If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples on the Lengths of Arcs

Samuel Dominic Chukwuemeka (SamDom For Peace) Verify your answers with these Calculators as applicable.

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
State the formulas and reasons for applicable steps
Negative angles are not allowed
Show all work
Except stated otherwise, use:

$ d = diameter \\[3ex] r = radius \\[3ex] L = arc\:\:length \\[3ex] A = area\;\;of\;\;sector \\[3ex] \theta = central\:\:angle \\[3ex] \pi = \dfrac{22}{7} \\[5ex] RAD = radians \\[3ex] ^\circ = DEG = degrees \\[5ex] \underline{\theta\;\;in\;\;DEG} \\[3ex] L = \dfrac{2\pi r\theta}{360} \\[5ex] \theta = \dfrac{180L}{\pi r} \\[5ex] r = \dfrac{180L}{\pi \theta} \\[5ex] \underline{\theta\;\;in\;\;RAD} \\[3ex] L = r\theta \\[5ex] \theta = \dfrac{L}{r} \\[5ex] r = \dfrac{L}{\theta} \\[5ex] Circumference\:\:of\:\:a\:\:circle = 2\pi r = \pi d \\[3ex] L = \dfrac{2A}{r} \\[5ex] r = \dfrac{2A}{L} \\[5ex] A = \dfrac{Lr}{2} $

(1.) Determine the radian measure of an angle in a circle when a point $3\dfrac{1}{5}$ feet from the center of rotation travels $4$ feet


This implies that:

$ r = 3\dfrac{1}{5} = \dfrac{16}{5} \\[5ex] L = 4 \\[3ex] \theta = \dfrac{L}{r} \\[5ex] \theta = 4 \div \dfrac{16}{5} \\[5ex] \theta = 4 * \dfrac{5}{16} \\[5ex] \theta = \dfrac{4 * 5}{16} \\[5ex] \theta = \dfrac{20}{16} = \dfrac{5}{4} RAD $
(2.) The outside diameter of the tire of a $2017$ Buggati has a diameter of $37$ inches.
Through what angle (in radians) does the tire turn while traveling $1$ mile
Round to the nearest hundredth as needed.


We need to have the same unit for the distance/length of arc and the radius.
This implies that:

$ 1\:\:mile = 63360\:\:inches\\[3ex] d = 37 \:\:inches \\[3ex] r = \dfrac{d}{2} = \dfrac{37}{2} \:\:inches \\[5ex] L = 1\:\:mile = 63360\:\:inches\\[3ex] \theta = \dfrac{L}{r} \\[5ex] \theta = 63360 \div \dfrac{37}{2} \\[5ex] \theta = 63360 * \dfrac{2}{37} \\[5ex] \theta = \dfrac{63360 * 2}{37} \\[5ex] \theta = \dfrac{126720}{37} \approx 3424.86 RAD $
(3.) The radius of the earth is approximately $6400$ kilometers.
How many kilometers is $12^\circ$ of latitude?
Round to the nearest tenth as needed.


$ L = \dfrac{2\pi r\theta}{360} ...\theta \:\:in\:\:DEG \\[5ex] \pi = \dfrac{22}{7} \\[5ex] r = 6400 km \\[3ex] \theta = 12^\circ \\[3ex] L = \left(2 * \dfrac{22}{7} * 6400 * 12\right) \div 360 \\[5ex] L = 2 * \dfrac{22}{7} * 6400 * 12 * \dfrac{1}{360} \\[5ex] L = \dfrac{2 * 22 * 6400 * 12}{7 * 360} \\[5ex] L = \dfrac{3379200}{2520} \\[5ex] L = 1340.952381 \\[3ex] L \approx 1341 \:km $
(4.) Through how many radians does the minute hand of a clock rotate from $12:40\:PM$ to $1:25\:PM$?


$ From\:\:12:40\:PM \:\:to\:\: 1:25\:PM = 45\:\:minutes \\[3ex] 1\:\:complete\:\:rotation = 1\:\:hour = 60\:\:minutes \\[3ex] L = 2\pi r \:\:for\:\: 1\:\:hour \\[3ex] \rightarrow 60\:\:minutes \rightarrow 2\pi r \\[3ex] \therefore 45\:\:minutes = \dfrac{45 * 2\pi r}{60} = \dfrac{3\pi r}{2} \\[5ex] \rightarrow L = \dfrac{3\pi r}{2} \\[5ex] r = r ...not\:\:given \\[3ex] \theta = \dfrac{L}{r} \\[5ex] \theta = \dfrac{3\pi r}{2} \div r \\[5ex] \theta = \dfrac{3\pi r}{2} * \dfrac{1}{r} \\[5ex] \theta = \dfrac{3\pi}{2} $
(5.) WASSCE An arc of a circle of radius 7.5 cm is 7.5 cm long.
Find, correct to the nearest degree, the angle which the arc subtends at the centre of the circle.

$ \left[Take\;\;\pi = \dfrac{22}{7}\right] \\[5ex] A.\;\; 29^\circ \\[3ex] B.\;\; 57^\circ \\[3ex] C.\;\; 65^\circ \\[3ex] D.\;\; 115^\circ \\[3ex] $

$ \theta \\[3ex] = \dfrac{180L}{\pi r} \\[5ex] = (180 * L) \div (\pi * r) \\[3ex] = (180 * 7.5) \div \left(\dfrac{22}{7} * 7.5\right) \\[5ex] = 1350 \div \dfrac{165}{7} \\[5ex] = 1350 * \dfrac{7}{165} \\[5ex] = \dfrac{9450}{165} \\[5ex] = 57.27272727 \\[3ex] \approx 57^\circ $
(6.) JAMB The chord $ST$ of a circle is equal to the radius, $r$, of the circle.
Find the length of arc $ST$

$ A.\;\; \dfrac{\pi r}{6} \\[5ex] B.\;\; \dfrac{\pi r}{2} \\[5ex] C.\;\; \dfrac{\pi r}{12} \\[5ex] D.\;\; \dfrac{\pi r}{3} \\[5ex] $

Let us draw the diagram
Number 6

$ Chord\;\;ST = r ...based\;\;on\;\;the\;\;question \\[3ex] \implies \triangle OST \;\;is\;\;an\;\;equilateral\;\;\triangle \\[3ex] \implies \theta = 60^\circ = \dfrac{\pi}{3} \\[5ex] \therefore Arc\;\;ST = r\theta \\[3ex] Arc\;\;ST = r * \dfrac{\pi}{3} \\[5ex] Arc\;\;ST = \dfrac{\pi r}{3} $