Calculate angle, given arc length and sector area
Problem: The length of arc of circle of radius $r$ which subtends an angle $\theta$ radians at the centre of the circle $r\theta$. The area of the sector formed by this arc and radii to either end of the arc is $\frac{1}{2}r^2\theta$. Find the angle $\theta$ if the magnitude of the arc length is $10\pi $ $cm$ and the area is $0.740\pi$ $cm^2$. (Source: Engineering Maths by Leslie Mustoe, Chapter 5, Exercise 5.1 on page 88.)
Result: $\frac{5\pi}{4}$ (According to the book.)
Question: My solution attempt yielded a very different number, and shame on me, but I do not understand what is wrong, in this seemingly simple calculation. So, could you please point out what is wrong in the below solution? (Please, don't tell me the right solution, as I want to arrive to it myself. I just want to understand where I'm making a mistake.)
Attempt:
$r\theta=10\pi$ $cm$ (1)
$\frac{1}{2}r^2\theta=0.74 \pi$ $cm^2$ (2)
Dividing (2) by (1):
$\frac{1}{2}r = \frac {0.74}{10}$ $cm$
$r = \frac{0.74}{5} $ $cm$ = $0.148$ $cm$.
From (1):
$\theta = \frac{10\pi}{r} = \frac{10\pi}{0.148} $
$\theta = 212.27 = 67.57 \pi$
Similar questions
--> This is basically the proof for a slightly different (degree-based) version of (2).
Find the Radius, Central Angle, and Arc Length Given Sector Area and Perimeter --> The answer gives me (1) and (2).
Central angle of a circular sector from area and arc length --> This is the same problem I have, with different numbers. Strange thing is, if I try to do it, I get the same (right) result as the OP.
$\endgroup$ 61 Answer
$\begingroup$I think that there are no doubts. Data are wrong. $\theta $ cannot be more than $2\pi$ and the picture below is the only possibility to "save" the exercise, otherwise is meaningless: a sector having perimeter $31$ cm and more CANNOT have an area of $2.3$ cm$^2$
Hope this is useful
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