Find the angle subtended at the centre of the circle of radius 5cm by an arc of length $\dfrac{{5\pi }}{3}cm$.
Answer
632.1k+ views
Hint: Here we solve the problem by comparing the given length of arc with the length of arc formula from which we can obtain the angle subtended at the centre of the given circle.
Complete step by step answer:
Here the given length of arc is $\dfrac{{5\pi }}{3}cm$.
Given a circle of radius (r) = 5cm.
We know that the formula of length of arc =$\dfrac{\theta }{{{{360}^ \circ }}}\left( {2\pi r} \right)$ where $\theta $ is angle subtended at centre and $2\pi r$ is the circumference of circle.
So, here let us equate the given length of arc with its formula to get the angle subtended at the centre of the circle.
Therefore
$ \Rightarrow \dfrac{{5\pi }}{3} = \dfrac{\theta }{{{{360}^ \circ }}} \times 2\pi r$
$ \Rightarrow \dfrac{{5\pi }}{3} = \dfrac{\theta }{{{{360}^ \circ }}} \times 2\pi \times 5$ $\left( {\because r = 5} \right)$
On further simplification we get
$ \Rightarrow \theta = \dfrac{{{{360}^ \circ }}}{6}$
$ \Rightarrow \theta = {60^ \circ }$
Therefore the angle subtended at the centre of the circle is ${60^ \circ }$.
NOTE: In the above problem we have compared the given value of length of arc with the formula of length of arc to get the theta value which is the angle subtended at the centre of the circle. Generally we forget to substitute the value of r with the radius of the circle which is mandatory to get the theta value.
Complete step by step answer:
Here the given length of arc is $\dfrac{{5\pi }}{3}cm$.
Given a circle of radius (r) = 5cm.
We know that the formula of length of arc =$\dfrac{\theta }{{{{360}^ \circ }}}\left( {2\pi r} \right)$ where $\theta $ is angle subtended at centre and $2\pi r$ is the circumference of circle.
So, here let us equate the given length of arc with its formula to get the angle subtended at the centre of the circle.
Therefore
$ \Rightarrow \dfrac{{5\pi }}{3} = \dfrac{\theta }{{{{360}^ \circ }}} \times 2\pi r$
$ \Rightarrow \dfrac{{5\pi }}{3} = \dfrac{\theta }{{{{360}^ \circ }}} \times 2\pi \times 5$ $\left( {\because r = 5} \right)$
On further simplification we get
$ \Rightarrow \theta = \dfrac{{{{360}^ \circ }}}{6}$
$ \Rightarrow \theta = {60^ \circ }$
Therefore the angle subtended at the centre of the circle is ${60^ \circ }$.
NOTE: In the above problem we have compared the given value of length of arc with the formula of length of arc to get the theta value which is the angle subtended at the centre of the circle. Generally we forget to substitute the value of r with the radius of the circle which is mandatory to get the theta value.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

