Answer

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Hint: Find the perimeter of the sector of circle and length of the arc of the semi-circle. They both have the same radius. Equate both the expressions to get the value of angle \[\theta \] and convert it into degrees, minutes and seconds by multiplying it with 60’ and 60’’.

“Complete step-by-step answer:”

A sector of a circle is a portion of a disk enclosed by 2 radii and an arc. The smaller area will be known as a minor sector and larger as a major sector.

Let us assume the radius of the circle as ‘r’ and \[\theta \] be the angle of the circle.

The perimeter of a sector of the circle is given as, \[2r+\dfrac{\pi r.\theta }{360}\], which is the 2 radii and measure of the arc.

\[\therefore \]Perimeter of sector \[=r+r+\dfrac{\pi r.\theta }{360}=2r+\pi r\dfrac{\theta }{360}\]

The length of the arc of the semicircle is given by \[\pi r\].

We have been told that the perimeter of a certain sector of a circle is equal to the length of the arc of a semicircle.

\[\therefore \]Perimeter of sector of circle = length of arc of semi-circle.

By substituting the values we get,

\[2r+\pi r\dfrac{\theta }{360}=\pi r\] (Let us take, \[\pi =\dfrac{22}{7}\])

\[\begin{align}

& 2r+\dfrac{22}{7}\times r\dfrac{\theta }{360}=\dfrac{22}{7}r \\

& r\left[ 2+\dfrac{22}{7}\times \dfrac{\theta }{360} \right]=\dfrac{22}{7}r \\

\end{align}\]

Cancel out ‘r’ on both sides, we get

\[2+\dfrac{22}{7}\times \dfrac{\theta }{360}=\dfrac{22}{7}\]

\[\dfrac{22}{7}\times \dfrac{\theta }{360}=\dfrac{22}{7}-2\]

\[\begin{align}

& \dfrac{22}{7}\times \dfrac{\theta }{360}=\left( \dfrac{22-14}{7} \right) \\

& \dfrac{\theta }{360}=\dfrac{8}{7}\times \dfrac{7}{22} \\

& \therefore \dfrac{\theta }{360}=\dfrac{8}{22}=\dfrac{4}{11} \\

& \theta =\dfrac{4}{11}\times 360=\dfrac{1440}{11} \\

\end{align}\]

\[\theta =130\dfrac{10}{11}\]or 130.91.

Now, we have to convert it into degrees, minutes, seconds.

\[\begin{align}

& \therefore \theta ={{130}^{\circ }}+{{0.91}^{\circ }} \\

& ={{130}^{\circ }}+0.91\left( 60' \right) \\

\end{align}\]

\[={{130}^{\circ }}+54.6'\] (Conversion in degree and minutes)

\[={{130}^{\circ }}+54'+0.6\left( 60'' \right)\](Convert to seconds)

\[={{130}^{\circ }}+54'+36''\]

\[={{130}^{\circ }}54'36''\](Conversion in degree, minutes and seconds)

Hence, we found the angle of sector \[={{130.91}^{\circ }}\].

The angle of the sector expressed in degrees, minutes and seconds is, \[\theta ={{130}^{\circ }}54'36''\].

Note: The conversion of angles to degree, minutes and seconds are important concepts. First convert it to minutes which is (60’) and then to seconds (60’’). You need to convert it to degree, minutes and seconds if its asked in the question or else you can let the answer be \[\theta =130\dfrac{10}{11}\]or \[{{130.91}^{\circ }}\].

“Complete step-by-step answer:”

A sector of a circle is a portion of a disk enclosed by 2 radii and an arc. The smaller area will be known as a minor sector and larger as a major sector.

Let us assume the radius of the circle as ‘r’ and \[\theta \] be the angle of the circle.

The perimeter of a sector of the circle is given as, \[2r+\dfrac{\pi r.\theta }{360}\], which is the 2 radii and measure of the arc.

\[\therefore \]Perimeter of sector \[=r+r+\dfrac{\pi r.\theta }{360}=2r+\pi r\dfrac{\theta }{360}\]

The length of the arc of the semicircle is given by \[\pi r\].

We have been told that the perimeter of a certain sector of a circle is equal to the length of the arc of a semicircle.

\[\therefore \]Perimeter of sector of circle = length of arc of semi-circle.

By substituting the values we get,

\[2r+\pi r\dfrac{\theta }{360}=\pi r\] (Let us take, \[\pi =\dfrac{22}{7}\])

\[\begin{align}

& 2r+\dfrac{22}{7}\times r\dfrac{\theta }{360}=\dfrac{22}{7}r \\

& r\left[ 2+\dfrac{22}{7}\times \dfrac{\theta }{360} \right]=\dfrac{22}{7}r \\

\end{align}\]

Cancel out ‘r’ on both sides, we get

\[2+\dfrac{22}{7}\times \dfrac{\theta }{360}=\dfrac{22}{7}\]

\[\dfrac{22}{7}\times \dfrac{\theta }{360}=\dfrac{22}{7}-2\]

\[\begin{align}

& \dfrac{22}{7}\times \dfrac{\theta }{360}=\left( \dfrac{22-14}{7} \right) \\

& \dfrac{\theta }{360}=\dfrac{8}{7}\times \dfrac{7}{22} \\

& \therefore \dfrac{\theta }{360}=\dfrac{8}{22}=\dfrac{4}{11} \\

& \theta =\dfrac{4}{11}\times 360=\dfrac{1440}{11} \\

\end{align}\]

\[\theta =130\dfrac{10}{11}\]or 130.91.

Now, we have to convert it into degrees, minutes, seconds.

\[\begin{align}

& \therefore \theta ={{130}^{\circ }}+{{0.91}^{\circ }} \\

& ={{130}^{\circ }}+0.91\left( 60' \right) \\

\end{align}\]

\[={{130}^{\circ }}+54.6'\] (Conversion in degree and minutes)

\[={{130}^{\circ }}+54'+0.6\left( 60'' \right)\](Convert to seconds)

\[={{130}^{\circ }}+54'+36''\]

\[={{130}^{\circ }}54'36''\](Conversion in degree, minutes and seconds)

Hence, we found the angle of sector \[={{130.91}^{\circ }}\].

The angle of the sector expressed in degrees, minutes and seconds is, \[\theta ={{130}^{\circ }}54'36''\].

Note: The conversion of angles to degree, minutes and seconds are important concepts. First convert it to minutes which is (60’) and then to seconds (60’’). You need to convert it to degree, minutes and seconds if its asked in the question or else you can let the answer be \[\theta =130\dfrac{10}{11}\]or \[{{130.91}^{\circ }}\].

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