Home fsc 1st year math solutions Arc length formula 15 Solved Questions

Arc length formula 15 Solved Questions

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Arc Length Formula
Arc Length Formula

The arc length formula is a fundamental concept in mathematics, specifically in geometry and trigonometry. It plays a crucial role in various fields, including physics, engineering, and computer graphics. This formula allows us to calculate the length of a curved segment or arc on a circle or a curve defined by a function. By comprehending the arc length formula, we gain powerful tools to analyze and measure curved shapes accurately. In this comprehensive guide, we will delve into the details of the arc length formula and its applications.

The arc length formula is a fundamental concept in mathematics, specifically in geometry and trigonometry. It plays a crucial role in various fields, including physics, engineering, and computer graphics. This formula allows us to calculate the length of a curved segment or arc on a circle or a curve defined by a function. By comprehending the arc length formula, we gain powerful tools to analyze and measure curved shapes accurately. In this comprehensive guide, we will delve into the details of the arc length formula and its applications.

The arc length formula is a fundamental concept in mathematics, specifically in geometry and trigonometry. It plays a crucial role in various fields, including physics, engineering, and computer graphics. This formula allows us to calculate the length of a curved segment or arc on a circle or a curve defined by a function. By comprehending the arc length formula, we gain powerful tools to analyze and measure curved shapes accurately. In this comprehensive guide, we will delve into the details of the arc length formula and its applications.

  1. Definition of Arc Length: Arc length refers to the distance along a curved line or a portion of a circle, measured from one endpoint to another. The arc length formula provides a precise way to determine this length, overcoming the challenges of measuring a non-linear path.
  2. Basic Arc Length Formula: For a circle of radius ‘r’ and a central angle ‘θ‘ (in radians), the arc length ‘s’ can be calculated using the basic arc length formula:

        \begin{equation*} l = r \cdot \theta \end{equation*}

  3. Converting Degrees to Radians: Since the arc length formula requires angles in radians, we often need to convert degrees to radians using the formula: \text{radians} = \left(\frac{\pi}{180}\right) \cdot \text{degrees}.
  4. Practical Applications: The arc length formula finds extensive applications in real-world scenarios such as measuring the length of curved tracks, calculating the distance traveled by moving objects along curves, and determining the length of wires used in construction.

Solved Questions with Arc length formula

Question 03: What is the circular measure of angle between the hands of a watch at 4 ‘O clock?

Circular measure of angle for one complete revolution =2 \pi. Circular measure of angle between hands of a watch at 4 ‘ \mathrm{O}

    \[\begin{aligned} \text { clock } & =2 \pi \times \frac{4}{12} \\ & =\frac{2 \pi}{3} \text { radians } \end{aligned}\]

What is the circular measure of angle between the hands of a watch at 4 'O clock?

Question 04: Find \theta when
\text { (i) } \quad l=1.5 \mathrm{~cm} \quad r=2.5 \mathrm{~cm}
\text { (ii) } l=3.2 \mathrm{~m} \quad \mathrm{r}=2 \mathrm{~m}

(i)

l={1 . 5 \mathrm { cm }} \quad {r}=2.5 \mathrm{~cm}
we Know 1=\mathrm{r} \theta
Now \theta =\frac{l}{\mathrm{r}}
\theta  =\frac{1.5}{2.5}=0.6 \text { radian. }

(ii)

l={3.2 \mathrm { cm }} \quad {r}=2 \mathrm{~cm}
we Know 1=\mathrm{r} \theta
Now \theta =\frac{l}{\mathrm{r}}
\theta  =\frac{3.2}{2}=1.2 \text { radian. }

Question 05: Find l when
(i) \theta=\pi \text { radian } r=6 \mathrm{~cm}
(ii) \theta=65^{\circ} 20^{\prime} \quad r=18 \mathrm{~mm}

(i)

we Know 1=\mathrm{r} \theta
and l=6 \times \pi =6 \times 3.1416=18.85 \mathrm{~cm}

(ii)

