Question
Find the trigonometric functions of : 240°

Answer

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Angle of measure $240^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=240^{\circ}$
Its terminal arm (ray $O A)$ intersects the standard unit circle at $P(x, y)$. Draw seg PM perpendicular to the $X$-axis.
$\triangle O M P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$O P =1$
$O M  =\frac{1}{2} \mathrm{OP}$
$ =\frac{1}{2}(1)$
$ =\frac{1}{2}$
$P M  =\frac{\sqrt{3}}{2} \mathrm{OP}$
$ =\frac{\sqrt{3}}{2}(1)$
$ =\frac{\sqrt{3}}{2} $
Since point $P$ lies in the 3 rd quadrant, $x<0, y<0$
$ \therefore \quad x=-\mathrm{OM}=-\frac{1}{2} \text { and } y=-\mathrm{PM}=-\frac{\sqrt{3}}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{-1}{2},-\frac{\sqrt{3}}{2}\right)$
$\sin 240^{\circ}=y=-\frac{\sqrt{3}}{2}$
$\cos 240^{\circ}=x=-\frac{1}{2}$
$\tan 240^{\circ}=\frac{y}{x}=\frac{\left(-\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=\sqrt{3}$
$\operatorname{cosec} 240^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{-\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}}$
$\sec 240^{\circ}=\frac{1}{x}=\frac{1}{-\frac{1}{2}}=-2$
$\cot 240^{\circ}=\frac{x}{y}=\frac{\left(\frac{-1}{2}\right)}{\left(\frac{-\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}} $

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