Two towers 10 meters apart, are 4 m and 6 m high respectively. What will be the height of pointof intersection of lines joining the top of each tower to the bottom of opposite tower?

Let AB and CD be the towers of length 4m and 6m respectively.

Let the length of BF be x therefore, the length of FC will be 10-x.

Triangle BEF is similar to triangle BDC

So, $$\frac{EF}{DC}=\frac{BF}{BC}=\frac{BE}{BD}$$

or, $$\frac{EF}{6}=\frac{x}{10}=\frac{BE}{BD}$$

Therefore, $$EF=\frac{3x}{5}$$

Similarly, triangle EFC is similar to triangle ABC

So, $$\frac{EF}{AB}=\frac{CF}{BC}=\frac{EC}{AC}$$

or, $$\frac{EF}{4}=\frac{10-x}{10}=\frac{EC}{AC}$$

or, $$EF=\ \frac{2\left(10-x\right)}{5}$$

Therefore, $$\frac{2\left(10-x\right)}{5}=\frac{3x}{5}$$

or, $$x=4$$

EF= $$\frac{3x}{5}=2.4\ m$$