Question

If in a triangle $\Delta ABC,a^2-b^2-c^2=0$, then

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

Answer 1

Answer: (C)

$A=\frac{\pi }{2}$

Given, in a triangle $\Delta ABC,a^2-b^2-c^2=0$.....(1).

We have,

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

or, $\cos A=0$ [ Using (1)]

or, $A=\frac{\pi }{2}$.