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.

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.

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TopNewestYou 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 · 10 months ago

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– すすべての すべての · 10 months ago

im not good at solving variables involving thetaLog in to reply

– Pi Han Goh · 10 months ago

Do you know numerical/approximation methods?Log in to reply

– すすべての すべての · 10 months ago

don't know what is that.Log in to reply

– Julian Poon · 10 months ago

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.Log in to reply

– Pi Han Goh · 10 months ago

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.Log in to reply

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 \(y\):

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 · 10 months ago

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– すすべての すべての · 10 months ago

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 \(90>\theta>70\)Log in to reply

– Pi Han Goh · 10 months ago

Your value of \(\theta\) is measured in degrees, whereas my value of \(\theta\) is measured in radians. \(1.4 \text{ radians} \approx 80^\circ \).Log in to reply

– すすべての すべての · 10 months ago

Yes. The \(\theta\) in the latter part of my statement was in degrees. In radian that would be \(\large \frac{\pi}{2}>\theta>\frac{7\pi}{18}\)Log in to reply

– Pi Han Goh · 10 months ago

Yes, that's what I'm saying, the value you've found matches mine!Log in to reply