Resultant of two vectors is $20 \sqrt{3}$ and makes an angle $30^{0}$ with one of them which has magnitude $20$ . The magnitude of other vector:

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10 \sqrt{3}

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20 \sqrt{3}

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Solution

The correct option is B 20
Let us consider that the $3$ vectors form the 3 sides of a triangle, with
$a = 20, b = 20 \sqrt{3}$ , c = ? $θ = 30$
Using cosine rule,
$c =$ ${(a^{2} + b^{2} - 2 a b c o s θ)}^{\frac{1}{2}}$
$c = \sqrt{20^{2} + (20 \sqrt{3})^{2} - 2 \times 20 \times 20 \sqrt{3} \times \frac{\sqrt{3}}{2}}$
$c = \sqrt{400 + 1200 - 1200} = 20$ .
Thus, $c = 20$

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Resultant of two vectors is 20√3 and makes an angle 300 with one of them which has magnitude 20. The magnitude of other vector:

The correct option is B 20Let us consider that the 3 vectors form the 3 sides of a triangle, witha=20,b=20√3, c = ? θ=30Using cosine rule,c= (a2+b2−2abcosθ)12c=√202+(20√3)2−2×20×20√3×√32c=√400+1200−1200=20.Thus, c=20

Resultant of two vectors is $20 \sqrt{3}$ and makes an angle $30^{0}$ with one of them which has magnitude $20$ . The magnitude of other vector: