The shadow of a tower standing on a level plane is found to be 50 m longer when the sun's altitude is 30° than when it is 60°. Find the height of the tower.

Let us draw a figure as per given question.
Let AB = 5 meter and CD = 2 meter be the heights of a lamp post and the men respectively.
At any time t BD = x meter and shadow of man ED = y meter.
Then, dx/dt = 6 m/min
Now, right triangles ABE and CDE are similar, then
AB/CD = BE/DE
? 5/2 = (x + y)/y
? 5y = 2x + 2y
? 5y - 2y = 2x
? 3y = 2x
? 3 dy/dt = 2 dx/dt
? 3 dy/dt = 2 x 6
? dy/dt = 2 x 6 / 3
? dy/dt = 2 x 2 = 4
Hence, length of his shadow increase at the rate of 4 m/min.

Let us draw a figure below as per given question.
Let AB = CD = h meter be the heights of the towers. E is a point such that DE = 100 meter;
?CED = 60° and ?AEB = 30°
Now, BE = x meter (say)
From right triangle CDE.
h = 100 tan 60°
? h = 100?3 meter
From right triangle ABE,
x = h cot 30° put the value of h, we will get
x = 100?3 X ?3
x = 100 X 3 = 300 meters
Distance between the tower = DE + EB = 100 + 300 = 400 meters
Height of the tower = h = 100?3 meter

Let us draw a figure below from given question.
Let AB = h meter be the height of the tower B and C are two points such that ?ACB = 30° ?ADB = 45° and CD = x meter (say)
From right triangle ABD,
tan 45° = h/BD
? BD = h meter;
Again from right triangle ABC
tan 30° = h/(h + x )
? h + x = ?3 h
? x = (1.73 - 1)h = 0.73h
Now, 0.73h meter covered in 12 min
Hence, h meter covered in 12/0.73 = 1200/73 min = 16 min 23 sec .