Chapter 9.8, Problem 35E

To determine

To show that the formula can be rewritten as $\frac{WQ}{NA}=\frac{70}{h\left(1+\tan \theta \right)}$ .

Explanation of Solution

Given:

The ratio of the image length $WQ$ to the length $NA$ of the actual object is given by the formula $\frac{WQ}{NA}=\frac{35\tan \left(\theta -t\right)+35\tan t}{h\tan \theta }$

Calculation:

Using the following formula

  $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

Now,

  $\begin{array}{l}\frac{WQ}{NA}=\frac{35\tan \left(\theta -t\right)+35\tan t}{h\tan \theta }\\ \frac{WQ}{NA}=\frac{35\left[\frac{\tan \theta -\tan t}{1+\tan \theta \tan t}\right]+35\tan t}{h\tan \theta }\\ \frac{WQ}{NA}=\frac{35\left[\frac{\tan \theta -\tan t}{1+\tan \theta \tan t}+\tan t\right]}{h\tan \theta }\end{array}$

  $\begin{array}{l}\frac{WQ}{NA}=\frac{35\left[\frac{\tan \theta -\tan t+\tan t+\tan \theta {\tan }^{2}t}{1+\tan \theta \tan t}\right]}{h\tan \theta }\\ \frac{WQ}{NA}=\frac{35\left[\frac{\tan \theta (1+{\tan }^{2}t)}{1+\tan \theta \tan t}\right]}{h\tan \theta }\end{array}$

  $\begin{array}{l}\frac{WQ}{NA}=35\left[\frac{\tan \theta (1+{\tan }^{2}t)}{h\tan \theta (1+\tan \theta \tan t)}\right]\\ \frac{WQ}{NA}=35\left[\frac{2}{h(1+\tan \theta )}\right]\{\because t={45}^{\circ }\}\\ \frac{WQ}{NA}=\frac{70}{h(1+\tan \theta )}\end{array}$