An observer 1.5m tall is 23.5m away from a tower 25m high. The angle of elevation of the top of the tower from the eye of the observer isa)30°b)45°c)60°d)none of theseCorrect answer is option 'B'. Can you explain this answer?

Question

An observer 1.5m tall is 23.5m away from a tower 25m high. The angle of elevation of the top of the tower from the eye of the observer isa)30&deg;b)45&deg;c)60&deg;d)none of theseCorrect answer is option 'B'. Can you explain this answer?

Answer

Answer

Answer

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

```
A
/|
/ |
/ | 25m
/ |
/ |
/θ |
/ |
/_______|
23.5m B
Observer Top of tower
```

In this diagram, A represents the top of the tower, B represents the observer, and θ represents the angle of elevation of the top of the tower from the eye of the observer.

We know that AB = 25m (the height of the tower) and OB = 23.5m (the distance from the observer to the tower). We want to find θ.

We can use the tangent function to find θ:

tan θ = opposite / adjacent

In this case, the opposite side is AB (25m) and the adjacent side is OB (23.5m):

tan θ = 25 / 23.5

Using a calculator, we find that:

θ ≈ 52.2°

However, this is the angle of elevation from the ground. Since the observer is 1.5m tall, their eye level is 1.5m above the ground. So we need to subtract the inverse tangent of 1.5/23.5 to find the angle of elevation from the eye of the observer:

θ' = θ - tan^-1(1.5/23.5)

Using a calculator, we find that:

θ' ≈ 30°

Therefore, the angle of elevation of the top of the tower from the eye of the observer is approximately 30 degrees.

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