The magnitude of scalar and vector products of two vectors are483 and 144 respectively. What is the angle between the two vectors

abcosθ=483 absinθ=144 absinθabcosθ=144483 tanθ=3×33 tanθ=3 tanθ=tan60° θ=60°

abcosθ=483 absinθ=144 absinθabcosθ=144483 tanθ=3×33 tanθ=3 tanθ=tan60° θ=60°

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The magnitude of scalar and vector products of two vectors are $48 \sqrt{3}$ and 144 respectively. What is the angle between the two vectors

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30 °

60 °

90 °

45 °

answer is C.

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ab \cos θ = 48 \sqrt{3}

$ab \sin θ = 144$

$\frac{ab \sin θ}{ab \cos θ} = \frac{144}{48 \sqrt{3}}$

$\tan θ = \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}$

$\tan θ = \sqrt{3}$

$\tan θ = \tan 60 °$

$θ = 60 °$

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