Question
Sat December 08, 2012 By:
 

on a straight coast there are 3 objects A,B and C such that AB=BC=2km a vessel approaches B in a line perpendicular to the coast,and at a certain point AC is found to subtend an angle of 60degrees afterr sailing a while in the same direction for 10 minutes AC is found to subtend an angle of 120degrees find the rate at which ship is going?

Expert Reply
Mon January 07, 2013
Answer: Given:on a straight coast there are 3 objects A,B and C such that AB=BC=2km a vessel approaches B in a line perpendicular to the coast,and at a certain point AC is found to subtend an angle of 60degrees afterr sailing a while in the same direction for 10 minutes AC is found to subtend an angle of 120degrees.
 
To find : the rate at which ship is going
 
Let AC subtends 60 degree on the pt D and 120 degree on pt E.
Now in triangle ABD and CBD
AB=BC {given}
Angle ABD =angle CBD  = 90 degree{ Given}
BD common
therefore BY RHS , triangle ABD is similar to triangle CBD
=> angle ADB and angle CDB = 30 degree each using similar triangle property
 
Similarily in triangle ABE and CBE
angle AEB = angle CEB = 60 degree
 
 
 
Now in triangle CBD
CB/BD = tan 30 degree { ryt angle at angle CBD }
=> BD = CB/ tan 30
=> BD = 2/ (1/31/2)   {tan 30 = 1/31/2 }
 => BD = 2 (31/2)................(1)
 
 
similarily in triangle CBE ryt angled at angle CBE
=> BC / BE = tan 60 degree
=> BE = BC/  tan 60   
=> BE = 2/31/2 ...................(2)
 
Now 
BE+ED = BD 
=> ED= BD-BE  
=> ED = 2 (31/2) - ( 2 / 31/2 )         { from eq 1 and 2 }
 => ED = 4 / 31/2
=> ED = 2.31 m { approx}....................(4)
 
As given , the distance covered from pt D to E is 10 mins = 10* 60 = 600 seconds
 
The speed = distance / time = ED/ 600 sec = 2.31 / 600 
= 3.85 * 10 -3 m sec-1 (approx)Answer
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