From the top of a light-hours, the angles of depression of two ships on the opposite sides are observed to be `30^(@)" and "45^(@)`. If the height of the light-house, is 90 m and the line joining the two ships passes through the light- house, then find the distance between the ships.

Answer

To solve the problem step by step, we will use trigonometric ratios and the properties of right triangles.

### Step 1: Understand the Problem
We have a lighthouse of height 90 m. The angles of depression to two ships on opposite sides are 30° and 45°. We need to find the distance between the two ships.

### Step 2: Draw the Diagram
Draw a vertical line representing the lighthouse. Mark the top of the lighthouse as point A and the base as point B. From point A, draw lines to the two ships, marking the angles of depression as 30° and 45°. Let the ship at 30° be point C and the ship at 45° be point D.

### Step 3: Identify the Right Triangles
From point A (top of the lighthouse), drop perpendiculars to the horizontal line from points C and D. This creates two right triangles:
1. Triangle ABC (with angle CAB = 30°)
2. Triangle ABD (with angle DAB = 45°)

### Step 4: Calculate Distance BC (from Lighthouse to Ship C)
Using the tangent function for triangle ABC:
\[ \tan(30°) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \]
Where AB = height of the lighthouse = 90 m.

Thus,
\[ \tan(30°) = \frac{90}{BC} \]
We know that \(\tan(30°) = \frac{1}{\sqrt{3}}\). Therefore:
\[ \frac{1}{\sqrt{3}} = \frac{90}{BC} \]
Cross-multiplying gives:
\[ BC = 90\sqrt{3} \]

### Step 5: Calculate Distance BD (from Lighthouse to Ship D)
Using the tangent function for triangle ABD:
\[ \tan(45°) = \frac{AB}{BD} \]
Again, AB = 90 m.

Thus,
\[ \tan(45°) = \frac{90}{BD} \]
We know that \(\tan(45°) = 1\). Therefore:
\[ 1 = \frac{90}{BD} \]
Cross-multiplying gives:
\[ BD = 90 \]

### Step 6: Calculate the Total Distance Between the Ships
The total distance between the two ships (CD) is the sum of distances BC and BD:
\[ CD = BC + BD = 90\sqrt{3} + 90 \]

### Step 7: Substitute the Value of \(\sqrt{3}\)
Using the approximate value of \(\sqrt{3} \approx 1.732\):
\[ CD = 90(1.732) + 90 \]
Calculating this gives:
\[ CD \approx 155.88 + 90 = 245.88 \text{ m} \]

### Final Answer
The distance between the two ships is approximately **245.88 meters**.

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