## Saturday, 19 November 2011

### Trigonometry-3 (CAT-2008)

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 C 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 600 to AB. Also C is located between south and southwest of A with AC at 300 to AB. The latest time by which Rahim must leave A and still catch the train is closest to
(1)6:15 am     (2) 6:30 am    (3) 6:45 am    (4) 7:00 am    (5) 7:15 am
Solution follows here:
Solution:
Plotting the points A,B and C is the key here:

AB = 500;
B = 600;          A = 300 => C = 900 ABC is a right triangle.
cosB = BC/AB => BC = AB cos60 = 500*(1/2) = 250
sinB = AC/AB => AC = AB sin60 = 500*(√3/2) = 250√3
speed of the train = 50 kmph
Time of travel from B to C = distance/speed = BC/70 = 250/50 = 5 hour
As the train starts at 8:00 am, it reaches C by 13:00.
“Rahim must reach C at least 15 minutes before the arrival of the train”
=> he must reach C at least by 12:45 pm              --(1)
speed of Rahim = 70 kmph
Time of travel from A to C = distance/speed = AC/70 = 250√3/70
= 6.18 hour = 6 hr 11 min                                       --(2)
From (1) and (2), Rahim must start latest by 12:45 – 6:11 = 6:34 am