Convert 40^{circ} 20^{prime} into radian meas | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We know that $180^{\circ} = \pi 	ext{ radians}$.First,convert $40^{\circ} 20^{\prime}$ into degrees:$20^{\prime} = (\frac{20}{60})^{\circ} = (\fr

Vedclass · Accepted Answer

$\frac{121 \pi}{540}$

Vedclass · Answer

(A) We know that $180^{\circ} = \pi 	ext{ radians}$.First,convert $40^{\circ} 20^{\prime}$ into degrees:$20^{\prime} = (\frac{20}{60})^{\circ} = (\frac{1}{3})^{\circ}$.So,$40^{\circ} 20^{\prime} = (40 + \frac{1}{3})^{\circ} = (\frac{120+1}{3})^{\circ} = (\frac{121}{3})^{\circ}$.Now,convert degrees to radians by multiplying by $\frac{\pi}{180}$:$	ext{Radian measure} = (\frac{121}{3}) 	imes \frac{\pi}{180} = \frac{121 \pi}{540} 	ext{ radians}$.Therefore,$40^{\circ} 20^{\prime} = \frac{121 \pi}{540} 	ext{ radians}$.

Convert $40^{\circ} 20^{\prime}$ into radian measure.

Similar Questions

$\frac{\sin 81^o + \cos 81^o}{\sin 81^o - \cos 81^o}$ is equal to

$A$ wire of length $44 \text{ cm}$ is bent into an arc of a circle of radius $12 \text{ cm}$. The angle (in degrees) subtended by the arc at the centre of the circle is

Prove that: $\cot ^{2} \frac{\pi}{6} + \csc \frac{5 \pi}{6} + 3 \tan ^{2} \frac{\pi}{6} = 6$

$\cos 36^{\circ} - \cos 72^{\circ}$ is equal to

$\cos ^2\left(\frac{7 \pi}{8}\right)+\cos ^2\left(\frac{5 \pi}{8}\right)+\cos ^2\left(\frac{3 \pi}{8}\right)+\cos ^2\left(\frac{\pi}{8}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module