The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be $$45^\circ$$. After walking a distance of $$80$$ meters towards the top, up a slope inclined at angle of $$30^\circ$$ to the horizontal plane the angle of elevation of the top of the hill becomes $$75^\circ$$. Then the height of the hill (in meters) is_____.

Let the foot of the hill be the point $$B$$ and let the top of the hill be the point $$T$$. We take the horizontal plane through $$B$$ as our reference ground level.

We first stand at a point $$P$$ on this horizontal plane such that the angle of elevation of $$T$$ from $$P$$ is $$45^\circ$$. If the horizontal distance $$BP$$ is denoted by $$x$$ metres and the vertical height of the hill $$BT$$ is denoted by $$h$$ metres, then by the definition of tangent we have

$$\tan 45^\circ=\frac{\text{opposite}}{\text{adjacent}}=\frac{h}{x}.$$

Since $$\tan 45^\circ=1$$, this at once gives

$$h=x.$$

Now we walk from $$P$$ directly towards the top, along a path that is inclined at $$30^\circ$$ to the horizontal, and we cover a distance of $$80$$ metres along this path. Let the new point of observation be $$Q$$. Because the path makes an angle of $$30^\circ$$ with the horizontal, we can resolve the $$80$$ m into horizontal and vertical components using the standard relations

$$\text{horizontal component}=80\cos30^\circ,\qquad\text{vertical component}=80\sin30^\circ.$$

Recalling that $$\cos30^\circ=\dfrac{\sqrt3}{2}$$ and $$\sin30^\circ=\dfrac12$$, we obtain

$$\text{horizontal component}=80\left(\frac{\sqrt3}{2}\right)=40\sqrt3,$$ $$\text{vertical component}=80\left(\frac12\right)=40.$$

Thus, compared with point $$P$$, the point $$Q$$ is $$40\sqrt3$$ m closer to the foot $$B$$ horizontally and $$40$$ m higher vertically. Consequently the horizontal distance from $$Q$$ to $$B$$ equals

$$x-40\sqrt3,$$

and the height of $$Q$$ above the ground plane equals

$$40\text{ m}.$$

From $$Q$$ the angle of elevation of the top $$T$$ is now $$75^\circ$$. The vertical separation between $$Q$$ and $$T$$ is

$$h-40,$$

while the horizontal separation is

$$x-40\sqrt3.$$

Applying the definition of tangent again, we write

$$\tan75^\circ=\frac{h-40}{\,x-40\sqrt3\,}.$$

But we already have $$h=x$$ from the first observation, so we substitute $$x=h$$ to get

$$\tan75^\circ=\frac{h-40}{\,h-40\sqrt3\,}.$$

To solve for $$h$$ we first recall the exact value of $$\tan75^\circ$$. Using the angle-addition formula

$$\tan(45^\circ+30^\circ)=\frac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ},$$

and the facts $$\tan45^\circ=1$$ and $$\tan30^\circ=\dfrac1{\sqrt3},$$ we find

$$\tan75^\circ=\frac{1+\dfrac1{\sqrt3}}{1-1\cdot\dfrac1{\sqrt3}} =\frac{\sqrt3+1}{\sqrt3-1}.$$

Introduce the shorthand $$t=\tan75^\circ=\dfrac{\sqrt3+1}{\sqrt3-1}.$$ Our equation becomes

$$\frac{h-40}{h-40\sqrt3}=t.$$

Cross-multiplying,

$$h-40=t(h-40\sqrt3).$$

Expand the right side:

$$h-40=th-40t\sqrt3.$$

Rearrange to collect the $$h$$ terms on the left and constants on the right:

$$h-th=40-40t\sqrt3.$$

Factor out $$h$$ on the left:

$$h(1-t)=40(1-t\sqrt3).$$

Finally, divide by the coefficient $$1-t$$ (noting that $$t\neq1$$) to obtain $$h$$ explicitly:

$$h=40\frac{1-t\sqrt3}{1-t}.$$

We now substitute the exact value of $$t$$. First compute $$1-t$$ and $$1-t\sqrt3$$ separately:

$$1-t=1-\frac{\sqrt3+1}{\sqrt3-1} =\frac{\sqrt3-1-(\sqrt3+1)}{\sqrt3-1} =\frac{-2}{\sqrt3-1},$$

$$1-t\sqrt3=1-\sqrt3\left(\frac{\sqrt3+1}{\sqrt3-1}\right) =\frac{\sqrt3-1-(3+\sqrt3)}{\sqrt3-1} =\frac{-4}{\sqrt3-1}.$$

Taking the ratio,

$$\frac{1-t\sqrt3}{1-t} =\frac{\dfrac{-4}{\sqrt3-1}}{\dfrac{-2}{\sqrt3-1}} =\frac{-4}{-2}=2.$$

Therefore

$$h=40\times2=80.$$

Thus the height of the hill is $$80\text{ metres}$$.

So, the answer is $$80$$.