Let QR be called x.
The length of the rope is:
\(\begin{aligned}{c}l &= PR + RS\\ &= \sqrt {P{Q^2} + {x^2}} + \sqrt {S{T^2} + {{\left( {QT - x} \right)}^2}} \end{aligned}\)
From equation (1), the length is minimum for:
\(\begin{aligned}{c}{\frac{{dl}}{{dx}}_{x = {x_{\min }}}} &= 0\\\frac{{{x_{\min }}}}{{\sqrt {P{Q^2} + x_{\min }^2} }} - \frac{{\left( {QT - {x_{\min }}} \right)}}{{\sqrt {S{T^2} + {{\left( {QT - {x_{\min }}} \right)}^2}} }} &= 0\\\frac{{{x_{\min }}}}{{\sqrt {P{Q^2} + x_{\min }^2} }} &= \frac{{\left( {QT - {x_{\min }}} \right)}}{{\sqrt {S{T^2} + {{\left( {QT - {x_{\min }}} \right)}^2}} }}\end{aligned}\)
But from the figure:
\(\begin{aligned}{c}\frac{{{x_{\min }}}}{{\sqrt {P{Q^2} + x_{\min }^2} }} &= \frac{{QR}}{{PR}}\\ &= \cos {\theta _1}\end{aligned}\)
and
\(\begin{aligned}{c}\frac{{\left( {QT - {x_{\min }}} \right)}}{{\sqrt {S{T^2} + {{\left( {QT - {x_{\min }}} \right)}^2}} }} &= \frac{{RT}}{{RS}}\\ &= \cos {\theta _2}\end{aligned}\)
Thus, the minimization condition becomes:
\(\cos {\theta _1} = \cos {\theta _2}\)
Since both the angles are acute angles, the condition becomes\({\theta _1} = {\theta _2}\).
It is proved that the length of the rope is minimum when \({\theta _1} = {\theta _2}\).
