Q) As observed from the top of a lighthouse, 100 m above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45°. Determine the distance travelled by the ship during the period of observation. (Use √3 = 1.732)
Ans:
Let’s start with the diagram for this question: 
Here we have tower AB of 100m height.
Angle of depression to D is 300 and to C is 450.
We need to find the length of DC.
Let’s make a simplified diagram of the same for our better understanding:

Step 1:
Let’s start from Δ ABC, tan C = tan 45° = 
 1 =
 1 = 
 AC = 100
 AC = 100
Step 2:
In Δ ABC, tan B = 
 tan 30 =
 tan 30 = 
 
 
 100 + DC = 100 √3
 100 + DC = 100 √3
 DC = 100 √3 – 100 = 100 (√3 – 1)
 DC = 100 √3 – 100 = 100 (√3 – 1)
 DC = 100 ( 1.732 – 1) = 100 x 0.732 = 73.2
 DC = 100 ( 1.732 – 1) = 100 x 0.732 = 73.2
Therefore, distance travelled by the ship is 73.2 m
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