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The angles of elevation of an aeroplane ...

The angles of elevation of an aeroplane from two places 10 km apart are found to be `60^(@) and 30^(@)` respectively. Find the height of the aeroplane.

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To find the height of the aeroplane given the angles of elevation from two points 10 km apart, we can follow these steps: ### Step 1: Understand the Problem We have two points A and B that are 10 km apart. From point A, the angle of elevation to the aeroplane is 60°, and from point B, the angle of elevation is 30°. We need to find the height (h) of the aeroplane. ### Step 2: Set Up the Diagram Let: - Point C be the position of the aeroplane. - Point D be the point directly below the aeroplane on the ground. - The distance between points A and B is 10 km. ### Step 3: Define the Right Triangles From point A, we can form a right triangle ACD: - The height of the aeroplane (h) is the opposite side. - The distance AD (the base) can be defined as x. From point B, we can form another right triangle BCD: - The height of the aeroplane (h) is the opposite side. - The distance BD (the base) can be defined as (10 - x). ### Step 4: Apply Trigonometric Ratios Using the tangent function for both triangles: 1. From triangle ACD: \[ \tan(60°) = \frac{h}{x} \] Therefore, \[ h = x \cdot \tan(60°) = x \cdot \sqrt{3} \] 2. From triangle BCD: \[ \tan(30°) = \frac{h}{10 - x} \] Therefore, \[ h = (10 - x) \cdot \tan(30°) = (10 - x) \cdot \frac{1}{\sqrt{3}} \] ### Step 5: Set the Equations Equal Since both expressions equal h, we can set them equal to each other: \[ x \cdot \sqrt{3} = (10 - x) \cdot \frac{1}{\sqrt{3}} \] ### Step 6: Solve for x Multiply both sides by \(\sqrt{3}\) to eliminate the fraction: \[ 3x = 10 - x \] Now, add x to both sides: \[ 3x + x = 10 \] \[ 4x = 10 \] \[ x = \frac{10}{4} = 2.5 \text{ km} \] ### Step 7: Find the Height (h) Now substitute x back into the equation for h: \[ h = x \cdot \sqrt{3} = 2.5 \cdot \sqrt{3} \] Calculating this gives: \[ h \approx 2.5 \cdot 1.732 \approx 4.33 \text{ km} \] ### Final Answer The height of the aeroplane is approximately **4.33 km**. ---
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