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A boy standing in the middle of a field, observes a flying bird in the north at an angle of elevation fo 30 degree. and after 2 min, he observes the same bird in the south at an angle of elevation o f60 degree. If the bird flies all along in a straight line at a height of 50root3 m, then its speed in km/h is?
Ans)
tan30=50√3/x1/√3=50√3/x150=x√3=50√3/yy=50total distance covered =x+y=200m.2kmtime=2/60hrspeed=6km/hr
2*8
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x 4 - 25 There is no greatest common factor here. Though, notice that it is the difference of two perfect squares. x 4 - 25 = ( x 2 ) 2 - (5) 2 Thus, we can employ
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ans) 6km/hr {take tan 60=50root 3/x , tan 30 =50 root 3/y , then add x+y ,this x+y is the distance travelled by it , given time t=2min=2/60 hrs you will get x+y=200m=0.2 km velocity =d/t=0.2*60/2 km/hr}
ans) 6km/hr
{take tan 60=50root 3/x , tan 30 =50 root 3/y , then add x+y ,this x+y is the distance travelled by it , given time t=2min=2/60 hrs
you will get x+y=200m=0.2 km
velocity =d/t=0.2*60/2 km/hr}
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