2{c}^2+4a^4}}{2a^2}$

$\cos A=\frac{bc\pm \sqrt{a^4+b^2{c}^2}}{a^2}\sqrt{a^4+b^2{c}^2}>bc\therefore \cos Awillhavea-vesign$

It will be in $(\frac{\pi }{2},\pi )$ interval.

Hence, $\frac{\pi }{2}<A<\pi$ is the correct answer.