Question
Find $\operatorname{cosec} 30^{\circ} \cos 60^{\circ}$ geometrically.
✓
Answer
Make an equilateral $\triangle PQR$
$\angle P=\angle Q=\angle R=60^{\circ}$
Draw a perpendicular $PS$ from $P$ to the side $QR$
As, a perpendicular bisector divides the equilateral triangle into two congruent triangles,
So, $\triangle P Q S \cong \triangle P R S \ ($by $\text{RHS}$ congruency criterion$)$
$\therefore Q S=R S \ (\text{CPCT})$
Here length of each side is $2 a$
$\Rightarrow Q S=R S=a$
In $, \triangle P Q S$
$\operatorname{cosec} 30^{\circ}=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{P Q}{Q S}$
$\Rightarrow \operatorname{cosec} 30^{\circ}=\frac{2 a}{a}=2$
$\text { And, } \cos 60^{\circ}=\frac{\text { base }}{\text { hypotenuse }}=\frac{Q S}{P Q}$
$\Rightarrow \cos 60^{\circ}=\frac{a}{2 a}=\frac{1}{2}$
Hence, $\operatorname{cosec} 30^{\circ}=2$ and $\cos 60^{\circ}=\frac{1}{2}$.
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