CBSE Class 10 Answered
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 | 08 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
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
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 = 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 | 07 Jan, 2013, 05:49: PM
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