Question #153fd

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Question #153fd

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Answer 1 · 2016-12-30T05:46:55

drawn

From the above figure

Given the coordinates of point Q (3,-9)
$O N = 3 and Q N = - 9$ $ON=3 and QN=-9$

$O Q = \sqrt{O N^{2} + Q N^{2}}$ $OQ=sqrt(ON^2+QN^2)$

Now

$sin θ = \frac{opposite}{hypotenuse} = \frac{Q N}{O Q} = - \frac{9}{3 \sqrt{10}} = - \frac{3}{\sqrt{10}}$ $sintheta="opposite"/"hypotenuse"=(QN)/(OQ)=-9/(3sqrt10)=-3/sqrt10$

So $csc θ = \frac{1}{sin θ} = - \frac{\sqrt{10}}{3}$ $csctheta=1/sintheta=-sqrt10/3$

$cos θ = \frac{adjacent}{hypotenuse} = \frac{O N}{O Q} = \frac{3}{3 \sqrt{10}} = \frac{1}{\sqrt{10}}$ $costheta="adjacent"/"hypotenuse"=(ON)/(OQ)=3/(3sqrt10)=1/sqrt10$

So $sec θ = \frac{1}{cos θ} = \sqrt{10}$ $sectheta=1/costheta=sqrt10$

$tan θ = \frac{opposite}{adajcent} = \frac{Q N}{O N} = - \frac{9}{3} = - 3$ $tantheta="opposite"/"adajcent"=(QN)/(ON)=-9/3=-3$

So $cot θ = \frac{1}{tan θ} = - \frac{1}{3}$ $cottheta=1/tantheta=-1/3$

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Question #153fd

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