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An observer standing on the deck of a ship just sees the top of a lighthouse which is 30 m above the sea level. If the height of the observer’s eye is 10 m above the sea level, then the distance of the observer from the lighthouse will be nearly
  • a)
    22.5 km
  • b)
    24.3 km
  • c)
    33.3 km
  • d)
    59.7 km
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
An observer standing on the deck of a ship just sees the top of a ligh...
Above the sea level is 10 m, then what is the distance between the ship and the lighthouse?

We can use trigonometry to solve the problem. Let's draw a diagram:

```
A (lighthouse)
|\
| \
| \
| \
| \
| \
| \
| \
| \
|θ \
| \
| \
| \
O (observer)
|
|
|
|
S (ship)
```

We want to find the distance OS. We know that the height of the lighthouse is 30 m, the height of the observer is 10 m, and we can assume that the angle θ is small enough that we can use the approximation tan(θ) ≈ θ. Then we have:

tan(θ) = OA / OA'
tan(θ) = 30 / OS
θ = tan^-1(30 / OS)

tan(θ) = OB / OA'
tan(θ) = 10 / OS
θ = tan^-1(10 / OS)

Setting these two expressions for θ equal to each other, we get:

tan^-1(30 / OS) = tan^-1(10 / OS)
30 / OS = 10 / OS
OS = 20

Therefore, the distance between the ship and the lighthouse is 20 meters.
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An observer standing on the deck of a ship just sees the top of a ligh...
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An observer standing on the deck of a ship just sees the top of a lighthouse which is 30 m above the sea level. If the height of the observer’s eye is 10 m above the sea level, then the distance of the observer from the lighthouse will be nearlya)22.5 kmb)24.3 kmc)33.3 kmd)59.7 kmCorrect answer is option 'C'. Can you explain this answer?
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