If the angles of triangle $A B C$ 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 A $b^{2} = a^{2} + c^{2} - a c$
Since $A, B, C$ are in $A . P .$

$\Rightarrow 2 B = A + C = π - B \Rightarrow B = \frac{π}{3}$

$\Rightarrow cos B = \frac{1}{2} \Rightarrow \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{1}{2}$

$\Rightarrow a^{2} + c^{2} - b^{2} = a c$

$\Rightarrow b^{2} = a^{2} + c^{2} - a c$

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Similar questions

In the adjoining figure, if $B C = a, A C = b, A B = c$ and $∠ C A B = 120^{\circ}$ , then the correct relation is

In the adjoining figure, if BC = a, AC = b, AB = c and $∠ C A B = 120^{\circ}$ , then the correct relation is-

△ 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}

\frac{a^{2} + b^{2} + c^{2}}{R^{2}}

8

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

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

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

In the adjoining figure, if $B C = a, A C = b, A B = c$ and $∠ C A B = 120^{\circ}$ ,
then which of the following is the correct relation?

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