**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 30^{0} and to C is 45^{0}.

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 =

AC = 100

**Step 2: **

In Δ ABC, tan B =

tan 30 =

100 + DC = 100 √3

DC = 100 √3 – 100 = 100 (√3 – 1)

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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