Question 810122
 What is sin if  is in standard position, its terminal side lies in quadrant IV, and sec=√3?
a.√2/2

b.-√6/3

c.√3/3

d.-√3/3


sec=√3_____cos θ = {{{1/sqrt(3)}}}____cos θ = {{{sqrt(3)/3}}} = {{{A/H}}} = {{{x/r}}}


Therefore, {{{x = sqrt(3)}}}, and {{{r = 3}}}


{{{y = sqrt(r^2 - x^2)}}}


{{{y = sqrt(3^2 - (sqrt(3))^2)}}}


{{{y = sqrt(9 - 3)}}}


{{{y = sqrt(6)}}}


sin &#952; = {{{O/H}}} = {{{y/r}}} = {{{sqrt(6)/3}}}, but since sin is < 0 in the 4th quadrant, we get: - sin &#952; = {{{highlight_green(- sqrt(6)/3)}}} (CHOICE b.)


You can do the check!! 


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