A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°, when he retires 40 meters from the bank then he find the angle to be 30°. Then the breadth of the river is?

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 speed in the return journey = V
? Speed in onward journey = (125/100) x V = 5V/4 km/h
Average speed = (2x 5V/4 x V) / (5V/4 + V) = 10V/9 km/h
? 1600 x (9/10V) = 30 [? halts for 2 h ]
? V = 1600 x 9/30 x 10 = 48 km/h
? Speed in onward journey = 5V/4 = 5/4 x 48 = 60 km/h

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.

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