If the angles of a triangle ABC be in A.P., then

c^{2} = a^{2} + b^{2} - a b

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b^{2} = a^{2} + c^{2} - a c

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a^{2} = b^{2} + c^{2} - a c

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b^{2} = a^{2} + c^{2}

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Solution

The correct option is B $b^{2} = a^{2} + c^{2} - a c$
A, B, C are in A. P. then angle B = $60^{o},$
$c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}, {\begin{matrix} s i n c e A + B + C = 180^{o} a n d A + C = 2 B \Rightarrow B = 60^{o} \end{matrix}} \Rightarrow \frac{1}{2} = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = a c \Rightarrow b^{2} = a^{2} + c^{2} - a c .$

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a^{2} = b^{2} + c^{2}

b^{2} = c^{2} + a^{2}

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△ A B C

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

cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}, cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

In a triangle ABC, $2 c a s i n \frac{A - B + C}{2}$ is equal to

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