The best seat for the viewing angle

Visual-angle geometryGiven viewing geometry: screen height5m, lower edge1m above the observer eye, distancex, visual angleθ. No optimal distance is supplied.5 m1 mx mθ
Question diagram 1
Visual-angle geometryGiven viewing geometry: screen height5m, lower edge1m above the observer eye, distancex, visual angleθ. No optimal distance is supplied.5 m1 mx mθ

Given screen and sight lines with dimensions5m,1m,x and angleθ.

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

(i) Show that θ=tan⁡−1(6/x)−tan⁡−1(1/x)\theta=\tan^{-1}(6/x)-\tan^{-1}(1/x). mark

(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.] marks

(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. marks

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

Total marks: 7

Solution

(i)

Modelling with angles

A screen 5 m high has its lower edge 1 m above an observer's eye. Show that the visual angle is θ=tan⁡−16x−tan⁡−11x\theta=\tan^{-1}\frac6x-\tan^{-1}\frac1x when the observer is xx m away.

Hint

  1. The top of the screen is 6 m above eye level and the bottom is 1 m above.

  2. θ\theta is the difference of the two angles of elevation.

Answer · (i)

θ=arctan⁡(6/x)−arctan⁡(1/x)\theta=\arctan(6/x)-\arctan(1/x).

x>0x>0.

(ii)

Optimisation with inverse trigonometry

Find the exact distance the observer should sit to obtain the largest visual angle.

Explore this idea

Hint

  1. Use ddxtan⁡−1u=u′1+u2\frac d{dx}\tan^{-1}u=\frac{u'}{1+u^2}.

  2. Set dθdx=0\frac{d\theta}{dx}=0 and solve.

Answer · (ii)

x=6x=\sqrt6 m.

Explore the idea

The visual angle against distance (in degrees)

The angle rises to a maximum at x = √6, then falls off as the observer sits further back.

Position along the curvex = 8
Try an example
Test a critical position

Buttons use full-precision reviewed positions; displayed coordinates are rounded.

f(x)
29.74488
Gradient f′(x)
-2.55627
negative

Beyond x=√6 the angle decreases: its gradient is negative.

Selected values are within the original drawing windows.

  • θ in degrees
  • Tangent
  • Dθ/dx, and where θ increases
  • Join P to Q; the dashed legs show Δx and Δy
Key idea

Sitting very close or very far both give small angles; the best seat is between, at x=6x=\sqrt6.

Ideas to try
Find the seat where the tangent is flat.
(iii)

Extreme values on a closed interval

The observer is between 2 m and 15 m from the screen. Find, to the nearest degree, the smallest visual angle.

Explore this idea

Hint

  1. On a closed interval the extreme values occur at stationary points or at the ends.

  2. The only stationary point is the maximum; so the minimum is at an endpoint.

Answer · (iii)

18∘18^\circ, attained at x=15x=15 m.

Explore the idea

On 2 ≤ x ≤ 15 the minimum is at the far end

Switch on the endpoints and the maximum.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • θ on the allowed interval
Ideas to try
Which end of the interval gives the smaller angle?