Question

A right triangle \(ABC\)is given with\(\angle A = \theta \)and\(|AC| = b\).\(CD\)is drawn perpendicular to\(AB,DE\)is drawn perpendicular to\(BC,EF \bot AB\),and this process is continued indefinitely as shown in the figure. Find the total length of all the perpendiculars.

\(\left| {CD} \right| + \left| {DE} \right| + \left| {EF} \right| + \left| {FG} \right| + ....\)

In terms of\(b\)and\(\theta \).

Short answer

The total length of all the perpendiculars is \(\frac{b}{{1 - \sin \theta }}\).

Step-by-step solution

Determine the values

Here, we have to find the sum of the sequence of perpendiculars of the given figure.

We have a side value \(b\) and angle \(\theta \).

We have to find \(\left| {CD} \right| + \left| {DE} \right| + \left| {EF} \right| + \left| {FG} \right| + ....\)

To find \(\left| {CD} \right|\)

In a triangle \(ADC\), \(\sin \left( \theta \right) = \frac{{\left| {CD} \right|}}{b}\)

\( \Rightarrow \left| {CD} \right| = b\sin \left( \theta \right)\)

Thus, the Value of \(\left| {CD} \right|\)in terms of \(b\)and\(\theta \).

To find \(\left| {DE} \right|\)

Now, notice that \(\theta \)appears many other places in the following figure.

Since \(AC\)and\(DC\)are parallel, ṣ\(\angle ACD = \angle EDC\). Then, we also have \(\angle EDC = \theta \).

And, we know that \(EDC\)is a right-triangle. So, we can fill in\(\theta \)for all the corresponding angles of the smaller triangles.

Consider the triangle \(CDE\), we have

\(\begin{aligned}\sin \left( \theta \right) = \frac{{\left| {DE} \right|}}{{\left| {CD} \right|}}\\\left| {DE} \right| = \left| {CD} \right|\sin \left( \theta \right)\\ = b{\sin ^2}\left( \theta \right)\end{aligned}\)

Since \(CD = b\sin \left( \theta \right)\)

Finding the geometric sequence

In general, If \({a_n}\)and \({a_{n + 1}}\) are the lengths of the \({n^{th}}\) and \({\left( {n + 1} \right)^{th}}\) perpendiculars, we have

\(\begin{aligned}{l}\sin \left( \theta \right) = \frac{{{a_{n + 1}}}}{{{a_n}}}\\{a_{n + 1}} = {a_n}\sin (\theta )\end{aligned}\)

From the above expression it is show that each perpendicular length has an increasing factor of \(\sin \left( \theta \right)\).

Thus, the geometric series will be in the form of

\(\begin{aligned}\left| {CD} \right| + \left| {DE} \right| + \left| {EF} \right| + .... = b\sin \left( \theta \right) + b{\sin ^2}\left( \theta \right) + b{\sin ^3}\left( \theta \right) + .....\\ \Rightarrow b\left( {\sin \theta + {{\sin }^2}\theta + {{\sin }^3}\theta + ....} \right)\\ = \sum\limits_{n = 1}^\infty {b.{{\sin }^{n - 1}}\left( \theta \right)} \end{aligned}\)

Here, the geometric series formula is\({s_\infty } = \frac{a}{{1 - r}}\)where\(a = b\)and\(r = \sin \theta \)

Substitute the values into the series formula, we get\(\frac{b}{{1 - \sin \theta }}\)

Therefore, the total length of all the perpendiculars is given by the geometric series\(\frac{b}{{1 - \sin \theta }}\).