Let AB=dIn △ABP1​∠BAP1​=60°⇒tan60°=dBP1​​⇒BP1​=3​dIn △ABP2​∠BAP2​=30°⇒tan30°=dBP2​​⇒BP2​=3​ d​P1​P2​=BP1​−BP2​=d3​−3​ d​=3​ 2d​So, car took 6 sec to travel 3​ 2d​ distance.So, to reach foot of tower the car should travel BP2​velocity=timedistance​Velocity=63​ 2d​ ​=33​ d​to travel BP2​ distancetime=speeddistance​          =33​ d​ 3​ d​ ​          =3sec