Q) A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h.

Ans:

Step 1: Let’s start with the diagram for this question:

Here, let’s consider AB is the cliff and the boat initially at point C, and then after 2 minutes, it is at point D.

Given that the angle of depression from point B at the cliff top to the boat at Point C is 60⁰, hence the elevation angle from C to point B will be 60⁰ (for being alternate interior angle). Similarly, the angle of depression from point B at the Cliff top to the boat at Point C is 45⁰, the elevation angle from C to point B will be 45⁰.

Let’s consider the distance covered by the boat from point C to point D is D₁ and the distance from C to point A at the foot of the cliff is D₂.

Step 2: Let’s start from Δ ABC,

tan ACB =

∴ tan 60 =

∴ √ 3 =

∴ D₂ =

∴ D₂ = 50√ 3  ….. (i)

Step 3: Next in Δ ABD, tan ADB =

∴ tan 45 =

∴ 1 =

∴ AD = 150……………….. (ii)

Step 4: We can see that CD = AD – AC or D₁ = AD – D₂

Substituting the values from equations (i) and (ii), we get

D₁ = 150 – 50 √ 3

∴ D₁ = 50√3 (√3 – 1) m

Step 5: Now, it is given that this distance D₁ is covered in 2 minutes

We know that: Speed =

∴ Speed =

= 25 √3 (√3 – 1) m / min

= 25 x 60 √3 (√3 – 1) m / hr   (∵ 1 min = 1/60 hr)

= 1500 √3 (√3 – 1) m / hr

Therefore, the speed of the boat is 1500 √3 (√3 – 1) m / hr

Note: Since we are not given value of √3 in the question, we can leave the answer here itself). 

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