Answer:

Explanation:
From the question we are told that
Angle of cable 2 
Weight of sculpture 
Generally the Tension from cable 2
is mathematically given by



Generally the Tension from Cable 1
is mathematically given by



Answer:
The smallest diameter is 
Explanation:
From the question we are told that
The resolution of the telescope is 
The wavelength is 
From the question we are told that

So 
Therefore


Now 
So 
=> 

The smallest diameter is mathematically represented as

substituting values

