Problem: The radius of the earth is approximately 4,000 mi. Approximate to within 10 mi the distance from the horizon to a plane flying at an altitude of 4 mi.

Solution: Since the distances from the airplane to the horizon, from the horizon to the earth's center, and from the earth's center to the airplane form a right triangle, we shall use the Pythagorean Theorem to find the unknown distance from the plane to the horizon.

From the above figure, we know that one of the legs of our triangle is 4000 mi. The figure also shows that the distance from the earth's center to the airplane is 4000 mi (earth's radius) + 4 mi (the airplane's altitude). Therefore, the hypotenuse is 4004 mi. On the other hand, we don't know the length of the other leg from the figure (which happens to be the quantity we're interested in).

We shall now use our known information along with the Pythagorean Theorem. We know that for any right triangle with legs $a$ and $b$, and hypotenuse $c$, that $$a^2+b^2=c^2.$$ Substituting the known information from the previous paragraph into the above equation, we have $$4000^2+b^2=4004^2,$$ where $b$ is the unknown distance from the airplane to the horizon.

We shall now solve this equation: $$4000^2+b^2=4004^2$$ $$b^2=4004^2-4000^2$$ $$b^2=(4004-4000)(4004+4000)$$ $$b^2=4 \cdot 8004$$ $$b^2=32016.$$ Taking the positive root we have that $$b=\sqrt{32016} \approx 178.9 \mbox{ mi.}$$ Therefore, the approximate distance^* from the plane to the horizon is 178.9 miles.

^* Since this is a straight-line distance, this is only an approximation of the curved distance that the plane will travel.

Print Tutor's Name Here:____________________________________

Tutor's Signature Here:_____________________________________