A tourist is traveling by car due east on a road through a forest. From point A on the road, a path heads due?

Question

northeast to an observation tower. The tower is 4 mi from A. and a ranger in the tower can see for a distance of 3.5 mi. For what length of road can the ranger see the tourist's car?

Answer 1

Assuming a forest of uniform height, there are several possible answers to this question. I'll limit myself to the two extremes. The car can be seen while it travels along the road for :

(i) Only a few yards or

(ii) About 4.1231... miles.

(i) If the trees are too tall (that is, several times taller than the car, which would not be unusual), the car could only be seen for the length of time that it crosses the clear line of sight down the straight NE-SW road of length 3.5 miles from the foot of the tower towards the road that the car is travelling on. The reason for this is that the average road is not particularly wide, and you couldn't even SEE cars on the "other" side of the road if the trees were too tall, from an observation tower of a realistic height.

(ii) If we ignore such realism (and the reason why the ranger's vision is limited to be 3.5 miles), so that the car can still be seen for ALL of the time it is within a range of 3.5 miles from the tower, the length of road for which it could be seen (with one more assumption) is 4.1231... miles. This comes about as follows:

Assuming the car's path CONTINUES on a STRAIGHT LINE east, the distance of the tower T from the road at point N is 4/sqrt(2) miles. Let a circle of radius R = 3.5 miles intersect the car's path for the first time at F. (There will be another, corresponding "last time" of intersection at L, beyond the point N.)

Then in triangle TFN, Pythagoras's Theorem gives

(FN)^2 = (TF)^2 - (NT)^2 = (3.5)^2 - (4/sqrt(2))^2 miles^2

= (12.25 - 8) miles^2.

So FN = (4.25)^(1/2) miles = 2.06155... miles.

But the car can be seen for a distance along the road of FL = 2 FN

= 4.1231... miles.

Answer 2

The best way to answer this question is to draw it!
The point is that the line joining point A and the tower have a 45 degree angle with the road the car is taking. That means if we draw a perpendicular line from the tower to the road, we have a triangle with two 45 degree angles. So the length of the line from tower to the road is the same as the one from point A to the point the lines meet.
Now since this triangle has a 90 degree angle (the line from tower is perpendicular to the road) then x^2 + x^2 = 4^2 (where x is the length of the other edges of the triangle).
so 2x^2 = 16 => x=sqrt8 = 2sqrt2

Now imagine a circle with radius 3.5 and center of the tower. This circle intersects with the road in two points. The length of intersection is what we want and we call it y.
Now again because of the triangle formed we have
(y/2)^2 + x^2 = 3.5^2
so y^2 = 4(3.5^2 - 8)

Answer 3

Let T be the location of the tower.
Drop a perpndicular from T to the road intersecting it at H.
Then TH = 2sqrt(2) since triangle is a right isosecles triangle.
Now with Tas the center and 3.5 as the radius draw an arc intersecting the road at poits B and C.
TC = 3.5 and TH = 2sqrt(2).
Thus HC = sqrt(3.5^2 -(2sqrt(2))^2) = 4.25
So BC = 2* HC = 9 miles ranger sees car.