# 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/ (1/3^{1/2}) {tan 30 = 1/3^{1/2} }
=> BD = 2 (3^{1/2})................(1)
similarily in triangle CBE ryt angled at angle CBE
=> BC / BE = tan 60 degree
=> BE = BC/ tan 60
=> BE = 2/3^{1/2} ...................(2)
Now
BE+ED = BD
=> ED= BD-BE

=> ED = 2 (3^{1/2}) - ( 2 / 3^{1/2} ) { from eq 1 and 2 }
=> ED = 4 / 3^{1/2}
=> 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

therefore BY RHS , triangle ABD is similar to triangle CBD

=> angle ADB and angle CDB = 30 degree each using similar triangle property

angle AEB = angle CEB = 60 degree

^{1/2}) {tan 30 = 1/3

^{1/2}}

^{1/2})................(1)

^{1/2}...................(2)

=> ED = 2 (3

^{1/2}) - ( 2 / 3

^{1/2}) { from eq 1 and 2 }

^{1/2}

**3.85 * 10**

^{-3}m sec^{-1 }(approx)Answer### Answered by | 7th Jan, 2013, 05:49: PM

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