A girl is observing a bird sitting at a height of 500 m from the ground. The angle of elevation of the bird from the girl's eyes is `45^(@)` at the initial position. After some time, the girl moves on the ground towards the bird by 50 m and bird flies away from her. At this point, the angle of elevation reduces to `30^(@)` . Find the distance travelled by the bird. Take `sqrt(3) = 1.73 `

Options

A. 415 m
B. 915 m
C. 315 m
D. 365 m

Answer

To solve the problem step by step, we will use trigonometric principles to find the distance traveled by the bird.

### Step 1: Initial Setup
Let’s denote:
- The height of the bird (AB) = 500 m
- The initial position of the girl (C) is at a distance (BC) from the base of the bird (B).
- The angle of elevation from the girl to the bird at the initial position (C) is 45°.

### Step 2: Using the Tangent Function
From triangle ABC, we can use the tangent of the angle of elevation:
\[ \tan(45°) = \frac{AB}{BC} \]
Since \(\tan(45°) = 1\):
\[ 1 = \frac{500}{BC} \implies BC = 500 \text{ m} \]

### Step 3: Girl Moves Towards the Bird
After moving 50 m towards the bird, the new position of the girl is F. Therefore, the new distance from the bird (BF) is:
\[ BF = BC - 50 = 500 - 50 = 450 \text{ m} \]

### Step 4: New Angle of Elevation
At this new position (F), the angle of elevation to the bird is now 30°. We can again use the tangent function:
\[ \tan(30°) = \frac{AB}{BF + y} \]
Where \(y\) is the distance the bird has moved away from its original position.

### Step 5: Substitute Values
Since \(\tan(30°) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{500}{450 + y} \]

### Step 6: Cross-Multiply
Cross-multiplying gives:
\[ 450 + y = 500\sqrt{3} \]

### Step 7: Substitute \(\sqrt{3}\)
Using the given value \(\sqrt{3} = 1.73\):
\[ 450 + y = 500 \times 1.73 \]
Calculating the right side:
\[ 450 + y = 865 \]

### Step 8: Solve for \(y\)
Now, solving for \(y\):
\[ y = 865 - 450 = 415 \text{ m} \]

### Conclusion
The distance traveled by the bird is **415 m**.

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