Question
Find the trigonometric functions of $: 30^\circ$
✓
Answer
Angle of measure $30^{\circ}$ :
Let $\mathrm{m} \angle \mathrm{XOA}=30^{\circ}$
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.
$\mathrm{OP}=1$
$\mathrm{OM} =\frac{\sqrt{3}}{2} \mathrm{OP}$
$ =\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 1st quadrant, $x>0, y>0$
$\therefore \mathrm{x}=\mathrm{OM}=\frac{\sqrt{3}}{2}$ and $\mathrm{y}=\mathrm{PM}=\frac{1}{2}$
$ \therefore \quad \mathrm{P} \equiv\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$\sin 30^{\circ}=y=\frac{1}{2}$
$\cos 30^{\circ}=x=\frac{\sqrt{3}}{2} $
$\tan 30^{\circ}=\frac{y}{x}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$
$\operatorname{cosec} 30^{\circ}=\frac{1}{y}=\frac{1}{\left(\frac{1}{2}\right)}=2$
$\sec 30^{\circ}=\frac{1}{x}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$
$\cot 30^{\circ}=\frac{x}{y}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
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