In Delta ABC, if a^{2} cos^{2} A - b^{2} - c^ | vedclass

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(B) Given, $a^{2} \cos^{2} A - b^{2} - c^{2} = 0$ $\Rightarrow a^{2} \cos^{2} A = b^{2} + c^{2}$ Using the Law of Cosines, $\cos A = \frac{b^{2} + c^{

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

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(B) Given, $a^{2} \cos^{2} A - b^{2} - c^{2} = 0$ $\Rightarrow a^{2} \cos^{2} A = b^{2} + c^{2}$ Using the Law of Cosines, $\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$. Substituting $b^{2} + c^{2} = a^{2} \cos^{2} A$, we get: $\cos A = \frac{a^{2} \cos^{2} A - a^{2}}{2bc} = \frac{-a^{2}(1 - \cos^{2} A)}{2bc} = \frac{-a^{2} \sin^{2} A}{2bc}$. Since $a, b, c > 0$ and $\sin^{2} A > 0$ for $0 Therefore, $A$ must lie in the second quadrant, i.e.,$\frac{\pi}{2} < A < \pi$.

In $\Delta ABC$, if $a^{2} \cos^{2} A - b^{2} - c^{2} = 0$, then

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