find the HCF of
a. 18,27
b. 32,72
c. 60,84
d. 44,66
e. 80,112
f. 21,306
g. 28,70
h. 35,105
Answers
Answer:
Radius of Circle = 100 cm
Length of Arc = 22 cm
\huge{\bf{\underline{\red{To\:Find:}}}}
ToFind:
Degree measure of the angle subtended at the centre of a circle.
\huge{\bf{\underline{\red{Formula\:Used:}}}}
FormulaUsed:
{\bf{\boxed{r=\dfrac{l}{θ}}}}
r=
θ
l
\huge{\bf{\underline{\red{Solution:}}}}
Solution:
Using Formula,
\sf :\implies\:r=\dfrac{l}{θ}:⟹r=
θ
l
Putting Values,
\sf :\implies\:100=\dfrac{22}{θ}:⟹100=
θ
22
\sf :\implies\:θ=\dfrac{22}{100}:⟹θ=
100
22
\sf :\implies\:θ=\dfrac{11}{50}\: radians:⟹θ=
50
11
radians
Now,
\sf :\implies\:θ=(\dfrac{11}{50}\times \dfrac{180}{\pi})°:⟹θ=(
50
11
×
π
180
)°
\sf :\implies\:θ=(\dfrac{11}{5}\times \dfrac{18\times 7}{22})°:⟹θ=(
5
11
×
22
18×7
)°
\sf :\implies\:θ=(\dfrac{11}{5}\times \dfrac{9\times 7}{11})°:⟹θ=(
5
11
×
11
9×7
)°
\sf :\implies\:θ=(\dfrac{693}{55})°:⟹θ=(
55
693
)°
\sf :\implies\:θ=(12\dfrac{33}{55})°:⟹θ=(12
55
33
)°
\sf :\implies\:θ=12°(\dfrac{3}{5}\times 60)':⟹θ=12°(
5
3
×60)
′
\sf :\implies\:θ=12°36':⟹θ=12°36
′
Hence, The Degree measure of the angle subtended at the centre of a circle is 12°36'.
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