Assuming the angles are measured from the same point on the ground, the distance is
3000tan60° - 3000tan30°
An aeoroplane flying at a ht 3000 mtr from the ground passes vertically above another plane at an instant angle of elevation from the ground are 60 and 30 degrees respectivly find distance between the planes
2 answers
tan 60o = (Y1+Y2)/X1 = 3000/X1
X1*tan 60 = 3000
X1 = 3000/tan60 = 1732 m.
tan 30o = (Y1+Y2)/(X1+X2)=3000/(1732+X2)
(1732+X2)*tan30 = 3000
1732+X2 = 3000/tan30 = 5196
X2 = 5196-1732 = 3464 m
tan30 = Y2/X2 = Y2/3464
Y2 = 3464*tan30 = 2,000 m.
Y1 = 3000-2000 = 1,000 m. = Distance between the planes.
X1*tan 60 = 3000
X1 = 3000/tan60 = 1732 m.
tan 30o = (Y1+Y2)/(X1+X2)=3000/(1732+X2)
(1732+X2)*tan30 = 3000
1732+X2 = 3000/tan30 = 5196
X2 = 5196-1732 = 3464 m
tan30 = Y2/X2 = Y2/3464
Y2 = 3464*tan30 = 2,000 m.
Y1 = 3000-2000 = 1,000 m. = Distance between the planes.