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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the radian measure of $-37^{\large\circ}30'$

$\begin{array}{1 1}(A)\;(\large\frac{-5\pi}{24})^c\\(B)\;(\large\frac{5\pi}{24})^c\\(C)\;(\large\frac{7\pi}{24})^c\\(D)\;(\large\frac{-7\pi}{24})^c\end{array} $

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1 Answer

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  • $1^{\large\circ}=(\large\frac{\pi}{180})^c$
We know that
$180^{\large\circ}=\pi^c$
$1^{\large\circ}=(\large\frac{\pi}{180})^c$
$\Rightarrow -37^{\large\circ}30'=\big(\large\frac{\pi}{180}\times (\frac{-75}{12})\big)^c$
Since $-37^{\large\circ}30'=(-37\large\frac{1}{2})^{\large\circ}=(-\frac{75}{2})^{\large\circ}$
$\Rightarrow \big(-\large\frac{5\pi}{24}\big)^c$
Hence $(-37^{\large\circ}30')=\big(-\large\frac{5\pi}{24}\big)^c$
Hence (A) is the correct answer.
answered May 20, 2014 by sreemathi.v
 
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