A drawbridge that is 104 ft in length is raised at its midpoint so that the uppermost points are 8 ft apart.?

Question

A drawbridge that is 104 ft in length is raised at its midpoint so that the uppermost points are 8 ft apart.?

Answer 1

What's your question?  Are you looking for the angle of elevation?

If so, then draw an isosceles trapezoid with bases of lengths 104 and 8, and the other "slanted" sides are 52 each.

Now drop a perpendicular from the tip of one of the raised roads to the base.  You now have a triangle that has a base length of 48 and a hypotenuse of 52.

Let A=angle of elevation

Then cos(A) = 48/52, and

A = arccos(48/52) = 22.61986495 degrees (or in radians, A = .3947911197 radians)

Or, if you're looking for the height of the tip of the raised road, then just use the Pythagorean Theorem:

height = sqrt(52^2 - 48^2) = sqrt(2704 - 2304) = sqrt(400) = 20 feet