The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is `60^(@)`. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is `45^(@)`. Find the height of the tower PQ and the distance PX. (Use `sqrt3=1.73`).

Explanation

To solve the problem, we will use trigonometric ratios and properties of triangles. Let's break down the solution step by step.

### Step 1: Set up the problem
Let the height of the tower \( PQ \) be \( h \) meters, and the distance from point \( X \) to the base of the tower \( P \) be \( d \) meters.

### Step 2: Use the angle of elevation from point X
From point \( X \), the angle of elevation to the top of the tower \( Q \) is \( 60^\circ \). According to the tangent function:
\[
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