You have the following information about the triangle formed between A, B and C (where the airplane is) points:
distance AB = 6 mi
m∠x = 25°
m∠y = 55°
In order to determine the distance from point B to C, first find the measure of the internal angles of the triangle.
Consider the line of the trajectory of the airplane and the line of the surface of the ground as parallel lines. Then, you can assume that both segments AC and BC are secant line which cross two parallel lines.
The previous conclusion allows you to determine the angles m∠ABC and m∠CAB as equal to m∠y and m∠x respectively.
m∠ABC = m∠y = 55°
m∠CAB = m∠x = 25°
Next, take into account the sum of all interioir angles of a triangle is equal to 180°, then for the angle m∠ACB you have:
m∠ACB = 180° - m∠ABC - m∠CAB = 180° - 55° - 25° = 100°
Next, take into account you have information about one side of the triangle and all interioir angles. Then, you can use the law of sines to find the value of the length of segmen AC, just as follow:
AC/sin(m∠CAB) = AB/sin(m∠ACB) = BC/sin(m∠ABC) law of sines
take the second equation:
AB/sin(m∠ACB) = BC/sin(m∠ABC) replace the values of the parameters
6/(sin100°) = BC/(sin55°) solve for BC
(6/(sin100°))(sin55°) = BC
4.99 = BC
BC ≈ 5.00
Hence, the distance from the point B to the airplane is 5 mi.