Let C and D be the position of two aeroplanes. The height of the aeroplane which is at point D be 3000 m and it passes another aeroplane vertically which is at point C. Let BC = x m. It is also given that the angles of elevation of two planes from the point A on the ground is 45° and 60° respectively.
In right triangle ABC, we have


In right triangle ABD, we have
]

Comparing (i) and (ii), we get


Hence, vertical distance between the aeroplane
= CD = BD - BC

Height of lighthouse = 100 m
Let distance travelled by the ship, when the angle of depression changes from 60° to 30° (DC) =y m and distance BC = x m Then, in right ∆ABC,


In right ABD, tan 

From (i) and (ii), we get


Hence, distance travelled by the ship = 115.473 m.

Tips: -

Let CD be the hill of height h km. Let A and B be two stones due east of the hill at a distance of 1 km. from each other. It is also given that the angles of depression of. stones A and B from the top of a hill be 30° and 45° respectively.
Let    BC = x km
In right triangle BCD, we have


In right triangle ACD, we have


Comparing (i) and (ii), we get


Hence, the height of hill is 1.365.

Let AB be the building such that AB = 9 m and CD is the tower. The angles of depression of the top and the bottom of the building from the tower are 30° and 60° respectively.


In 
Comparing (i) and (ii), we get

⇒    3h = h + 9
⇒    3h - h = 9
⇒    2h = 9
⇒    h = 4.5 m
Now, height of the tower
= (h + 9) met.
= (4.5 + 9) met.
= 13.5 met.
Difference between the building and tower (x)