The angle of elevation of the highest point P of a vertical tower at point A on the horizontal ground is `45^(@)`. The height of tower is 'h'/ The angle of a elevation of the tower becomes `60^(@)` from B on moving a distance 'd' at `30^(@)` angle from the horizontal. Prove that: `d=h(sqrt(3)-1)`

Explanation

To solve the problem step by step, we will analyze the given information and apply trigonometric principles.

### Step 1: Understanding the Setup
We have a vertical tower of height \( h \) and two points \( A \) and \( B \) on the ground. The angle of elevation from point \( A \) to the top of the tower (point \( P \)) is \( 45^\circ \). The angle of elevation from point \( B \) to the top of the tower is \( 60^\circ \), and point \( B \) is reached by moving a distance \( d \) at an angle of \( 30^\circ \) from the horizontal.

### Step 2: Finding Distance \( AO \)
From point \( A \), we can use the tangent of the angle of elevation:
\[
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