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So this problem was on our textbook and it seems that the infos given are not enough.

A sector of a circle has an arc of 16.76 cm and with chord 15.43 cm. Find the area of the sector.

Note by すすべての すべての
7 months, 3 weeks ago

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You have \(r\theta = 16.76\) and by cosine rule, \(2r^2 - 2r^2 \cos\theta = 15.43^2\). So you have two equations and two variables, and you're instructed to find \(\dfrac12r^2\theta\). Pi Han Goh · 7 months, 3 weeks ago

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@Pi Han Goh im not good at solving variables involving theta すすべての すべての · 7 months, 3 weeks ago

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@すすべての すべての Do you know numerical/approximation methods? Pi Han Goh · 7 months, 3 weeks ago

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@Pi Han Goh don't know what is that. すすべての すべての · 7 months, 3 weeks ago

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@すすべての すべての Alternatively, you can use Newton's Method for estimating the value of y. Though there's something to note, newton's method would not work on certain functions. Julian Poon · 7 months, 3 weeks ago

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@Julian Poon That's not the best way to teach someone about numerical methods. You should teach them about bisection method first. Newton-Raphson is way too complicated especially for someone who don't know calculus yet. Pi Han Goh · 7 months, 3 weeks ago

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@すすべての すべての So you want to find \( \dfrac12 r^2 \theta = \dfrac12 \cdot \dfrac{(r \theta)^2}r = \dfrac{16.76^2}{2\theta} \).

From the second equation, you have \(2r^2(1-\cos\theta) =2r^2 \cdot 2\sin^2\left(\dfrac \theta2 \right) = 15.43^2 \Rightarrow 2r \sin \dfrac{\theta}2 = 15.43 \).

\[ (r \theta) \div \left(2r \sin\dfrac\theta2 \right) = \dfrac{16.76}{15.43} \Rightarrow \dfrac{\theta /2}{\sin(\theta /2)} = \dfrac{1676}{1543} \]

Let \(y = \dfrac\theta2 \), then you're left to solve for \(\dfrac y{\sin y} = \dfrac{1676}{1543} \approx 1.08619 \) for positive \(y\) only as \(\theta > 0\) as well.

Now for the numerical methods: Let \(f(y) = \dfrac y{\sin y} \), and we want to find the best estimate of \(y\) which gives \( f(y) \approx 1.08619 \).

Let's try for some value of \(y\):

y0.50.60.70.80.9
f(y)1.0429141.06261931.0865891.1152062561.1489455

Comparing all these values of \(f(y) \) against the value \(1.08619\), we can see that the closest value of \(y\) that approximates to \(f(y) \approx 1.08619\) is at \(y = 0.7 \). (Note that the more accurate reading is \(y = 0.698497366525694\ldots \))

Thus we can make a rough estimate that \(y \approx 0.7\) or \(\dfrac \theta2 \approx 0.7 \Rightarrow \theta \approx 1.4 \).

Our desired answer is approximately \( \dfrac{16.76^2}{2\times1.4} = 100\).

Note that the actual value is \(100.536\ldots \). Pi Han Goh · 7 months, 3 weeks ago

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@Pi Han Goh thank you for the solution, my bad actually it is the area of the segment that I want to find and was needed (not sector). I used those two equations \(2r^2-2r^2\cos\theta=15.43^2\) and \(r\theta=16.76\) and I came up with \(\large \frac{1-cos\theta}{\theta^2}=a\) where \(\theta\) is in radian. (I forgot what the value of \(a\) was) and used approximation. Result gives me \(90>\theta>70\) すすべての すべての · 7 months, 3 weeks ago

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@すすべての すべての Your value of \(\theta\) is measured in degrees, whereas my value of \(\theta\) is measured in radians. \(1.4 \text{ radians} \approx 80^\circ \). Pi Han Goh · 7 months, 3 weeks ago

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@Pi Han Goh Yes. The \(\theta\) in the latter part of my statement was in degrees. In radian that would be \(\large \frac{\pi}{2}>\theta>\frac{7\pi}{18}\) すすべての すべての · 7 months, 3 weeks ago

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@すすべての すべての Yes, that's what I'm saying, the value you've found matches mine! Pi Han Goh · 7 months, 3 weeks ago

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