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?

Asked by  | 8th Dec, 2012, 12:20: PM

Expert Answer:

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

Answered by  | 7th Jan, 2013, 05:49: PM

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