A straight highway leads to the …

In right triangle ABP,
{tex}\tan 30 ^ { \circ } = \frac { \mathrm { AB } } { \mathrm { BP } }{/tex}
{tex}\Rightarrow \frac { 1 } { \sqrt { 3 } } = \frac { \mathrm { AB } } { \mathrm { BP } }{/tex}
BP = AB{tex}\sqrt3{/tex} ........ (i)
In right triangle ABQ,
{tex}tan\;60^0\;={AB\over BQ}{/tex}
{tex}\Rightarrow \sqrt { 3 } = \frac { A B } { B Q }{/tex}
{tex}\Rightarrow _ { B Q } = \frac { A B } { \sqrt { 3 } }{/tex}....... (ii)
{tex}\because{/tex} PQ = BP - BQ
{tex}\therefore{/tex} PQ = AB{tex}\sqrt { 3 } - \frac { A B } { \sqrt { 3 } } = \frac { 3 A B - A B } { \sqrt { 3 } } = \frac { 2 A B } { \sqrt { 3 } }{/tex} = 2BQ [From eq. (ii)]
{tex}\Rightarrow{/tex} BQ = {tex}\frac12{/tex}PQ
{tex}\because{/tex} Time taken by the car to travel a distance PQ = 6 seconds.
{tex}\therefore{/tex} Time taken by the car to travel a distance BQ, i.e. {tex}\frac12{/tex}PQ = {tex}\frac12{/tex} {tex}\times{/tex} 6 = 3 seconds.
Hence, the further time taken by the car to reach the foot of the tower is 3 seconds.