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As observed from the top of a lighthouse...

As observed from the top of a lighthouse, 45 m high above the sea-level, the angle of depression of a ship, sailing directly towardsit, changes from `30^(@)` to `45^(@)` . The distance travelled by the ship during the period of observation is: (Your answer should be correct to one decimalplace.)

A

32.9 m

B

33.4 m

C

36.9 m

D

24.8 m

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the distance traveled by the ship as observed from the top of a lighthouse. Here are the steps to arrive at the solution: ### Step-by-Step Solution: 1. **Understanding the Problem**: - The height of the lighthouse is given as 45 meters. - The angle of depression changes from \(30^\circ\) to \(45^\circ\). 2. **Drawing the Diagram**: - Draw a vertical line representing the lighthouse of height 45 m. - Mark the point at the top of the lighthouse as point A. - Mark the position of the ship when the angle of depression is \(30^\circ\) as point B and when it is \(45^\circ\) as point C. 3. **Using Trigonometry**: - For angle \(30^\circ\): - In triangle AOB (where O is the point on the sea level directly below A), we can use the tangent function: \[ \tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{45}{OB} \] - Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we can set up the equation: \[ \frac{1}{\sqrt{3}} = \frac{45}{OB} \implies OB = 45\sqrt{3} \] 4. **Calculating Distance OB**: - Calculate \(OB\): \[ OB = 45\sqrt{3} \approx 45 \times 1.732 = 77.0 \text{ meters} \] 5. **For angle \(45^\circ\)**: - In triangle AOC, using the tangent function: \[ \tan(45^\circ) = \frac{45}{OC} \] - Since \(\tan(45^\circ) = 1\): \[ 1 = \frac{45}{OC} \implies OC = 45 \text{ meters} \] 6. **Calculating Distance BC**: - The distance traveled by the ship (BC) is the difference between distances OB and OC: \[ BC = OB - OC = 77.0 - 45 = 32.0 \text{ meters} \] 7. **Final Answer**: - The distance traveled by the ship during the period of observation is approximately \(32.0\) meters.
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