Support Beams

It is possible to see that, from the dimensions given:

$\tan\theta = \frac{4}{5}$ where $\theta$ is the angle between the two lines.

Since the bottom 5m is split into 23 intervals, each interval must be $\frac{5}{23}m$ long. It is better to consider the distance from the vertical support to the point where the two lines meet, and this distance is given by $\frac{5n}{23}$ where n is a number from 1 to 23. The length of support $x_{n}$ is given by $\frac{5n}{23} \times \frac{4}{5} = \frac{20n}{115}$ (since opposite = adjacent $\times \tan\theta$ .

The total heights of all the supports is therefore $\sum_{n=1}^{23} \frac{20n}{115} = \frac{20}{115} \sum_{n=1}^{23} n$ .

Therefore:

Total height (metres) = $\frac{20}{115} \times \frac{23(23+1)}{2} = \boxed{48}$

Interestingly, this can be simplified to give:

Total height = $\frac{a(b+1)}{2}$ where $a$ = height of the tallest support and $b$ = the number of divisions (supports). Therefore, the total height is independent of the width over which the supports are spread (the 5m was not needed for this problem).