An aircraft is flying along a horizontal path PQ directly towards an observer on the ground at O and maintaining an altitude of 3000m . When the aircraft is at P, the angle of depression is 30 degree and when at Q, the angle of depression is 60 degree. Find the distance of PQ.
Answers
Answer:
Hey!!
Step-by-step explanation:
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)
Tʜx..
Answer:
Step-by-step explanation:
