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: From the top of a building 60 m high, the angles of depression of the top and bottom CBSE 2024

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:

As observed from the top of a lighthouse, 100 m above sea level

Step 1:

Let’s start from Δ ABC, tan C = tan 45° = \frac{AB}{AC}

\Rightarrow 1 = \frac{100}{AC}

\Rightarrow AC = 100

Step 2:

In Δ ABC, tan B = \frac{AB}{AD}

\Rightarrow tan 30 = \frac{AB}{AC + DC}

\Rightarrow \frac{1}{\sqrt 3} = \frac{100}{100 + DC}

\Rightarrow 100 + DC = 100 √3

\Rightarrow DC = 100 √3 – 100 = 100 (√3 – 1)

\Rightarrow 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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