A man on the top of a vertical towers observes a car moving at a uniform speed coming directly towards it. if it takes 12 minute for the angle of depression to change from 30° to 45°, how soon after this will the car reach the tower?

Let OP be the tower of height h (say) and A and B be the two positions on the horizontal line through O, such that
?OAP = ?, ?OBP = ? and OB = x
In ?OBP, Use the trigonometry formula
Tan? = P/B = Perpendicular distance / Base distance
Tan? = OP/OB
? OB = OP/Tan?
? OB = OP Cot?
Put the value of OB and OP , We will get
x = h Cot ?...............(1)
In ?OAP, Similarly
Tan? = OP/OA
? OA = OP/ Tan?
? OA = OP Cot ?
Put the value of OA and OP
? a + x = h Cot ?
? x = h Cot ? - a ............(2)
From equation (1) and (2)
? h Cot ? = h Cot ? - a
? a = h Cot ? - h Cot ?
? a = h (Cot ? - Cot ?)
? a = h (Cos ?/ Sin ? - Cos ? / Sin ? )
? a = h( (Cos ? Sin ? - Cos ? Sin ? ) /Sin ? Sin ? )
? a = h( Sin(? - ?) / Sin ? Sin ?)
? h = a Sin ? Sin ?/ Sin(? - ?)

Let CD = h unit be the height of the tower and A and B be the two points on the ground, such that DA = a; DB = b;
? DAC = ? and ?DBC = 90° - ?
From right triangle ADC, CD = h = a tan ? ...(i)
From right triangle BDC, CD = h = b tan (90° - ? ) = b cot ? .......(ii)
Multiplying equations (i) and (ii), we get
h² = a tan ? X b cot?
Hence, h = ?ab

Let us draw a figure below as per given question.
Let AB = h meter be the height of cliff and CD = x meter be the height of the tower and also ? ADB = 60° and ?ACE = 30° ,
Now, from the figure, AE = (h - x) meter
From right triangle ABD, BD = h cot 60° = h/?3 m
From right triangle CEA. (h - x) = EC tan 30° = BD tan 30°
? h - x = h/?3 X 1/?3
? x = h - h/3 = 2h/3 meter

Let AB be the wall and BC be the ladder.
Then, ? ACB = 60° and AC = 4.6 m.

AC/BC = cos 60° = 1/2
BC = 2 x AC
= (2 x 4.6) m
= 9.2 m.

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 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.