# [SOLVED]Find the central angle

#### karush

##### Well-known member
In the figure, the length of the chord $AB$ is $4 \text { cm}$ and the length of the arc is $5\text{ cm}$
View attachment 1713

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

(b) Give the answer to the nearest degree

this 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....

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#### MarkFL

Staff member
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 $$\displaystyle 0<\theta$$.

#### MarkFL

Staff member
I would have looked at a graph of the function:

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-known member
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.