Chapter 8.2, Problem 58E

To determine

To find: the total length of all the perpendiculars;

  $\left|CD\right|+\left|DE\right|+\left|EF\right|+\left|FG\right|+\mathrm{.......}$ in terms of b and $\theta$ .

Answer to Problem 58E

  $\left|CD\right|+\left|DE\right|+\left|EF\right|+\left|FG\right|+\mathrm{.......}=\frac{b(1+\cos \theta )}{\sin \theta }.$

Explanation of Solution

Given information: A right triangle ABC is given with $\angle A=\theta \text{and}\left|AC\right|=b$ . CD is

drawn perpendicular to AB ,DE is drawn perpendicular to BC, EF $\perp AB$ and this process is continued infinitely as shown below figure.

  

Calculation:

Because they are all perpendicular to the same segment, $\overline{AB}$ ,

  $\overline{CD}\parallel \overline{EF}\parallel \overline{GH}\parallel \mathrm{...........}$

Similarly, because they are all perpendicular to $\overline{BC}$

  $\overline{DE}\parallel \overline{FG}\parallel \mathrm{...........}$

Let $m\angle ACD=\beta$ , Because the sum of the interior angles of $△ACD={180}^{\circ }$

  $\begin{array}{c}m\angle ADC+m\angle ACD+m\angle CAD={180}^{\circ }\\ {90}^{\circ }+\beta +\theta ={180}^{\circ }.\\ \beta +\theta ={90}^{\circ }\end{array}$

Since $\begin{array}{c}m\angle BCA={90}^{\circ }\text{}\\ \text{}m\angle BCA=m\angle BCD+m\angle DCA,\\ {90}^{\circ }=m\angle BCD+\beta \\ m\angle BCD={90}^{\circ }-\beta \\ m\angle BCD=\theta \end{array}$

Therefore, by Angle-Angle similarity $△ACD\sim △CDE$ similarly,

  $△ABC\sim △ACD\sim △CDE\sim △DEF\sim △EFG\sim △FGH\mathrm{........}$

Because of similarity, find a scale factor s between $△ABC\text{and}△ACD$ .

  $\begin{array}{l}s=\frac{AD}{AC}\\ \cos \theta =\frac{AD}{AC}\\ AD=AC\cos \theta .\\ s=\frac{AC\cos \theta }{AC}=\cos \theta \end{array}$

Similarity, find another scale factor r between $△ACD\text{and}△CDE$ .

  $\begin{array}{l}r=\frac{CD}{DE}\\ \cos \theta =\frac{CD}{DE}\\ CD=DE\cos \theta .\\ r=\frac{DE\cos \theta }{DE}=\cos \theta \end{array}$

So,

  $\cos \theta =\frac{CD}{BC}=\frac{DE}{CD}=\frac{EF}{DE}=\mathrm{.......}$

By trigonometry

  $\begin{array}{l}\tan \theta =\frac{BC}{AC}.\\ BC=AC\tan \theta \\ BC=b\tan \theta \\ \cos \theta =\frac{CD}{BC}\\ CD=BC\cos \theta \\ \cos \theta =\frac{DE}{CD}\\ DE=CD\cos \theta .\\ DE=(BC\cos \theta )cos\theta \\ DE=BC{\cos }^2\theta .\\ \mathrm{....}\\ \mathrm{.....}\end{array}$

Therefore,

  $\begin{array}{c}\left|CD\right|+\left|DE\right|+\left|EF\right|+\left|FG\right|+\mathrm{.......}=\left|BC\cos \theta \right|+\left|BC{\cos }^2\theta \right|+\left|BC{\cos }^3\theta \right|+\mathrm{....}\\ =\left|(b\tan \theta )\cos \theta \right|+\left|(b\tan \theta ){\cos }^2\theta \right|+\left|b\tan \theta {\cos }^3\theta \right|+\mathrm{....}\\ =\frac{a}{1-r}.\\ =\frac{(b\tan \theta )\cos \theta }{1-\cos \theta }.\\ =\frac{(b\frac{\sin \theta }{\cos \theta })\cos \theta }{1-\cos \theta }.\\ =\frac{(b\sin \theta )}{1-\cos \theta }.\\ =\frac{(b\sin \theta )}{1-\cos \theta }(\frac{1+\cos \theta }{1+\cos \theta }).\\ =\frac{(b\sin \theta )(1+\cos \theta )}{1-{\cos }^2\theta }.\\ =\frac{(b\sin \theta )(1+\cos \theta )}{{\sin }^2\theta }.\\ \left|CD\right|+\left|DE\right|+\left|EF\right|+\left|FG\right|+\mathrm{.......}=\frac{b(1+\cos \theta )}{\sin \theta }.\end{array}$