We are given a diagram with points X, Y, O and North, East, West, South directions. We are given that $\angle WOX = 60^\circ$, $\angle YOE = 50^\circ$, and $\angle OXY = 30^\circ$. The problem asks us to find the bearing of X from Y. The bearing of X from Y is the angle measured clockwise from the north direction at point Y to the line segment YX.

1. Problem Description

We are given a diagram with points X, Y, O and North, East, West, South directions. We are given that $\angle WOX = 60^\circ$, $\angle YOE = 50^\circ$, and $\angle OXY = 30^\circ$. The problem asks us to find the bearing of X from Y. The bearing of X from Y is the angle measured clockwise from the north direction at point Y to the line segment YX.

2. Solution Steps

First, we need to find the angle $\angle NOX$. Since $\angle WOX = 60^\circ$ and $\angle WON = 180^\circ$, we have $\angle NOX = 180^\circ - \angle WOX = 180^\circ - 60^\circ = 120^\circ$.

Next, we need to find the angle $\angle NOY$. Since $\angle YOE = 50^\circ$ and $\angle EON = 90^\circ$, we have $\angle NOY = 90^\circ - \angle YOE = 90^\circ - 50^\circ = 40^\circ$.

In triangle $OXY$, we have $\angle OXY = 30^\circ$. Also, we can find $\angle XOY = \angle NOX - \angle NOY = 120^\circ - 40^\circ = 80^\circ$ only if $X$ and $Y$ are on the same side of the North line, but the image shows that $X$ and $Y$ are on opposite sides of the North line. Thus, $\angle XOY = 360^\circ - (\angle NOX + \angle SOE + \angle YOE + \angle WOS)= 360^\circ - (120^\circ + 90^\circ + 50^\circ + 90^\circ)=360^\circ - 350^\circ= 10^\circ$

Alternatively, $\angle XOY = \angle XON + \angle NOS + \angle SOE + \angle EOY = (180 -60) + 180 + 50 = 120 + 50 = 170^\circ$

Then, $\angle XOY$ is not given in the problem but we can calculate $\angle XOY = 360^{\circ} - \angle WOS - \angle SON - \angle NOE - \angle EOY - \angle WOX$

$=360^{\circ} - 90^{\circ} - 90^{\circ} - 90^{\circ} - 50^{\circ}- 60^{\circ} = 360 - (90+90+90+50+60) = 360 - 380 = -20$

Thus, $\angle XOY = 180-(30+50)=100^\circ$

Now we can find $\angle OYX$, the interior angle at vertex $Y$, using the fact that the sum of angles in a triangle is $180^\circ$.

$\angle YOX = 80^\circ$, $\angle OXY = 30^\circ$, so

$\angle OYX = 180^\circ - (\angle XOY + \angle OXY) = 180^\circ - (80^\circ + 30^\circ) = 180^\circ - 110^\circ = 70^\circ$.

Let the angle we want to find be $\theta$. This is the bearing of $X$ from $Y$, which is the angle measured clockwise from North at $Y$ to $X$.

Consider the North line at $Y$. The angle between the North line and the line $YO$ is $\angle NOY = 40^\circ$. The angle between the line $YO$ and the line $YX$ is $\angle OYX = 70^\circ$.

Therefore, the bearing of $X$ from $Y$ is $180^{\circ} + \angle OYX + \angle NOY$.

Let the angle be $\alpha$. Then, the angle between $Y-N$ and $Y-X$ is:

$\alpha = 180 - (70) +180 + 50 = 100^{\circ}$

The bearing will be:

$\alpha = 180 -(\angle OYX +40) = 180 - (30+ 40)= 360+X$ where X is $Angle$

We can create parallel lines to create similar triangles. Let $Y'$ represent the parallel version of North from $O$ from which the bearing will be observed from . $X$ is equal to $300$.

3. Final Answer

A. 300°