\(\triangle ABC\) and \(\triangle PQR\) are equilateral triangles with vertices \(B\) and \(P\) on line segment \(MN\). The triangles intersect at two points, \(X\) and \(Y\), as shown.
If \(\angle NPQ = 75\degree\) and \(\angle MBA = 65\degree\), determine the measure of \(\angle CXY\).
In any equilateral triangle, all sides are equal in length and each angle measures \(60\degree\).
Since \(\triangle ABC\) and \(\triangle PQR\) are equilateral, \(\angle ABC = \angle ACB = \angle CAB = \angle QPR = \angle PRQ = \angle RQP = 60\degree\).
Since the angles in a straight line sum to \(180\degree\), we have \(180\degree = \angle MBA + \angle ABC + \angle YBP
= 65\degree + 60\degree + \angle YBP\).
Rearranging, we have \(\angle YBP = 180\degree
- 65\degree - 60\degree = 55\degree\).
Similarly, since angles in a straight line sum to \(180\degree\), we have \(180\degree = \angle NPQ + \angle QPR + \angle YPB
= 75\degree + 60\degree + \angle YPB\).
Rearranging, we have \(\angle YPB = 180\degree
- 75\degree - 60\degree = 45\degree\).
Since the angles in a triangle sum to \(180\degree\), in \(\triangle BYP\) we have \(\angle YPB + \angle YBP + \angle BYP =
180\degree\), and so \(45\degree +
55\degree + \angle BYP = 180\degree\).
Rearranging, we have \(\angle BYP = 180\degree
- 45\degree - 55\degree = 80\degree\).
When two lines intersect, vertically opposite angles are equal. Since \(\angle XYC\) and \(\angle BYP\) are vertically opposite angles, we have \(\angle XYC = \angle BYP = 80\degree\).
Again, since angles in a triangle sum to \(180\degree\), in \(\triangle XYC\) we have \(\angle XYC + \angle XCY + \angle CXY = 180\degree\). We have already found that \(\angle XYC = 80\degree\), and since \(\angle XCY = \angle ACB\), we have \(\angle XCY = 60\degree\). So, \(\angle XYC + \angle XCY + \angle CXY = 180\degree\) becomes \(80\degree + 60\degree + \angle CXY = 180\degree\). Rearranging, we have \(\angle CXY = 180\degree - 80\degree - 60\degree = 40\degree\).
Therefore, \(\angle CXY = 40\degree\).