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.

#Geometry

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Comments

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 - 5 years, 4 months ago

im not good at solving variables involving theta

Phak Mi Uph - 5 years, 4 months ago

Do you know numerical/approximation methods?

Pi Han Goh - 5 years, 4 months ago

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

Phak Mi Uph - 5 years, 4 months ago

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

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

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 $y$ :

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

Note that the actual value is $100.536\ldots$ .

Pi Han Goh - 5 years, 4 months ago

@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 - 5 years, 4 months ago

@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 - 5 years, 4 months ago

@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 - 5 years, 4 months ago

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

Pi Han Goh - 5 years, 4 months ago

@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 - 5 years, 4 months ago

@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 - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$