A man observes a car from …

Let AB be the tower of height h
{tex}\angle A Q B = 45 ^ { \circ }{/tex}
Now, in {tex}\triangle {/tex}ABQ,
{tex}\tan 45 ^ { \circ } = \frac { A B } { B Q }{/tex}

{tex}\Rightarrow \quad 1 = \frac { h } { B Q }{/tex}
{tex}\Rightarrow {/tex}  BQ = h
In {tex}\triangle A P B{/tex}  
{tex}\tan 30 ^ { \circ } = \frac { A B } { P B }{/tex}  
{tex}\Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { h } { x + h }{/tex}
{tex}\Rightarrow \quad x + b = h \sqrt { 3 }{/tex}
i.e.,{tex} \quad x = h ( \sqrt { 3 } - 1 ){/tex}
So speed of car
{tex}V=\frac{distance}{time}=\frac{x}{12}=\frac{h(√3-1)}{12}{/tex}
Hence time taken by car to reach upto the tower 
{tex}t=\frac{BQ}{V}=\frac{h}{\frac{h}{12}(√3-1)}{/tex} 
{tex}=\frac{12}{√3-1}=\frac{12(√3+1)}{(√3+1)(√3-1)}{/tex}
= 6(√3 + 1) = 6 {tex}\times{/tex} (1.73 + 1)
= 6 {tex}\times{/tex} 2.73 = 16.38 minutes

{tex}{/tex}