If - A -21- direction of A is East- direction of B is 37- West of North and direction of C is North-East- then the magnitude of C is Given- C - A - B A- 10 -2B- 12 -2C- 12D- 6 -2

The correct option is B 12√(2)The given condition of the vectors can be shown as Let |B| be x, nbsp;|C| be y Given, C=A+B Now, resolving the vectors along t ...

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Byju's Answer

Standard XII

Physics

Vector Addition

If | A |=21, ...

Question

If $| \to A | = 21$ , direction of $\to A$ is East, direction of $\to B$ is $37^{\circ}$ West of North and direction of $\to C$ is North-East, then the magnitude of $\to C$ is (Given, $\to C = \to A + \to B$ )

6 \sqrt{2}

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10 \sqrt{2}

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12 \sqrt{2}

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12

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Solution

The correct option is C $12 \sqrt{2}$
The given condition of the vectors can be shown as

Let $| \to B |$ be $x$ , $| \to C |$ be $y$
Given, $\to C = \to A + \to B$
Now, resolving the vectors along the axes we get
$|_{x} | = 21 - x sin 37^{\circ}$
$\Rightarrow |_{x} | = 21 - \frac{3 x}{5}$
Similarly, $|_{y} | = x cos 37^{\circ}$
$\Rightarrow |_{x} | = \frac{4 x}{5}$

Thus, $tan 45^{\circ} = \frac{|_{y} |}{|_{x} |} = \frac{\frac{4 x}{5}}{21 - \frac{3 x}{5}} = 1$
$\Rightarrow x = 15$
So, $|_{x} | = 21 - 9 = 12$ and $|_{y} | = 12$
Thus, $y = \sqrt{{|_{x} |}^{2} + {|_{y} |}^{2}} = \sqrt{12^{2} + 12^{2}} = 12 \sqrt{2}$

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If |→A|=21, direction of →A is East, direction of →B is 37∘ West of North and direction of →C is North-East, then the magnitude of →C is (Given, →C=→A+→B )

If $| \to A | = 21$ , direction of $\to A$ is East, direction of $\to B$ is $37^{\circ}$ West of North and direction of $\to C$ is North-East, then the magnitude of $\to C$ is (Given, $\to C = \to A + \to B$ )