Pre-cal question. Please help!!!?

Question

In traveling across flat land, you notice a mountain directly in front of you. Its angle of elevation( to the peak) is 3.5 degrees. After you drive13 miles closer to the mountain, the angle of elevation is 9 degrees. Approxiamte the height of the mountain.

Answer 1

Imagine 2 right triangles, ABC and XYZ.
TRI. ABC has a 3.5 degree angle with X as its bottom leg and Y as its side leg. TRI. XYZ has a 9 degree angle with (X - 13) and its bottom leg and Y as its side leg. Using trig. ratios,
tan(3.5) = Y/X, and
tan(9) = Y/(X - 13).
Make both equations = to Y and plug them into your calculator....
ABC : Y = (tan(3.5))X
XYZ : Y = (tan(9))(X - 13)
then intersection ( 2nd trace, 5, enter, enter, enter)...
the Y value = height, aprox. 1.2953234

Answer 2

Tan(angle) =O/A where O is the height of the mountan, A is the distance to the peak at the first measurement...
A(tan(3.5)=(A-13)tan(9) solve for A...

Answer 3

Let
x + 13 = initial horizontal distance to peak
x = second horizontal distance to peak
h = height of mountain

tan(3.5°) = h/(x + 13)
tan(9°) = h/x

Solve for x.

x + 13 = h/tan(3.5°)
x = h/tan(3.5°) - 13

x = h/tan(9°)

Set the two equations equal and solve for h.

x = h/tan(3.5°) - 13 = h/tan(9°)
h/tan(3.5°) - h/tan(9°) = 13

h[1/tan(3.5°) - 1/tan(9°)] = 13

h = 13 / [1/tan(3.5°) - 1/tan(9°)]
h ≈ 1.29532337 miles = 6839.307391 ft