Answer
The plane is a distance of 182 miles from $X$
Work Step by Step
Suppose that the plane flies from point $X$ to point $Y$ and then to point $Z$. We need to find the distance between $X$ and $Z$. First we can find the angle formed at the point $Y$:
The angle from the vertical formed at the point $X$ is $180^{\circ}-125^{\circ} = 55^{\circ}$
The angle $Y = (270^{\circ}-230^{\circ})+(90^{\circ}-55^{\circ}) = 75^{\circ}$
We can use the law of cosines to find the distance $XZ$:
$XZ^2 = XY^2+YZ^2-2(XY)(YZ)~cos~Y$
$XZ = \sqrt{XY^2+YZ^2-2(XY)(YZ)~cos~Y}$
$XZ = \sqrt{(180~mi)^2+(100~mi)^2-(2)(180~mi)(100~mi)~cos~75^{\circ}}$
$XZ = \sqrt{33082.5~mi^2}$
$XZ = 182~mi$
The plane is a distance of 182 miles from $X$
