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Comments (9)
) makes an angle
with P, such that,
= Q sin
/ ( P + Q cos
)
/ ( v + w cos
)
/ ( f + cos
) where, f = v / w.........eqn. (1)
/ ( 2 v + w cos
) = sin
( 2 f + cos
)...........eqn. (2)
, so I've converted the two unknowns v,w into one unknown f = v / w. Now we have two unknows f &
and just two equations.
= 5 sin 
= - 1 / 5
= 180 + tan -1 ( - 1 / 5 ) degrees = 180 - tan -1 (1/5) degrees tan 45=qsin$/p-qcos$=1===>q sin$=p-q cos$--------1
now the speed of the man is doubled ==> speed is '2p'
the angle of the resultant =arc tan 2 eastof north ==> angle between the direction of the man and resultant=arc tan2 -arc tan 1=q sin(180 -$)/2p-q cos(180-$)=1/3 [since the angle arc tan 2 is greater than 45 the resultant comes nearer to east..]===>3q sin$=2p+q cos$-----2
from 1 and 2,
3qsin$-q cos$=2q(sin$+cos$)
3 sin$-cos$=2sin$+2cos$
==>sin$=3 cos$
==>tan$=3
==>$=arc tan 3 [from the direction of the man towards north..] or arc tan 2 west of north...
Let the wind velocity be w = x i + y j.
There fore the relative velocity = w - v = (x - u) i + (y - u) j
But the direction of the relative velocity is from north. Therefore x - u = 0. or x = u.
Now when the man velocity is doubled 2v = 2u i + 2u j.
therefore w - v = relative velocity.
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