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

rate if useful

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

rate if useful

let the triangle be ABC right angled at A as said

here sicC=c/a

cosC=b/a

hence tanC=c/b

 

   -1/tanC

=root of c²+b²   -b/c

which is a-c/b

rate if useful

00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000