A man is observing from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man's eye when at a distance of 60 m from the tower. After 5 second, the angle of depression becomes 30°, Find the speed of the boat, assuming that it is running in still water.

Let us draw a figure below as per given question.
Let A be the position of a person on the bank of a river and OP the tree on the opposite bank and ?OAP = 60°. When the person retires to the position B, then AB = 40 meter and ?OBP = 30°
Let us assume OA(Breadth of the river) = x meter and height of tree OP = h meter
In ?OAP, Use the trigonometry formula
Tan60° = P/B = Perpendicular distance / Base distance
? Tan60° = OP / OA
? OP = OA Tan60°
Put the value of OP and OA, We will get
? h = x?3 ..............(1)
Now in the triangle ?OBP
Tan30° = OP / OB
? OP = OB Tan30°
? OP = (x + 40)/?3
? h = (x + 40)/?3 ...................(2)
From Equation (1) and (2), We will get
? (x + 40)/?3 = x?3
? (x + 40) = x?3 X ?3
? (x + 40) = 3x
? 3x - x = 40
? x = 20 m

Consider the diagram is shown above where PR represents the ladder and RQ represent the wall.
Cos 60° = PQ / PR
? 1 / 2 = 12.4 / PR
? PR = 2 × 12.4 = 24.8 m

Let the length of slower train be L meters and the
length of faster train be (L/2) meters
Their relative speed = (36 + 54) km/hr
= 90 x (5/18)
= 25 m/sec

? 3L / (2 x 25) = 12
? 3L = 600
? L = 200
? Length of slower train = 200 meters
Let the length of platform by M meters
Then, 200 + M/[36 x (5/18)] = 90 sec.
? 200 + M = 900
? M = 700 meters
Length of platform = 700 meters.

Given, a = 45 km/h, y = ?, t₁ = 4 h 48 min and t₂ = 3 h 20 min
Using y = a?t₁ / t₂= 45?4 h and 48 min /3 h and 20 min
= 45?(24/5h) / (10/3 h) = 45?24 x 3/5 x 10
= 45 x ?1.44 = 45 x 12 = 54 km/h

Speed of the person = 4 km/h = 4.5 x (5/18) = m/s = 5/4 m/s = 1.25 m/s
Speed of 2nd person = 5.4 km/h = 5.4 x (5/18) m/s = 3/2 m/s = 1.5 m/s

Let speed of train be x m/s.
Then, (x - 1.25) x 8.4 = (x - 1.5) x 8.5
? 8.4x - 10.5 = 8.5x - 12.75
? 0.1x = 2.25
? x = 22.5
? Speed of the train = 225 x (18/5 ) = 81 km/h .

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 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 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 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 from the given question.
Let AB = h meter be the height of the towers. B and C are two points such that BC = 20 m; ?ADB = 30° and ?ACB = 60° BC = x meter (let us assume)
Now, from right triangle ABC,
x = h cot 60°
? x = h/?3 meter
Again, from right triangle ABD,
h = (20 + x) tan 30°
put the value of x in above equation.
? x = h/?3
? h = (20 + h/?3) x 1/?3 ( ? tan 30° = 1/?3 )
? h - h/3 = 20/?3 ? 2h/3 = 20/?3
? h = 20 x 3/2?3 = 10?3 m