Question
What is the distance between the points $(5\sin60^\circ,0)$ and $(0,5\sin30^\circ)?$
✓
Answer
Distance between the given points, $=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$ $=\sqrt{(0-5\sin60^\circ)^2+(5\sin30^\circ-0)^2}$ $=\sqrt{\Big(-5\times\frac{\sqrt{3}}{2}\Big)^2+\Big[5\Big(\frac{1}{2}\Big)\Big]^2}$ $\begin{Bmatrix}\because\ \sin60^\circ=\frac{\sqrt{3}}{2}\\\ \ \ \ \ \sin30^\circ=\frac{1}{2}\end{Bmatrix}$$=\sqrt{\frac{25\times3}{4}+\frac{25\times1}{4}}$
$=\sqrt{\frac{75}{4}+\frac{25}{4}}=\sqrt{\frac{100}{4}}$
$=\sqrt{25}=5\text{ units}$
$=\sqrt{\frac{75}{4}+\frac{25}{4}}=\sqrt{\frac{100}{4}}$
$=\sqrt{25}=5\text{ units}$
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