Chapter 4.7, Problem 32E

To determine

To calculate: The central angle $\theta$ in radians.

Answer to Problem 32E

The angle’s are

  ${\theta }_1\approx 2.2$

  ${\theta }_2\approx 2.2662$

  ${\theta }_3\approx 2.2622$

Explanation of Solution

Given information:

The length of the chord AB is 4 cm.

The length of the arc AB is 5cm.

Formula used:

Newton’s Method:

We seek a solution of $f(x)=0$ , starting from an initial estimate $x=x_1$ .

For $x=x_n$ , compute the next approximation $x_{n+1}$ by

  $x_{n+1}=x_n-\frac{f(x_n)}{f\text{'}(x_n)}$ and so on.

Formula: $s=r\theta$ where s is arc length, r is the radius of the circle and $\theta$ is the central angle.

Law of cosine $c^2=a^2+b^2-2ab\cos \theta$where a , b , c are sides of triangle .

Calculation:

Consider the,

Length of the chord is 4cm.

Length of the arc is 5 cm.

Therefore, $s=5$

  $\begin{array}{l}=s=r\theta \\ =5=r\theta \\ =r=\frac{5}{\theta }\end{array}$ (i)

Now , using law of cosine we have ,

  $\begin{array}{l}c=4\\ a=r\\ b=r\end{array}$ .

  $\begin{array}{l}$ ={(4)}^2={(r)}^2+{(r)}^2-2(r)(r)\cos \theta $=16=r^2+r^2-2r^2\cos \theta \\ =16=2r^2-2r^2\cos \theta \\ =16=2r^2(1-\cos \theta )\end{array}$ (ii)

Putting (i) in (ii) we get,

  $\begin{array}{l}16=2{\left[\frac{5}{\theta }\right]}^2(1-\cos \theta )\\ 16=2\left(\frac{25}{{\theta }^2}\right)(1-\cos \theta )\\ 16{\theta }^2=50(1-\cos \theta )\\ 16{\theta }^2=50-50\cos \theta \\ 16{\theta }^2-50+50\cos \theta =0\end{array}$

Hence, we get

  $\begin{array}{l}g(\theta )=16{\theta }^2-50+50\cos \theta \\ g\text{'}(\theta )=32\theta -50\sin \theta \end{array}$

Now, let initial approximation be ${\theta }_1=2.2$

For $n=1$

  ${\theta }_2={\theta }_1-\frac{g({\theta }_1)}{g\text{'}({\theta }_1)}$

  ${\theta }_2={\theta }_1-\frac{16{\theta }^2{}_1-50+50\cos {\theta }_1}{32{\theta }_1-50\sin {\theta }_1}$

  $\begin{array}{l}{\theta }_2=(2.2)-\frac{16{(2.2)}^2-50+50\cos (2.2)}{32(2.2)-50\sin (2.2)}\\ {\theta }_2=(2.2)-\frac{77.44-50-29.42505586}{70.4-40.42482019}\end{array}$

  $\begin{array}{l}{\theta }_2=2.2-\frac{(-1.98505586)}{29.97517981}\\ {\theta }_2=2.2+\frac{1.98505586}{29.97517981}\\ {\theta }_2=2.2+0.066223317\\ {\theta }_2=2.266223318\end{array}$

The second approximation is ${\theta }_2=2.2662$ .

For $n=2$

  ${\theta }_3={\theta }_2-\frac{g({\theta }_2)}{g\text{'}({\theta }_2)}$

  ${\theta }_3={\theta }_2-\frac{16{\theta }_2^2-50+50\cos {\theta }_2}{32{\theta }_2-50\sin {\theta }_2}$

  $\begin{array}{l}{\theta }_3=(2.2662)-\frac{16{(2.2662)}^2-50+50\cos (2.2662)}{32(2.2662)-50\sin (2.2662)}\\ {\theta }_3=(2.2662)-\frac{82.17059904-50-32.0347715}{72.5184-38.38975664}\end{array}$

  $\begin{array}{l}{\theta }_3=2.2662-\frac{(0.13582754)}{34.12864336}\\ {\theta }_3=2.2662-0.003979869301\\ {\theta }_3=2.262220131\end{array}$

The third approximation is ${\theta }_3=2.2622$ .

Hence , the central angle in radians are :-

  ${\theta }_1\approx 2.2$

  ${\theta }_2\approx 2.2662$

  ${\theta }_3\approx 2.2622$