If A- and B- denotes sides of the parallelogram and its area is 12 ABA and B are the magnitude of A- and B- respectively -the angle between A- and B- is

The magnitude of vector $\to A, \to B$ and $\to C$ are respectively 12, 5 and 13 units and $\to A + \to B = \to C$ then the angle between $\to A$ and $\to B$ is

Q. The magnitude of vectors

\to A

\to B

and

\to C

are

3

4

and

5

units respectively. if

\to A + \to B = \to C

, find the angle between

\to A

and

\to B

Q. If

\to A . \to B = | \to A \times \to B |

then the angle between

\to A a n d \to B

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If →A and →B denotes sides of the parallelogram and its area is 12AB(A and B are the magnitude of →A and →B respectively ).the angle between →A and →B is

The correct option is A 30∘Area of the parallelogram is given by S=→A×→B=ABsinθGiven : Area =12AB∴ ABsinθ=12ABSo, sinθ=12⟹θ=30o

If $\to A$ and $\to B$ denotes sides of the parallelogram and its area is $\frac{1}{2} A B$ (A and B are the magnitude of $\to A$ and $\to B$ respectively ).the angle between $\to A$ and $\to B$ is

The correct option is A $30^{\circ}$
Area of the parallelogram is given by $S = \to A \times \to B = A B sin θ$
Given : Area $= \frac{1}{2} A B$
$∴$ $A B sin θ = \frac{1}{2} A B$
So, $sin θ = \frac{1}{2}$
$⟹ θ = 30^{o}$