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Trigonometry – I — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – I1 Mark
Question
Find the polar co-ordinates of points whose Cartesian co-ordinates are : $(-\sqrt{3}, 1)$
✓

Answer

$(x, y)=(-\sqrt{3}, 1)$
$\therefore r=\sqrt{x^2+y^2}=\sqrt{3+1}=\sqrt{4}=2$
$\tan \theta=\frac{y}{x}=\frac{1}{-\sqrt{3}}=-\tan \frac{\pi}{6} $
Since the given point lies in the $2$ nd quadrant,
$ \tan \theta=\tan \left(\pi-\frac{\pi}{6}\right) \ldots[\because \tan (\pi-x)=-\tan \mathrm{x}]$
$\therefore \tan \theta=\tan \left(\frac{5 \pi}{6}\right)$
$\therefore \theta=\frac{5 \pi}{6}=150^{\circ}$
$\therefore$ the required polar co-ordinates are $\left(2,150^{\circ}\right)$

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