The angle of elevation of the top of a hill is 30° from a point on the ground. On walking 1 km towards the hill, angle is found to be 45°. Calculate the height of the hill.

The correct option is 1.366 km.

Let us define the variables and setup the geometric representation of the problem.
Let the height of the hill be $h$ km.
Let $A$ be the initial point on the ground where the angle of elevation to the top of the hill, say $T$, is $30°$.
Let $B$ be the second point on the ground, which is 1 km closer to the hill from $A$. The angle of elevation from $B$ to $T$ is $45°$.
Let $C$ be the foot of the hill directly below the top $T$. The triangle $TCA$ and triangle $TCB$ are right-angled triangles at $C$.
The distance $AB=1$ km. Let the distance $BC=x$ km. Therefore, the total distance $AC=AB+BC=1+x$ km.

First, let's consider the right-angled triangle $TCB$:
Using the definition of the tangent function:
$\tan (45°)=\frac{TC}{BC}$

Since $\tan (45°)=1$, we have:
$1=\frac{h}{x}$
This simplifies directly to:
$x=h$

Now, let's consider the larger right-angled triangle $TCA$:
Using the definition of the tangent function for the $30°$ angle:
$\tan (30°)=\frac{TC}{AC}$

We know that $\tan (30°)=\frac{1}{\sqrt{3}}$, $TC=h$, and $AC=1+x$.
Substituting $x=h$ into the equation, we get:
$\frac{1}{\sqrt{3}}=\frac{h}{1+h}$

Cross-multiplying to solve for $h$:
$1+h=h\sqrt{3}$
$1=h\sqrt{3}-h$
$1=h(\sqrt{3}-1)$
$h=\frac{1}{\sqrt{3}-1}$

To rationalize the denominator, multiply the numerator and denominator by $(\sqrt{3}+1)$:
$h=\frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt{3}+1)}$
$h=\frac{\sqrt{3}+1}{3-1}$
$h=\frac{\sqrt{3}+1}{2}$

Using the approximation $\sqrt{3}\approx 1.732$:
$h\approx \frac{1.732+1}{2}$
$h\approx \frac{2.732}{2}$
$h\approx 1.366$ km

Thus, the height of the hill is approximately 1.366 km.