If the angle of elevation of a cloud from a point 200 m above a lake is $30^\circ$ and the angle of depression of its reflection in the lake is $60^\circ$, then the height of the cloud above the lake is ______.

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If the angle of elevation of a cloud from a point 200 m above a lake is $30^\circ$ and the angle of depression of its reflection in the lake is $60^\circ$, then the height of the cloud above the lake is 400 m. Explanation: Let the height of the point of observation above the lake be H = 200 m. Let the height of the cloud above the lake be h. Let the horizontal distance between the observer and the cloud be d. Angle of Elevation ($30^\circ$): The height of the cloud above the point of observation is (h – 200) m. Using the definition of tangent: `tan(30^\circ) = "Opposite"/"Adjacent" = (h - 200)/d = 1/sqrt3 = (h - 200)/d` d = (h – 200) $\sqrt$3 ...(Eqn. 1) Angle of Depression ($60^\circ$): The reflection of the cloud in the lake is at a depth equal to its height above the water, which is h meters below the surface. The observer is 200 m above the lake, so the total vertical distance from the observer to the reflection is (200 + h) m. `tan(60^\circ) = "Opposite"/"Adjacent" = (h - 200)/d sqrt3 = (200+h)/d d = (200+h)/sqrt3` ...(eqn. 2) Solving for h: Equating the two expressions for d: `(h - 200) sqrt3 = (200+h)/sqrt3` Multiply both sides by $\sqrt$3 3 ( h - 200) = 200 + h 3h - 600 = 200 + h 3h - h = 200 + 600 2h = 800 h = 400

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