Chapter 9.4, Problem 40PS

Solution

To calculate:Photograph a painting with a camera mounted on a tripol.Calculate the value of $x$ that maximum value of $\theta$

The necessary separation is $2 ft.$

Given information:The smaller triangle implies that since the sum of a triangle's angle measurements is $180°.$

Calculation:

  $\alpha +\gamma +{90}^{°}={180}^{°}$

Isolate $\gamma$

  $\gamma ={180}^{°}-\alpha -{90}^{°}$

Calculate the right side.

  $\gamma ={90}^{°}-\alpha$

  

Given that the total of a triangle's angle measurements is $180°.$ , it follows from the larger triangle

  $\beta +(\theta +\gamma )+{90}^{°}={180}^{°}$

Isolated $\theta$

  $\theta ={180}^{°}-\beta -\gamma -{90}^{°}$

Calculate the right side.

  $\theta ={90}^{°}-\beta -\gamma$

  

Substitute $\gamma ={90}^{°}-\alpha$ into the last equation:

  $\theta ={90}^{°}-\beta -\left({90}^{°}-\alpha \right)$

Calculate the right side.

  $\theta =\alpha -\beta$

To find

  $\alpha$ and $\beta .$

The smaller triangle is

  $\begin{array}{r}\hfill \tan \alpha =\frac{x}{1}\\ \hfill \tan \alpha =x\end{array}$

Apply inverse tangent.

  $\alpha ={\tan }^{-1}x$

The bigger triangle

  $\tan \beta =\frac{x}{4}$

Apply inverse tangent.

  $\beta ={\tan }^{-1}\left(\frac{x}{4}\right)$

Substitute $\alpha ={\tan }^{-1}x$ and $\beta ={\tan }^{-1}\left(\frac{x}{4}\right)$ into $\theta =\alpha -\beta$ .

  $\theta ={\tan }^{-1}x-{\tan }^{-1}\left(\frac{x}{4}\right)$

By using a graphing calculator, determine that the angle $\theta$ is greatest when $x=2.$ Therefore, a gap of $\text{2 ft}\text{.}$

The necessary separation is $\text{2 ft}\text{.}$