In a cyclic quadrilateral $A B C D$ such that $A B = a, B C = b, C D = c, D A = d$ , then $cos B$ equals

\frac{a^{2} + b^{2} - c^{2} - d^{2}}{2 (a b + c d)}

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\frac{c^{2} + d^{2} - a^{2} + b^{2}}{2 (a b + c d)}

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\frac{a^{2} + b^{2} - c^{2} - d^{2}}{2 (a b - c d)}

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\frac{b^{2} + c^{2} - d^{2} - a^{2}}{2 (a b + c d)}

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Solution

The correct option is A $\frac{a^{2} + b^{2} - c^{2} - d^{2}}{2 (a b + c d)}$
In $△ A B C$ , we have
${A C}^{2} = a^{2} + b^{2} - 2 a b c o s B$ ...(1)
In $△ A C D$
${A C}^{2} = c^{2} + d^{2} - 2 c d c o s D = c^{2} + d^{2} - 2 c d c o s (π - B)$
$= c^{2} + d^{2} + 2 c d c o s B$ ...(2)
From (1) and (2)
$a^{2} + b^{2} - 2 a b c o s B = c^{2} + d^{2} + 2 c d c o s B \Rightarrow \frac{a^{2} + b^{2} - c^{2} - d^{2}}{2 (a b + c d)} = c o s B$

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Similar questions

2 (a b + c d) - a^{2} - b^{2} + c^{2} + d^{2}

2 (a b + c d) - a^{2} - b^{2} + c^{2} + d^{2}

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

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

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

\frac{a b + c d}{a d + b c}

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

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