Question

In a circle of 5 feet radius, what is the length of the arc which subtends an angle of ${33}^{\text{o}}{15}^{\prime }$ at the centre?

Answer 1

Hint: We will apply the formula to find the length of an arc. The formula is given by $L={\displaystyle \frac{\theta }{{360}^{\text{o}}}}\times 2\pi R$, where R is the radius of the circle and $\theta$ is the angle that is measured for the arc of the circle.

Complete step-by-step answer:
The required diagram for the question is shown below.

Now, first we consider the angle which is given to us as ${33}^{\text{o}}{15}^{\prime }$. Here, we will apply the formula to convert minutes into degrees as ${\left(1\right)}^{\text{o}}={60}^{\prime }$ or 1 minute is divided into 1 by 60 degrees. Numerically, this is represented as ${\left(1\right)}^{\prime }={\left({\displaystyle \frac{1}{60}}\right)}^{\text{o}}$. As we can write ${33}^{\text{o}}{15}^{\prime }$ as ${33}^{\text{o}}{15}^{\prime }={33}^{\text{o}}+{15}^{\prime }$. Therefore, after substituting the ${\left(1\right)}^{\prime }={\left({\displaystyle \frac{1}{60}}\right)}^{\text{o}}$ in ${33}^{\text{o}}{15}^{\prime }={33}^{\text{o}}+{15}^{\prime }$ we will have,
$\begin{array}{rl}& {33}^{\text{o}}{15}^{\prime }={33}^{\text{o}}+{15}^{\prime }\\ & \Rightarrow {33}^{\text{o}}{15}^{\prime }={33}^{\text{o}}+{(15\times {\displaystyle \frac{1}{60}})}^{\text{o}}\\ & \Rightarrow {33}^{\text{o}}{15}^{\prime }={(33+{\displaystyle \frac{15}{60}})}^{\text{o}}\\ & \Rightarrow {33}^{\text{o}}{15}^{\prime }={\left({\displaystyle \frac{1980+15}{60}}\right)}^{\text{o}}\\ & \Rightarrow {33}^{\text{o}}{15}^{\prime }={\left({\displaystyle \frac{1995}{60}}\right)}^{\text{o}}\\ & \Rightarrow {33}^{\text{o}}{15}^{\prime }={\left({\displaystyle \frac{133}{4}}\right)}^{\text{o}}\end{array}$
Now, we have degree of the circle as $\theta ={\left({\displaystyle \frac{133}{4}}\right)}^{\text{o}}$ and we are already given the value of radius of the circle as R = 5 feet. Therefore, we can now apply the formula to find the length of an arc. The formula is given by $L={\displaystyle \frac{\theta }{{360}^{\text{o}}}}\times 2\pi R$, where R is the radius of the circle and $\theta$ is the angle that is measured for the arc of the circle.
Therefore, the length of the arc is given by,
$\begin{array}{rl}& L={\displaystyle \frac{\theta }{{360}^{\text{o}}}}\times 2\pi R\\ & \Rightarrow L={\displaystyle \frac{{\left({\displaystyle \frac{133}{4}}\right)}^{\text{o}}}{{360}^{\text{o}}}}\times 2\pi 5\\ & \Rightarrow L={\displaystyle \frac{{\left(133\right)}^{\text{o}}}{{360}^{\text{o}}\times 4^{\text{o}}}}\times 2\pi 5\\ & \Rightarrow L={\displaystyle \frac{133}{72\times 2}}\times \pi \\ & \Rightarrow L={\displaystyle \frac{133}{144}}\times \pi \\ & \Rightarrow L={\displaystyle \frac{133}{144}}\pi \end{array}$
After this we will substitute $\pi ={\displaystyle \frac{22}{7}}$ in $L={\displaystyle \frac{133}{144}}\pi$. Therefore, we get
$\begin{array}{rl}& L={\displaystyle \frac{133}{144}}\pi \\ & \Rightarrow L={\displaystyle \frac{133}{144}}\times {\displaystyle \frac{22}{7}}\\ & \Rightarrow L={\displaystyle \frac{19}{72}}\times {\displaystyle \frac{11}{1}}\\ & \Rightarrow L={\displaystyle \frac{209}{72}}\end{array}$
Hence, the length of the arc of the circle is given by $L={\displaystyle \frac{209}{72}}$.

Note: If we don’t want our answer in fraction then we will not stop with $L={\displaystyle \frac{209}{72}}$. We will convert it into decimals after dividing the numerator by denominator. Thus, we get $L=2.90$ which is an approximate value. We could have also substituted $\pi =3.14$ instead of $\pi ={\displaystyle \frac{22}{7}}$ in $L={\displaystyle \frac{133}{144}}\pi$. Therefore, we get
$\begin{array}{rl}& L={\displaystyle \frac{133}{144}}\pi \\ & \Rightarrow L={\displaystyle \frac{133}{144}}\times 3.14\\ & \Rightarrow L=0.92\times 3.14\\ & \Rightarrow L=2.8888\end{array}$