A wire of length 44 text{ cm} is bent into an | vedclass

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

Question

(B) Given,length of the arc $l = 44 	ext{ cm}$.Radius of the circle $r = 12 	ext{ cm}$.The angle $	heta$ subtended by the arc at the centre in radi

Vedclass · Accepted Answer

$\left(\frac{660}{\pi}\right)^{\circ}$

Vedclass · Answer

(B) Given,length of the arc $l = 44 	ext{ cm}$.Radius of the circle $r = 12 	ext{ cm}$.The angle $	heta$ subtended by the arc at the centre in radians is given by $	heta = \frac{l}{r}$.$	heta = \frac{44}{12} = \frac{11}{3} 	ext{ radians}$.To convert the angle from radians to degrees,we multiply by $\frac{180}{\pi}$.$	heta_{	ext{deg}} = \frac{11}{3} 	imes \frac{180}{\pi} = \frac{11 	imes 60}{\pi} = \left(\frac{660}{\pi}ight)^{\circ}$.

$A$ 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

Similar Questions

The trigonometric function $y = \tan x$ in the $II$ quadrant

$\cos \theta(\operatorname{cosec} \theta - \sec \theta) - \cot \theta =$

$\tanh (\log x) = $

If $\tan \theta = - \frac{1}{\sqrt{10}}$ and $\theta$ lies in the fourth quadrant,then $\cos \theta = $

$\frac{{\cot^2 15^\circ - 1}}{{\cot^2 15^\circ + 1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module