The best seat for the viewing angle · source reference

Printed question

A movie theatre screen which is 5 m high has its lower edge 1 m above an observer’s eye. The visual angle $\theta$ of the observer seated $x$ m away is as shown in the diagram below.

(i) Show that $\theta=\tan^{-1}(6/x)-\tan^{-1}(1/x)$. [1]

(ii) Find the exact distance the observer should sit to obtain the largest visual angle. [You need not establish that the distance gives the largest visual angle.] [4]

(iii) Suppose that the observer is situated between 2 m and 15 m from the screen. Find, to the nearest degree, the smallest visual angle. [2]

[2008/RJC/Prelim/I/11(b)]

Printed working

(i) In the labelled figure, $\tan\angle ABD=1/x$, $\tan\angle ABC=6/x$.

$$\theta=\angle ABC-\angle ABD=\tan^{-1}(6/x)-\tan^{-1}(1/x)$$

(shown).

(ii)

$$\frac{d\theta}{dx}=\frac1{1+(6/x)^2}\left(\frac{-6}{x^2}\right)-\frac1{1+(1/x)^2}\left(\frac{-1}{x^2}\right)=\frac{-6}{x^2+36}+\frac1{x^2+1}.$$

For maximum/minimum, $d\theta/dx=0$:

$$\frac{-6}{x^2+36}+\frac1{x^2+1}=0\Rightarrow-6(x^2+1)+x^2+36=0\Rightarrow5x^2=30.$$

$x=\sqrt6$ or $-\sqrt6$ (rejected since $x>0$). Required distance $\sqrt6$ m.

(iii) Printed calculator display: Y1=tan⁻¹(6/X)−tan⁻¹(1/X), X=15, Y=17.987335…. Its curve falls toward the right after the maximum. From the graph the minimum on $[2,15]$ is at $x=15$.

$$\theta=\tan^{-1}(6/15)-\tan^{-1}(1/15)=18^\circ$$

(to nearest degree). A direct endpoint check gives a larger angle at $2$, consistent with the graph.

Teacher notes

Inverse tangent chain rule: $\frac d{dx}\tan^{-1}(f(x))=\frac1{1+(f(x))^2}f'(x)$.

Sketch the graph of $\theta=\tan^{-1}(6/x)-\tan^{-1}(1/x)$ in (iii).

Answer key

(ii) $\sqrt6$ m. (iii) $18^\circ$.

Native diagrams