Question

Find the degree measure of the angle subtended at the center of a circle of radius 100 cm by an arc of length 22 cm. (Useπ= 22/7 )/

Answer 1

Answer:

We know that in a circle of radius r unit, if an arc of length l unit

subtends an angle θ radian at the centre, then θ=

r

l

Therefore, for r=100 cm, l=22 cm, we have

θ=

100

22

radian=

π

180

×

100

22

degree=

22×100

180×7×22

degree

=

10

126

degree=12

5

3

degree=12

∘

36

′

,[∵1

∘

=60

′

]

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'.