How do I find the magnitude of the vector $v=21(cos18°i+sin18°j)?$ ?

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Answer 1 · 2018-02-22T15:21:12

Given, $vec v =21(cos 18 i + sin 18 j)$

Angle between $21 cos 18 i$ and $21 sin 18 j$ is $90$

So, $|vec v| = sqrt((21 cos 18)^2 + (21 sin 18)^2) = sqrt(21^2(cos ^2 18 + sin ^2 18))=sqrt(21^2×1)=21$

Answer 2 · 2018-02-22T15:21:19

$|| bbv ||=21$

If $bbv =v_1 bbi +v_2 bbj$ , then the magnitude of $bbv$ is $|| bbv ||=sqrt(v_1^2+v_2^2)$

$bbv =21cos18 bbi +21sin18 bbj$ .

So

$|| bbv ||=sqrt((21cos18)^2+(21sin18)^2)=sqrt(441cos^2 18+441sin^2 18)=sqrt(441(cos^2 18+sin^18))=sqrt441=21$

How do I find the magnitude of the vector v=21(cos18°i+sin18°j)?v=21(cos18°i+sin18°j)??