# Solve cos 30 * sin(-50) / tan 70

Solve $\mathrm{cos}30\cdot \frac{\mathrm{sin}\left(-50\right)}{\mathrm{tan}70}$
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Derrick

$\mathrm{sin}\left(-50\right)=-\mathrm{sin}\left(50\right)$
so the expression becomes $-\mathrm{cos}\left(30\right)\frac{\mathrm{sin}\left(50\right)}{\mathrm{tan}\left(70\right)}=-\mathrm{sin}\left(50\right)\frac{\sqrt{3}}{2}\mathrm{tan}\left(70\right)=-0.2415$ .
$\left[\mathrm{cos}\left(30\right)=\frac{\sqrt{3}}{2}=0.8660$ because in right-angled triangle ABC where B=90, C=30, A=60, the sides are in the ratio $1,\sqrt{3},2$ by Pythagoras),
$\mathrm{tan}\left(70\right)=2.7475\approx ,\mathrm{sin}\left(50\right)=0.7660\approx .$