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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 $\dfrac12r^2\theta$.

@Phak Mi Uph
–
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 $2r^2(1-\cos\theta) =2r^2 \cdot 2\sin^2\left(\dfrac \theta2 \right) = 15.43^2 \Rightarrow 2r \sin \dfrac{\theta}2 = 15.43$.

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 $y$:

y

0.5

0.6

0.7

0.8

0.9

f(y)

1.042914

1.0626193

1.086589

1.115206256

1.1489455

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 $\dfrac{16.76^2}{2\times1.4} = 100$.

@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 $90>\theta>70$

@Phak Mi Uph
–
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
–
Yes. The $\theta$ in the latter part of my statement was in degrees. In radian that would be $\large \frac{\pi}{2}>\theta>\frac{7\pi}{18}$

@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
–
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.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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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 $\dfrac12r^2\theta$.

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im not good at solving variables involving theta

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Do you know numerical/approximation methods?

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$\dfrac12 r^2 \theta = \dfrac12 \cdot \dfrac{(r \theta)^2}r = \dfrac{16.76^2}{2\theta}$.

So you want to findFrom 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 $100.536\ldots$.

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$2r^2-2r^2\cos\theta=15.43^2$ 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 $90>\theta>70$

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 equationsLog in to reply

$\theta$ is measured in degrees, whereas my value of $\theta$ is measured in radians. $1.4 \text{ radians} \approx 80^\circ$.

Your value ofLog in to reply

$\theta$ in the latter part of my statement was in degrees. In radian that would be $\large \frac{\pi}{2}>\theta>\frac{7\pi}{18}$

Yes. TheLog in to reply

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