From the top of a vertical tower, 319 feet above the the surface of the earth, the angle of depression to a doghouse is How far is it from the doghouse to the foot of the tower?

Answer 1

call the angle of depression A

then there's a right triangle with height of the tower as height of the right triangle

tanA = the height divided by the base (distance to the doghouse from the tower)

tanA =319/b

b = 319/tanA

0<A<90, given curvature of the earth and limited horizon, it's closer to 30<A<60

then tanA is between 1/sqr3 and sqr3

1/sqr3 = .577

.577 < tanA < 1.732

553 > b > 184

take an average as best guess or estimate, guestimate b=about 369 feet

average likely angle is near 45 degrees. tangent of 45 = 1

then base = 319/1 = 319 feet.

I'd go with guessing 319 as the mean estimate, with least error.

If it were noon at the equator, the angle of depression could be near 90 degrees with no shadow, and the doghouse right next to the base of the tower.

If it were sunset or sunrise, the shadow would be nearly infinite with the dog house infinitely far away from the tower. Half of infinity plus zero is still infinity. But that's poor logic given clear limits to how far it could be given the doghouse is on earth.

Anyway, the distance you want is = 319/tanA where A is the angle of depression.