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)
#ON=3 and QN=-9#

#OQ=sqrt(ON^2+QN^2)#

Now

#sintheta="opposite"/"hypotenuse"=(QN)/(OQ)=-9/(3sqrt10)=-3/sqrt10#

So #csctheta=1/sintheta=-sqrt10/3#

#costheta="adjacent"/"hypotenuse"=(ON)/(OQ)=3/(3sqrt10)=1/sqrt10#

So #sectheta=1/costheta=sqrt10#

#tantheta="opposite"/"adajcent"=(QN)/(ON)=-9/3=-3#

So #cottheta=1/tantheta=-1/3#