A and b are two adjacent sides of a parallelogram and c and d are its diagonals- If angle between a and b is - then cos- is equal to -

The correct option is D (c^2 d^2)/4ab We know, a+ b=c a b=d a^2+b^2+2abcos(θ)=c^2 a^2+b^2 2abcos(θ)=d^2 c^2 d^2=4abcos(θ) cos ...

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Multiplication with Vectors

a and b are...

Question

$\to a$ and $\to b$ are two adjacent sides of a parallelogram and $\to c$ and $\to d$ are its diagonals. If angle between $\to a$ and $\to b$ is $θ$ , then $c o s θ$ is equal to ?

\frac{(a^{2} + b^{2})}{4 c d}

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\frac{(a^{2} - b^{2})}{4 c d}

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\frac{(c^{2} - d^{2})}{4 a b}

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\frac{(c^{2} + d^{2})}{4 a b}

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Similar questions

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, b, c, d

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

a^{2} - b^{2}, b^{2} - c^{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}

2 A^{⊝} + B_{2} \to 2 B^{⊝} + A_{2}; 2 C^{⊝} + B_{2} \to N o r e a c t i o n; 2 D^{⊝} + A_{2} \to 2 A^{⊝} + D_{2}

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