π < θ < 3pi/2, so θ ∈ 3 quadrant,
so cosθ<0, tg θ>0
we know that sin²θ+cos²θ=1
cosθ=-√(1-sin²θ)=-√1-[tex] (\frac{ \sqrt{3} }{2}) [/tex]²=-√(1-[tex] \frac{3}{4}) [/tex]=- √0.25=- 0.5
tgθ=[tex] \frac{sin \alpha }{cos \alpha } [/tex]= - [tex] \frac{ \sqrt{3} }{2}: (- 0.5)[/tex]=√3
P.s. θ=[tex] \alpha [/tex], i can't put θ in formula, sorry
