Persons A and B are at the beach, their eyes are 5 ft and 6 ft, respectively, above sea level. How many miles farther out is Person B’s horizon than Person A’s? Hello!
I'm trying to solve this problem.
I have difficulty understanding this math problem.. I've tried to solve the problem using the symmetry of the triangles but I didn't get the right answer, and I can't seem to understand the "concept" of the horizon here.
So I'll be grateful if you give me some hints.

The radius of the earth is something like 4000 miles= about 20,000,000 feet. (Look up the exact radius of the earth to calculate the correct number yourself, I'm too lazy!)

The horizon is where you're looking at a tangent to the earth's surface. Since the tangent to a circle is perpendicular to the radius (if we assume the earth is a perfect sphere), we can set up a right angled triangle:

H...P

.

.

.

.

C

Where C is the centre of the earth, H is the horizon, and P is the person. CH = 20,000,000 feet. CP = 20,000,005 feet (in the case of the 5' person, that'd be me :) And there is a right angle at H.

So using Pythagoras' theorem, if d = PH, then:

d^2 = 20,000,005^2 - 20,000,000^2

=> d^2 = 200,000,025

=> d = 14142 feet (just under 3 miles)

For a 6' person:

d^2 = 20,000,006^2 - 20,000,000^2

=> d^2 = 240,000,036

=> d = 15491 feet

So the difference is 1349 feet or 0.256 miles

Repeat this exercise with the real radius of the earth (in feet) to get the correct answer, but it'll be pretty close to what I've given here

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

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im not sure