Question

an aircraft is flying along a horizontal path PQ directly towards an observer on the ground at home and maintaining an altitude of 3000 m when the aircraft is at P the angle of depression is 30 and 1 sq the angle of depression is 60 find the distance of PQ

Answer 1

The aeroplane is at P point
Let PA be the constant height at which the aeroplane is flying
PA= 3000m (Given)

Let B be the observer
In Triangle APB
Angle PBA = 30 degree
Angle PAB = 90 degree
tan 30° = PA / AB
1/√3 = 3000 / ABAB = 3000*√3
      = 3 * 1000 * √3
      = 3 * 1000 * 1.732 (Since √3  is equal to 1.732)
      =5196m


The aeroplane comes at Q point
Let QC be the constance height of 3000m
Angle QBC = 60 degree
QCB = 90 degree
tan 60° = QC / BC
√3 = 3000/BC
BC = 3000/√3 

On rationalizing the denominator
BC = 1000 *√3 m
      = 1000 * 1.732 (Since √3 is equal to 1.732)
      = 1732m

AC = AB - BC
      = 5196 - 1732
      = 3464m

Now ,
we find,
PQAC is a rectangle each with angles 90 degree
Hence,
PQ = AC (Opposite sides of rectangle are equal
      =3464m (Ans)