Question

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. (Useπ=
7
22
​
)

Answer 1

Answer:

We khow in a circle of radius r unit , if arc of length 1unit substends an angle theta....then theta=1/r....therefore r=100cm and l=22cm... theta=22/100=180/π x 22/100=(180*7*22/22*100)°i.e.= (126/10)°=12°36'...[1°=60']hope this answer help you....

Answer 2

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

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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