Find the degree measure of the angle subtended at the centre of a circle of radius $100 \,cm$ by an arc of length $22\, cm$ ( Use $\pi=\frac{22}{7}$ ).
Find the degree measure of the angle subtended at the centre of a circle of radius $100 \,cm$ by an arc of length $22\, cm$ ( Use $\pi=\frac{22}{7}$ ).
We know that in a circle of radius $r$ unit, if an are of length $l$ unit subtends an angle $\theta$ radian at the centre, then
$\theta=\frac{1}{r}$
Therefore, for $r=100 \,cm , l=22 \,cm ,$ we have
$\theta=\frac{22}{100}$ radian
$=\frac{180}{\pi} \times \frac{22}{100}$ degree
$=\frac{180 \times 7 \times 22}{22 \times 100}$ degree
$=\frac{126}{10}$ degree
$=12 \frac{3}{5}$ degree
$=12^{\circ} 36^{\prime} \quad \quad\left[1^{\circ}=60^{\prime}\right]$
Thus, the required angle is $12^{\circ} 36^{\prime}$
Similar Questions
$(m + 2)\sin \theta + (2m - 1)\cos \theta = 2m + 1,$ if
$(m + 2)\sin \theta + (2m - 1)\cos \theta = 2m + 1,$ if
Prove that $\cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)=\frac{3}{2}$
Prove that $\cos ^{2} x+\cos ^{2}\left(x+\frac{\pi}{3}\right)+\cos ^{2}\left(x-\frac{\pi}{3}\right)=\frac{3}{2}$
$tan\,\, 20^o + tan\,\, 40^o + \sqrt 3\,\, tan\,\, 20^o tan\,\, 40^o$ is equal to
$tan\,\, 20^o + tan\,\, 40^o + \sqrt 3\,\, tan\,\, 20^o tan\,\, 40^o$ is equal to
If $(1 + \sin A)(1 + \sin B)(1 + \sin C)$$ = (1 - \sin A)(1 - \sin B)(1 - \sin C),$ then each side is equal to
If $(1 + \sin A)(1 + \sin B)(1 + \sin C)$$ = (1 - \sin A)(1 - \sin B)(1 - \sin C),$ then each side is equal to
If $\tan \theta + \sec \theta = {e^x},$ then $\cos \theta $ equals
If $\tan \theta + \sec \theta = {e^x},$ then $\cos \theta $ equals