First, you need to draw the situation: you have three points on a straight line, A is the first kilometer stone, B is the second one and C is the foot of the hill. Let's call H the top of the hill. Connecting H with both A and B you get two triangles rectangle (BCH and ACH).
The triangle rectangle BCH has one other angle that is 45°, which means that it is an isosceles triangle, therefore BC = HC = height of the hill (h).
Now, consider the triangle ACH: it has an angle of 30° (which means that the other angle is 60°) and AC = BC + AB = h + 1 (where we considered h measured in km).
In such a particular triangle (30°- 60°- 90°), the ratio between the long and the short leg is √3, which means that:
√3·HC = AC
We can then write:
√3·h = (h + 1)
which can be solve for h:
√3·h - h = 1
h(√3 - 1) = 1
h =1/ (√3 - 1)
We can rationalize:
h = [tex] \frac{1}{ \sqrt{3} - 1} \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } [/tex]
= (√3 + 1) / 2
= 1.366 km
Therefore the correct answer is: the hill is 1.366km high
