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