If the angles $A, B, C$ of $△ A B C$ are 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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c^{2} = a^{2} + b^{2}

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Solution

The correct option is B $b^{2} = a^{2} + c^{2} - a c$
Ans. $(b)$ We have $A + B + C = 180^{o}$ .
Also $A, B, C$ are in $A . P . \Rightarrow 2 B = A + C$ .
$∴ 3 B = 180^{o}$ or $B = 60^{o}$
Now $b^{2} = a^{2} + c^{2} - 2 a c cos B$
$= a^{2} + c^{2} - 2 a c cos 60^{o}$
or $b^{2} = a^{2} + c^{2} - a c$

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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-

|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

A B C

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

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