Question

From a boat which is coming towards a bridge, at any instant the angle of elevation of the bridge is 30°. After travelling 4 minutes with the same velocity, the angle of elevation of the bridge becomes 60°. How much more time will be taken by the boat to reach the bridge?

Hint:

When dealing with angles of elevation, use the tangent function to relate the height and distance, and set up equations based on the change in distance over time.

Answer 1

Let the initial distance of the boat from the bridge be \( d \) and the velocity of the boat be \( v \). The time taken to travel the distance \( d \) to the bridge is \( \frac{d}{v} \).

Step 1: At the first position, the angle of elevation of the bridge is 30°, so we have: \[ \tan 30^\circ = \frac{h}{d}. \] Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), the equation becomes: \[ \frac{1}{\sqrt{3}} = \frac{h}{d} \quad \Rightarrow \quad h = \frac{d}{\sqrt{3}} \quad \text{(Equation 1)}.
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Step 2: After 4 minutes, the angle of elevation becomes 60°, so we have: \[ \tan 60^\circ = \frac{h}{d - v \times 4}. \] Since \( \tan 60^\circ = \sqrt{3} \), the equation becomes: \[ \sqrt{3} = \frac{h}{d - v \times 4} \quad \Rightarrow \quad h = \sqrt{3}(d - v \times 4) \quad \text{(Equation 2)}.
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Step 3: Now, substitute \( h = \frac{d}{\sqrt{3}} \) from Equation 1 into Equation 2: \[ \frac{d}{\sqrt{3}} = \sqrt{3}(d - v \times 4). \] Simplify: \[ d = 3(d - 4v) \quad \Rightarrow \quad d = 3d - 12v \quad \Rightarrow \quad 2d = 12v \quad \Rightarrow \quad d = 6v.
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Step 4: Thus, the remaining distance the boat has to travel is \( d - 4v \). The time taken to cover this distance is: \[ \frac{d - 4v}{v} = \frac{6v - 4v}{v} = \frac{2v}{v} = 2 \text{ minutes}.
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Conclusion: The boat will take 2 more minutes to reach the bridge.