let the nearest location be point A.
let the farthest location be point B.
let the location underneath the airplane be point D.
2 right triangles are formed with the first triangle being within the second triangle.
the right triangles are CAD and CBD.
angle CAD is equal to 65 degrees.
angle CBD is equal to 50 degrees.
the line DB contains the line DA and AB.
DA is the base of triangle CAD.
DB is the base of triangle CBD.
the length of line AB is equal to 3.5.
the length of line DA is equal to x.
the length of line CD is equal to y.
the tangent of angle CAD is equal to y/x.
since angle CAD is 65 degrees, then the tangent of 65 degrees is equal to y/x.
the tangent of angle CBD is equal to y/(x+3.5).
since angle CBD is equal to 50 degrees, then the tangent of 50 degrees is equal to y/(x+3.5)
solve both of these equations for y and you get:
y = x * tan(65)
y = (x + 3.5) * tan(50)
replace y with x * tan(65) in the second equation to get:
x * tan(65) = (x + 3.5) * tan(50)
solve this equation for x to get:
x = (3.5 * tan(50)) / (tan(65) - tan(50).
you will get x = 4.377982686.
since y = x * tan(65), solve for y using this value of x and you will get:
y = 9.388614169
this is actually as far as you need to go because you were asked the height of the plane which is the value of y.
the rest is more than you need but can be viewed if you wish, since it used to confirm everything is correct based on the values of x and y.
solve for x + 3.5 using this value of x to get:
x+3.5 = 7.877982686
the altitude of the plan is equal to y which is equal to 9.388614169.
tan(65) = 2.144506921
y/x = 9.388614169 / 4.377982686 = 2.144506921.
the values for tan(65) and y/x are the same so the measures of triangle CAD are correct.
tan(50) = 1.191753593
y/(x+3.5) = 9.388614169 / 7.877982686 = 1.191753593.
the values for tan(50) and y/(x+3.5) are the same so the measures of triangle CBD are correct.
a picture of your triangle is shown below:
in the calculations below the picture, i abbreviated tan to T as i got forther down into them.
anywhere you see T(...), that's the same as tan(...).
