Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to

According to given conditions angle between AC and AB is 30 degrees and between AB and BC is 60 degrees. So the triangle formed is a 30-60-90 triangle. 

Hence, $$\angle\ ABC=60^{\circ\ }$$, $$\angle\ CAB=30^{\circ\ }$$, which implies $$\angle\ ACB=180^{\circ\ }-\left(30^{\circ\ }+60^{\circ\ }\right)=90^{\circ\ }$$

Therefore, The triangle is a right-angled triangle with base = BC, height = AC, and hypotenuse = AB.

We know that $$\cos\ \angle\ ABC\ =\ \frac{\ BC}{AB}=>\ \cos60^{\circ\ }=\frac{BC}{500}\ =>\ 500\times\ \frac{1}{2}=250\ km$$

Similarly, $$\sin\ \angle\ ABC\ =\ \frac{\ AC}{AB}=>\ \sin60^{\circ\ }=\frac{AC}{500}\ =>\ 500\times\ \frac{\sqrt{\ 3}}{2}=250\sqrt{\ 3}\ km$$

So, total time taken by train to travel B to C is is (250/50) = 5 hrs, hence the train reaches at 1 pm. Accordingly, Rahim has to reach C fifteen minutes before i.e. at 12:45 PM.

Time taken by Rahim to travel by car is around $$\ \frac{\ 250\sqrt{\ 3}}{70}=6.19$$ hrs (Around 6.2 hrs = 6 hrs 12 minutes). So, the latest time by which Rahim must leave A and still be able to catch the train is 6:30 am.