Question

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what is the distance to the earth’s horizon from point p?
enter your answer as a decimal in the box. round only your final answer to the nearest tenth.
x = \boxed{} mi
(image: a circle labeled earths center with radius 3959 mi, a point p outside the circle 3.1 mi from the circles edge, and a tangent from p to the circle labeled horizon, forming a right triangle with the radius and the line from earths center to p)

Explanation:

Step1: Apply the Pythagorean theorem

The line from the horizon to the Earth's center is perpendicular to the line from the horizon to point \(P\) (tangent - radius property). Let the distance from the Earth's center to point \(P\) be \(r + h\) (\(r = 3959\) mi, \(h=3.1\) mi), the radius of the Earth be \(r = 3959\) mi, and the distance to the horizon be \(x\). By the Pythagorean theorem \(x^{2}+r^{2}=(r + h)^{2}\).
Expanding \((r + h)^{2}=r^{2}+2rh+h^{2}\), then \(x^{2}+r^{2}=r^{2}+2rh + h^{2}\). Subtracting \(r^{2}\) from both sides gives \(x^{2}=2rh+h^{2}\).

Step2: Substitute the values

Substitute \(r = 3959\) mi and \(h = 3.1\) mi into the formula \(x^{2}=2rh+h^{2}\).
\(x^{2}=2\times3959\times3.1+(3.1)^{2}\).
First, calculate \(2\times3959\times3.1 = 2\times3959\times3.1=24545.8\) and \((3.1)^{2}=9.61\).
Then \(x^{2}=24545.8 + 9.61=24555.41\).

Step3: Solve for \(x\)

Take the square - root of \(x^{2}\), \(x=\sqrt{24555.41}\approx156.7\)

Answer:

\(156.7\)