\theta=65^{\circ}+\frac{20^{\circ}}{60}=\left(65+\frac{20}{60}\right)^{\circ}=\frac{196^{\circ}}{3}
\theta=\frac{196}{3} \times \frac{\pi}{180} \text { radian }=\frac{49}{3 \times 45}(3.1415)
l  =\mathrm{r} \theta
l=18 \times \frac{49}{3 \times 45}(3.1415)=20.53 \mathrm{~mm}

Question 06: Find r when
(i) l=5 \mathrm{~cm} ,\quad \theta=\frac{1}{2} \text { radian }
(ii) l=56 \mathrm{~cm} \quad \theta=45^{\circ}

(i) l=5 \mathrm{~cm} \quad \theta=\frac{1}{2} radian
We know that l=\mathrm{r} \theta
\mathrm{r}=\frac{l}{\theta}=\frac{5}{\frac{1}{2}}=10 \mathrm{~cm}
(ii) l=56 \mathrm{~cm} \quad \theta=45^{\circ}

Now
\theta=45^{\circ}= 45 \times \frac{\pi}{180} radian
we know that l=\mathrm{r} \theta
\mathrm{r}=\frac{l}{\theta} = \frac{56}{45} \times \frac{\pi}{180} radian
=\frac{56}{0.785}=71.33 \mathrm{~cm}

Question 07: What is the length of the arc intercepted on a circle of radius 14 \mathrm{~cm} by the arms of central angle of 45^{\circ} ?

    \[\begin{aligned} & l=? \quad \mathrm{r}=14 \mathrm{~cm} \quad \theta=45^{\circ} \\ & \theta=45 \times \frac{\pi}{180} \text { radian } \\ & \theta=0.785 \end{aligned}\]

we know that l=\mathrm{r} \theta
l=14 \times 0.785=10.99 or 11 \mathrm{~cm} approximately.

Question 08: Find the radius of the circle, in which the arms of a central angle of measure 1 radian cut off an arc of length 35 \mathrm{~cm}.

\mathrm{r}=? \quad \theta=1 \text { radian } \quad l=35 \mathrm{~cm}
we know that l=\mathrm{r} \theta
\mathrm{r}=\frac{l}{\theta}=\frac{35}{1}=35 \mathrm{~cm}

Question 09: Railway train is running on a circular track of radius 500 meters at the rate of 30km/h. Through what angle will it turn in 10sec.

\theta=?, \quad \mathrm{r}=500 \mathrm{~m}, \quad Speed =30 \mathrm{~km} / \mathrm{h}
Speed =\frac{30 \times 1000}{3600}=\frac{25}{3} \mathrm{~m} / \mathrm{sec}
Distance covered in 10 \mathrm{sec}
Distance = speed \times time
l=\frac{25}{3} \times 10 \mathrm{~m}=\frac{250}{3} \mathrm{~m}
We know that l=\mathrm{r} \theta
\theta =\frac{l}{\mathrm{r}}
\theta=\frac{250}{3} \times \frac{1}{500}=\frac{1}{6} \text { radian } Railway train is running on a circular track of radius 500 meters at the rate of 30km/h. Through what angle will it turn in 10sec.

Question 10: A horse is tethered to a peg by a rope of 9m in length and it can move in a circle with the peg as centre. If the horse moves along the circumference of the circle, keeping the rope tight, how far will it have gone when the rope has turned through an angle of 70^{\circ}

135^{\circ}=135^{\circ} \times \frac{\pi}{180} radian =\frac{3 \pi}{4} radian

Question 11: The pendulum of a clock is 20cm long and it swings through an angle of 20^{\circ} Solution: each second. How far does the tip of the pendulum moves in 1 second?

\mathrm{r}=20 \mathrm{~cm}, \quad \theta=20^{\circ}=20 \times \frac{\pi}{180} radian, l=?
we know that l=\mathrm{r} \theta
l=20 \times 20 \times \frac{3.1416}{180}
l=6.89 \mathrm{~cm}
The pendulum of a clock is 20cm long and it swings through an angle of 20 degree

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