Chapter 8.3, Problem 54AYU

To determine

To show: That the given distance formula between the given points in the figure.

Answer to Problem 54AYU

  $x=r\cos \theta +\sqrt{r^2{\cos }^2\theta +L^2-r^2}$

Explanation of Solution

Given information:

The given figure is,

  

Concept used:

Law of Cosines:

If ABC is a triangle with sides $a$ , $b$ , and $c$ , then

  $\begin{array}{l}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos C=\frac{a^2+b^2-c^2}{2ab}\end{array}$

The given figure is,

  

The distance between point O and Bfrom triangle OABis calculated as,

  $\begin{array}{l}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos \theta =\frac{x^2+r^2-L^2}{2xr}\\ x^2-2xr\cos \theta +\left(r^2-L^2\right)=0\\ x=\frac{2r\cos \theta \pm \sqrt{{\left(-2r\cos \theta \right)}^2-4\left(1\right)\left(r^2-L^2\right)}}{2\left(1\right)}\end{array}$

Simplify further,

  $x=r\cos \theta +\sqrt{r^2{\cos }^2\theta +L^2-r^2}$

Hence, given equation is proved.