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You should really try to solve this yourself first, in order to maximize the value in doing that. If you read this solution, please understand each step before proceeding to the next step, otherwise the lesson will be lost to you.

To determine the required elevation of an observer in order that he may be able to see an object on the earth thirty miles away, assuming the earth's radius is 3956 mi and the earth is a smooth sphere, first draw the triangle involved.

This is a right triangle, where the hypotenuse is the radius of the earth plus the elevation of the observer. One side is the radius of the earth to the object. The other side is the line of sight distance from the observer to the object, which is greater than 30 miles. That line of sight is tangent to the earth's circumference, at the point of the object, so the angle of line of sight relative to the radius at the object is 90 degrees.

The angle at the center of the earth is 360 degrees times 30 miles divided by the circumference of the earth, 2 pi 3956, or 24856, which is an angle of 0.4345 degrees.

The hypotenuse of a right triangle with one angle of 0.4345 degrees and side of 3956 is 3956 divided by cosine 0.4345 degrees, or 3956.1138 miles.

Subtract the radius, 3956 miles, and you get 0.1138 miles, or 601 feet.

Not asked, but answered for completeness, is that the line of sight distance is hypotenuse times sine 0.4345 degrees, or 30.0005751 miles, or 3.04 feet more than 30 miles.

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Q: What must be the elevation of an observer in order that he may be able to see an object on the earth thirty miles away assuming earth's radius is 3956 mi and earth is smooth sphere?
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