An aircraft flies at 400km/hr in still water.a wind of 200^2km/hr is blowing from the south.the piolot wishes to travel from A to a point B north east of a.find the direction he must steer nd time of his journey if AB=1000km.

^=SQUARE ROOT OF 2.

GIVE ME SOLUTION OF THIS QN.

ANSWER IS    1.83 hr.

v_w=sooroot2 km/hr(BC)

v_aw=400km/hr (AC)

v_a = speed of plane alone (AB)

AC/sin45 =BC/sinX

X=30⁰

now byformula

V_a/sin(180-45-30) = 400/sin45

hence solving we get V_a=546.47km/hr

so now  t=AB/ V_a=1000/546.47 = 1.83 hrs

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V_aw = 400km/hr and V_a should be along AB or in the north-est direction. THUS, the direction of V_aw should be such as the resultant of V_W and V_aw is along AB or in north -east direction .

Let V_aw makes an angle X with ab is shown in the figure . APPLYING sine law in triangle ABC we get the solution as posted

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