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.

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 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)

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