greatwhiteshark said:
A forest ranger is walking on a path inclined at 5 degrees to the horizontal directly toward a 100-foot-tall fire observation tower. The angle of elevation of the top of the tower is 40 degrees. How far is the ranger from the tower at this time?
My first impression was to bug you about submitting so many of the exact same problem type. A quick read shows this problem to be just a hair more interesting.
What we know:
Gound is inclined at 5º
Tower is 100 ft
Angle of elevation to top of tower is 40º
That last one is the interesting part. It is an "angle of elevation", which is ALWAYS measured from the horizontal. This makes the angle from the ground only 35º.
Now we have a little problem. What does "how far" actually mean? Does it mean the horizontal distance, even though the ranger would have to dig through the ground? Does it mean the distance along the ground being walked. I tend toward the latter, but I'm only guessing.
Draw a picture:
I started with a right triangle, ABC
-- A is the location of the ranger
-- B is the Right angle (which will be underground)
-- C is the top of the tower
Draw another line from the ranger, sloping up from the base of the Right triangle. Extend it until it hits the opposite side. Lable the intersection P
Label angle PAB = 5º
Label angle PAC = 35º
Label segment PC = 100 ft
We don't know segment BP, label it 'z'.
We don't know segment AP, label it 't'.
We don't know segment AB, label it 's'.
Note: The height of the right triangle is z + 100.
Time for Trigonometry & Geometry to help us out.
tan(40º) = (100 + z)/s
tan(5º) = z/s
s<sup>2</sup> + z<sup>2</sup> = t<sup>2</sup>
That should be enough.
I get 133.048 ft horizontal and 133.556 ft along the inclined ground.
