Math, asked by Amank5840, 1 year ago

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


Amank5840: we know that in a circle of radius R unit if an Arc of length L units of 10 and angle theta Radian at the centre then theta is equal to one upon our therefore 4 is equal to 100 cm is equal to 22 CM we have theta is equal to 22 by 100 Radian is equal to 180 upon Pi

Answers

Answered by Anonymous
29
length =theta/360×2πr
22=l/360×2×22/7×100
l=18×7/10=12.6° hope this is the answer..

Amank5840: no
Answered by αηυяαg
118

 \huge{\bf{\underline{\red{Given:}}}}

Radius of Circle = 100 cm

Length of Arc = 22 cm

 \huge{\bf{\underline{\red{To\:Find:}}}}

Degree measure of the angle subtended at the centre of a circle.

 \huge{\bf{\underline{\red{Formula\:Used:}}}}

{\bf{\boxed{r=\dfrac{l}{θ}}}}

 \huge{\bf{\underline{\red{Solution:}}}}

Using Formula,

\sf :\implies\:r=\dfrac{l}{θ}

Putting Values,

\sf :\implies\:100=\dfrac{22}{θ}

\sf :\implies\:θ=\dfrac{22}{100}

\sf :\implies\:θ=\dfrac{11}{50}\: radians

Now,

\sf :\implies\:θ=(\dfrac{11}{50}\times \dfrac{180}{\pi})°

\sf :\implies\:θ=(\dfrac{11}{5}\times \dfrac{18\times 7}{22})°

\sf :\implies\:θ=(\dfrac{11}{5}\times \dfrac{9\times 7}{11})°

\sf :\implies\:θ=(\dfrac{693}{55})°

\sf :\implies\:θ=(12\dfrac{33}{55})°

\sf :\implies\:θ=12°(\dfrac{3}{5}\times 60)'

\sf :\implies\:θ=12°36'

Hence, The Degree measure of the angle subtended at the centre of a circle is 12°36'.

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