Question #0d666

1 Answer
Jim G.
Dec 13, 2017

#theta=(2pi)/3" or "theta=(5pi)/3#

Explanation:

#"given "tantheta<0" then "theta" is in second/fourth quadrant"#

#tantheta=-sqrt3#

#rArrtheta=tan^-1(sqrt3)=pi/3larrcolor(blue)"related acute angle"#

#rArrtheta=(pi-pi/3)=(2pi)/3larrcolor(blue)"second quadrant"#

#rArrtheta=(2pi-pi/3)=(5pi)/3larrcolor(blue)"fourth quadrant"#

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