In Delta ABC,if cos frac{A}{2} = sqrt{frac{b | vedclass

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(A) We know the formula for $\cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}$.Given $\cos \frac{A}{2} = \sqrt{\frac{b + c}{2c}}$,we equate the two:$\sqr

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$a^2 + b^2 = c^2$

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(A) We know the formula for $\cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}$.Given $\cos \frac{A}{2} = \sqrt{\frac{b + c}{2c}}$,we equate the two:$\sqrt{\frac{s(s - a)}{bc}} = \sqrt{\frac{b + c}{2c}}$Squaring both sides:$\frac{s(s - a)}{bc} = \frac{b + c}{2c}$$\frac{s(s - a)}{b} = \frac{b + c}{2}$$2s(s - a) = b(b + c)$Substitute $2s = a + b + c$ and $s - a = \frac{b + c - a}{2}$:$(a + b + c) \left( \frac{b + c - a}{2} \right) = b^2 + bc$$((b + c) + a)((b + c) - a) = 2(b^2 + bc)$$(b + c)^2 - a^2 = 2b^2 + 2bc$$b^2 + c^2 + 2bc - a^2 = 2b^2 + 2bc$$c^2 - a^2 = b^2$$c^2 = a^2 + b^2$.

In $\Delta ABC$,if $\cos \frac{A}{2} = \sqrt{\frac{b + c}{2c}}$,then:

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