Question
Find the trigonometric functions of $: 210^\circ$
✓
Answer
Angle of measure $210^{\circ}$ :
Let $m \angle X O A=210^{\circ}$
Its terminal arm (ray $O A$ ) intersects the standard unit circle at $P(x, y)$.
Draw seg PM perpendicular to the X-axis.
$\therefore \triangle \mathrm{OMP}$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$\mathrm{OP}=1$,
Since point $P$ lies in the 3 rd quadrant, $x<0, y<0$
$\therefore \mathrm{x}=-\mathrm{OM}=\frac{-\sqrt{3}}{2}$ and $\mathrm{y}=-\mathrm{PM}=\frac{-1}{2}$
$\therefore P \equiv\left(\frac{-\sqrt{3}}{2}, \frac{-1}{2}\right)$
$ \sin 210^{\circ}=y=\frac{-1}{2}$
$\cos 210^{\circ}=x=\frac{-\sqrt{3}}{2}$
$\tan 210^{\circ}=\frac{y}{x}=\frac{\left(-\frac{1}{2}\right)}{\left(-\frac{\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}} $ $\tan 210^{\circ}=\frac{y}{x}=\frac{\left(-\frac{1}{2}\right)}{\left(-\frac{\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}}$ $ \operatorname{cosec} 210^{\circ}=\frac{1}{y}=\frac{1}{\left(-\frac{1}{2}\right)}=-2$
$\sec 210^{\circ}=\frac{1}{x}=\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}}$
$\cot 210^{\circ}=\frac{x}{y}=\frac{\left(-\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=\sqrt{3} $
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