If the x component of a vector a ,in the xy plane , is half as large as he magnitude of the vector , find the tangent of the angle between the vector and the x axis ?

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Answer 1 · 2018-01-31T14:07:24

$\sqrt{3}$ $sqrt 3$

Suppose,the $a$ $a$ vector makes an angle $θ$ $theta$ w.r.t the $X$ $X$ axis,

so,its X component becomes $a cos θ$ $a cos theta$

Now,given, $a cos θ = \frac{a}{2}$ $acos theta=a/2$

so, $θ = 60$ $theta = 60$

$tan 60 = \sqrt{3}$ $tan 60 = sqrt3$

Answer 2 · 2018-01-31T14:13:38

$details has been showed below.$ $"details has been showed below."$

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${(O E)}^{2} = {(E B)}^{2} + {(O B)}^{2}$ $(OE)^2=(EB)^2+(OB)^2$
${(2 x)}^{2} = {(E B)}^{2} + x^{2}$ $(2x)^2=(EB)^2+x^2$

${(E B)}^{2} = 4 x^{2} - x^{2}$ $(EB)^2=4x^2-x^2$

${(E B)}^{2} = 3 x^{2}$ $(EB)^2=3x^2$
$\sqrt{{(E B)}^{2}} = \sqrt{3 x^{2}}$ $sqrt((EB)^2)=sqrt(3x^2)$

$E B = x \sqrt{3}$ $EB=x sqrt(3)$

$tan θ = \frac{¯ ¯¯¯¯ ¯ E B}{¯ ¯¯¯¯ ¯ O B}$ $tan theta=bar (EB)/bar(OB)$

$tan theta=(cancel(x) sqrt (3))/cancel(x)$

$tan theta=sqrt (3)$