Find Central Angle: Newton's Method

[karush]

In the figure, the length of the chord $AB$ is $4 \text { cm}$ and the length of the arc is $5\text{ cm}$
https://www.physicsforums.com/attachments/1713

(a) Find the central angle $\theta$, in radians, correct to four decimal places.

(b) Give the answer to the nearest degreethis problem is intended to be solved by Newton's Method, so I am have ?? as to how to set it up. I thot that using law of cosines would be part of it since

$$\cos{\theta} = \frac{4^2}{a^2+b^2-2ab}$$

or since $a=b=r$

$$\cos{\theta} = \frac{16}{2r^2-1}$$
and
$$\cos^{-1}{\left(\frac{16}{2r^2-1}\right)}=\theta$$

and also $$S=\theta\cdot r$$ for arc length

so this is where I have a flat tire without a spare...

 

[MarkFL]

The Law of Cosines gives you:

\(\displaystyle 4^2=r^2+r^2-2r^2\cos(\theta)\)

\(\displaystyle 8=r^2\left(1-\cos(\theta) \right)\)

Form the arc-length, we find:

\(\displaystyle 5=r\theta\implies r=\frac{5}{\theta}\)

And so, by substitution, we obtain:

\(\displaystyle 8=\left(\frac{5}{\theta} \right)^2\left(1-\cos(\theta) \right)\)

\(\displaystyle f(\theta)=8\theta^2+25\left(\cos(\theta)-1 \right)=0\)

Now use Newton's method to find the root for which \(\displaystyle 0<\theta\).

 

[karush]

View attachment 1724

(a) $2.2622$ rad
(b) $130^o$

I hope this got it...

 

[MarkFL]

I would have looked at a graph of the function:

View attachment 1726

From this we see the root we seek is about \(\displaystyle x\approx2.25\).

Newton's method gives us the recursive algorithm:

\(\displaystyle x_{n+1}=x_n-\frac{8x_n^2+25\left(\cos\left(x_n-1 \right) \right)}{16x_n-25\sin\left(x_n \right)}\)

Now, on my TI-89, I enter the following:

Input	Output
2.25 [ENTER]	2.25
ans(1)-(8ans(1)^2+25(cos(ans(1))-1))/(16ans(1)-25sin(ans(1))) [ENTER]	2.26234822745
[ENTER]	2.2622051906
[ENTER]	2.2622051713
[ENTER]	2.2622051713

And in degrees, this is about \(\displaystyle 129.614808708^{\circ}\).

 

[karush]

well that a good thing to know, tried on my TI-nspire cx cas and got the same thing.
good use of the ans feature.

 

[MarkFL]

well that a good thing to know, tried on my TI-nspire cx cas and got the same thing.
good use of the ans feature.

Yes, that function allows for easy evaluation of the terms in a recursive algorithm. For example, you can get the Fibonacci numbers with:

0 [ENTER]
1 [ENTER]
ans(1)+ans(2) [ENTER]
Now press [ENTER] for each new number in the sequence. :D