Two trees

You can solve this problem in many ways. One way involves using similar triangles, leading to a quartic equation which has to be solved by numerical methods. Sue Liu of Madras College used similar triangles and then the Newton Raphson to give a numerical solution 
 

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Let $AB = a$, $BC = b$, $CD = c$, $AC = 20$, BD = 30$, EF = 8$. By similar triangles:$$\frac{x}{b}=\frac{8}{c},\frac{b-x}{b}=\frac{8}{a}.$$ Adding these two expressions: $$\frac{8}{a}+\frac{8}{c}=1.(1)$$ We can write $c$ in terms of $a$ using Pythagoras Theorem which gives: $$b^2=900-c^2=400-a^2.(2)$$ and hence $$c^2=500+a^2.$$ So (1) becomes $$\frac{8}{a}+\frac{8}{\sqrt{(}500+a^2)}=1(3)$$ and we have to solve equation (3) to find $a$ and then we can find $b$ from equation (2). Rearranging (3) and squaring we get $$\frac{64}{(a^2+500)}=\frac{(a-8{)}^2}{a^2}.$$ Rearranging this equation gives: $$a^4-16a^3+500a^2-8000a+32000=0.$$ Using the Newton Raphson method, or alternatively 'interval halving' gives the solution: $a = 11.712$ because the other roots of the quartic equation are less than 8 or complex. From this we get $b=16.212$ and $c=25.242$ so the distance between the bases of the trees is 16.212 metres (to 3 decimal places). Another approach is to set up a spreadsheet to give the height of the intersection of the ladders $h$ in terms of the distance $b$ between the trees at ground level using the equivalent of equation (1), namely: $$\frac{1}{\sqrt{(}400-b^2)}+\frac{1}{\sqrt{(}900-b^2)}=\frac{1}{h}.$$ Use the spreadsheet calculation, and interpolation, until $h = 8$ as nearly as required. This method will then give the corresponding value of $b$.