Question
Find the trigonometric functions of : 150°
✓
Answer
Angle of measure $150 :$ Let $\mathrm{m} \angle \mathrm{XOA}=150^{\circ}$ 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.
$ O P=1_r$
$\mathrm{OM}=\frac{\sqrt{3}}{2} \mathrm{OP}$
$=\frac{\sqrt{3}}{2}(1)$
$=\frac{\sqrt{3}}{2}$
$\mathrm{PM}=\frac{1}{2} \mathrm{OP}$
$=\frac{1}{2}(1)$
$=\frac{1}{2}$
$ \mathrm{OM} =\frac{\sqrt{3}}{2} \mathrm{OP}$
$ =\frac{\sqrt{3}}{2}(1)$
$ =\frac{\sqrt{3}}{2}$
$\mathrm{PM} =\frac{1}{2} \mathrm{OP}$
$ =\frac{1}{2}(1)$
$ =\frac{1}{2} $
Since point $P$ lies in the $2$ nd quadrant, $x<0, y>0$
$ \therefore \quad x=-\mathrm{OM}=\frac{-\sqrt{3}}{2} \text { and } y=\mathrm{PM}=\frac{1}{2}$
$\therefore \quad \mathrm{P} \equiv\left(\frac{-\sqrt{3}}{2}, \frac{1}{2}\right)$
$\sin 150^{\circ}=y=\frac{1}{2}$
$\cos 150^{\circ}=x=-\frac{\sqrt{3}}{2}$
$\tan 150^{\circ}=\frac{y}{x}=\frac{\frac{1}{2}}{-\left(\frac{\sqrt{3}}{2}\right)}=-\frac{1}{\sqrt{3}}$
$\operatorname{cosec} 150^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{1}{2}\right)}=2$
$\sec 150^{\circ}=\frac{1}{x}=\frac{1}{-\left(\frac{\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}}$
$\cot 150^{\circ}=\frac{x}{y}=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\frac{1}{2}}=-\sqrt{3} $
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