Question

what is the value of x?

x = \boxed{\space}^\circ

Explanation:

Step1: Recognize straight angle sum

Angles around a point on a straight line sum to \(180^\circ\). So, \(71^\circ + 3x + (x + 31)^\circ + (x + 83)^\circ = 360^\circ\)? Wait, no—wait, the angles at point \(K\) on a straight line (since \(IJ\) and \(GH\) are straight lines? Wait, actually, the sum of angles around a point is \(360^\circ\), but the adjacent angles on a straight line sum to \(180^\circ\). Wait, looking at the diagram, the angles \(71^\circ\), \(3x\), \(x + 31^\circ\), and \(x + 83^\circ\) are around point \(K\), but actually, the straight line (like \(IJ\) is a straight line, so the angles on one side of \(IJ\) should sum to \(180^\circ\). Wait, maybe I misread. Let's check again: the angles at \(K\) are \(71^\circ\), \(3x\), \(x + 31^\circ\), and \(x + 83^\circ\). Wait, no—actually, the sum of angles around a point is \(360^\circ\), but maybe the two pairs of vertical angles? Wait, no, let's look at the straight line. For example, the line \(IJ\) is straight, so the angles on one side of \(IJ\) (at point \(K\)) should sum to \(180^\circ\). Wait, the angles given are \(71^\circ\), \(3x\), \(x + 31^\circ\), and \(x + 83^\circ\). Wait, maybe the correct approach is that the sum of all angles around \(K\) is \(360^\circ\), but actually, the straight lines (like \(JG\) and \(IH\) intersect at \(K\)), so vertical angles are equal, but here we have four angles: \(71^\circ\), \(3x\), \(x + 31^\circ\), \(x + 83^\circ\). Wait, no—wait, the correct equation is that the sum of angles around a point is \(360^\circ\), but actually, the angles on a straight line sum to \(180^\circ\). Wait, maybe the angles \(71^\circ\), \(3x\), \(x + 31^\circ\), and \(x + 83^\circ\) are arranged such that \(71^\circ + 3x + (x + 31)^\circ + (x + 83)^\circ = 360^\circ\)? No, that can't be. Wait, no—actually, the straight line (like \(IJ\) is horizontal, and \(JH\) is a line, so the angles adjacent to \(71^\circ\) on the straight line \(IJ\) would be \(71^\circ + (x + 83)^\circ + (x + 31)^\circ = 180^\circ\)? Wait, no, I'm confused. Wait, let's re-express: the four angles at \(K\) are \(71^\circ\), \(3x\), \(x + 31^\circ\), and \(x + 83^\circ\). Since they are around a point, their sum is \(360^\circ\). So:

\(71 + 3x + (x + 31) + (x + 83) = 360\)? Wait, no, that would be four angles, but maybe two pairs of angles. Wait, no, let's check the diagram again. The angles are: \(71^\circ\), \(3x\), \(x + 31^\circ\), \(x + 83^\circ\). So adding them up:

\(71 + 3x + x + 31 + x + 83 = 360\)

Combine like terms:

\(71 + 31 + 83 + 3x + x + x = 360\)

\(185 + 5x = 360\)

Subtract 185:

\(5x = 360 - 185 = 175\)

Then \(x = 175 / 5 = 35\). Wait, but let's verify. If \(x = 35\), then \(3x = 105\), \(x + 31 = 66\), \(x + 83 = 118\). Then sum: \(71 + 105 + 66 + 118 = 71 + 105 = 176; 176 + 66 = 242; 242 + 118 = 360\). Yes, that works. So the equation is \(71 + 3x + (x + 31) + (x + 83) = 360\).

Step2: Solve the equation

Combine like terms:

\(71 + 31 + 83 + 3x + x + x = 360\)

\(185 + 5x = 360\)

Subtract 185 from both sides:

\(5x = 360 - 185 = 175\)

Divide both sides by 5:

\(x = 175 / 5 = 35\)

Answer:

\(35\)