How to solve this?

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 Phak Mi Uph
3 years, 8 months ago

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You have rθ=16.76r\theta = 16.76 and by cosine rule, 2r22r2cosθ=15.4322r^2 - 2r^2 \cos\theta = 15.43^2. So you have two equations and two variables, and you're instructed to find 12r2θ\dfrac12r^2\theta.

Pi Han Goh - 3 years, 8 months ago

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im not good at solving variables involving theta

Phak Mi Uph - 3 years, 8 months ago

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Do you know numerical/approximation methods?

Pi Han Goh - 3 years, 8 months ago

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@Pi Han Goh don't know what is that.

Phak Mi Uph - 3 years, 8 months ago

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@Phak Mi Uph So you want to find 12r2θ=12(rθ)2r=16.7622θ \dfrac12 r^2 \theta = \dfrac12 \cdot \dfrac{(r \theta)^2}r = \dfrac{16.76^2}{2\theta} .

From the second equation, you have 2r2(1cosθ)=2r22sin2(θ2)=15.4322rsinθ2=15.432r^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 .

(rθ)÷(2rsinθ2)=16.7615.43θ/2sin(θ/2)=16761543 (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=θ2y = \dfrac\theta2 , then you're left to solve for ysiny=167615431.08619\dfrac y{\sin y} = \dfrac{1676}{1543} \approx 1.08619 for positive yy only as θ>0\theta > 0 as well.

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

Let's try for some value of yy:

y0.50.60.70.80.9
f(y)1.0429141.06261931.0865891.1152062561.1489455

Comparing all these values of f(y)f(y) against the value 1.086191.08619, we can see that the closest value of yy that approximates to f(y)1.08619f(y) \approx 1.08619 is at y=0.7y = 0.7 . (Note that the more accurate reading is y=0.698497366525694y = 0.698497366525694\ldots )

Thus we can make a rough estimate that y0.7y \approx 0.7 or θ20.7θ1.4\dfrac \theta2 \approx 0.7 \Rightarrow \theta \approx 1.4 .

Our desired answer is approximately 16.7622×1.4=100 \dfrac{16.76^2}{2\times1.4} = 100.

Note that the actual value is 100.536100.536\ldots .

Pi Han Goh - 3 years, 8 months 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 2r22r2cosθ=15.4322r^2-2r^2\cos\theta=15.43^2 and rθ=16.76r\theta=16.76 and I came up with 1cosθθ2=a\large \frac{1-cos\theta}{\theta^2}=a where θ\theta is in radian. (I forgot what the value of aa was) and used approximation. Result gives me 90>θ>7090>\theta>70

Phak Mi Uph - 3 years, 8 months ago

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@Phak Mi Uph Your value of θ\theta is measured in degrees, whereas my value of θ\theta is measured in radians. 1.4 radians801.4 \text{ radians} \approx 80^\circ .

Pi Han Goh - 3 years, 8 months ago

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@Pi Han Goh Yes. The θ\theta in the latter part of my statement was in degrees. In radian that would be π2>θ>7π18\large \frac{\pi}{2}>\theta>\frac{7\pi}{18}

Phak Mi Uph - 3 years, 8 months ago

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@Phak Mi Uph Yes, that's what I'm saying, the value you've found matches mine!

Pi Han Goh - 3 years, 8 months ago

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@Phak Mi Uph 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 - 3 years, 8 months 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 - 3 years, 8 months ago

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