On the ground level the angle of elevation of the top of a tower is 30°, On moving 20 m nearer the tower, the angle of elevation found to be 60°. The height of the tower is?

Let AB and CD be two towers of height h₁ and h₂ respectively and O the mid-point of the line joining the foots A and C of the towers.
Let OA = OC = x
Then h₁ = x tan 60° = x?3
and h₂ = x tan 30° = x/?3
? h₁ /h₂= 3/1.
Hence, h₁ : h₂ = 3 : 1

Let us draw a figure below as per given question.
Let AB = h m be the height of the tower, BC = x m be the breadth of the river and also ?ACB = 45°
Now from right triangle ABC
tan 45° = h/x ? 1 = h/x
? x = h
Hence, breadth of the river = height of the tower

Let us draw a figure below as per given question.
Let AB = 75 m be the height of pole and C is a point on the ground such that BC = 75m
Now, from right triangle ABC.
tan ? = AB/BC = 75/75
? tan? = 1
? ? = 45°

Let us draw a figure from given question.
Let, AB = 100 m be the height of a tower, P is a point on the ground such that ∠APB = 30°
From right triangle ABP,
BP = 100 cot 30°
⇒ BP = 100 √3
⇒ BP = 100 X 1.73
⇒ BP = 173 meter

Let us draw the figure from the given question.
Let AB = h meter be the height of the tower; C and D are the two points on the ground such that BC = 60 m; ?ACB = 45° and ?ADB = 30°
Now from right triangle ABC,
tan 45° = h/60
? 1 = h/60
? h = 60 m;
Again from right triangle ABD;
tan 30° = h/(x + 60)
? 1/?3 = 60/(x + 60)
? x + 60 = 60?3
? x = 60(1.73 - 1) = 43.8 meter
Hence, speed of boat = 43.8/5 m/s = 43.8/5 x 18/5 = 31.5 km/hr.