Question

find ( mangle i ) in rhombus ( fghi ).
( mangle i = square^circ )
submit

Explanation:

Step1: Recall rhombus angle property

In a rhombus, adjacent angles are supplementary (sum to \(180^\circ\)), and opposite angles are equal. Here, \(\angle I\) and \(\angle G\) are adjacent? Wait, no, actually in rhombus \(FGHI\), \(\angle I\) and \(\angle G\) – wait, no, let's check the labels. The rhombus is \(FGHI\), so the vertices are \(F, G, H, I\) in order. So \(\angle I\) and \(\angle G\) – wait, no, actually, in a rhombus, consecutive angles (adjacent) are supplementary. Wait, but also, opposite angles are equal. Wait, maybe \(\angle I\) and \(\angle G\) – no, wait, the angles given: \(\angle I = 9s + 11^\circ\) and \(\angle G = 7s + 37^\circ\). Wait, in a rhombus, opposite angles are equal? No, wait, no: in a parallelogram (rhombus is a parallelogram), opposite angles are equal, and consecutive angles are supplementary. So \(FGHI\) is a parallelogram, so \(\angle I\) and \(\angle G\) – wait, no, \(\angle F\) and \(\angle H\) are opposite, \(\angle I\) and \(\angle G\) are opposite? Wait, no, the order is \(F, G, H, I\), so sides \(FG\) and \(HI\) are parallel, \(FI\) and \(GH\) are parallel. So \(\angle I\) and \(\angle G\) are consecutive angles? Wait, no, \(\angle I\) is at vertex \(I\), between \(FI\) and \(HI\); \(\angle G\) is at vertex \(G\), between \(FG\) and \(HG\). So \(FI \parallel HG\) and \(FG \parallel HI\), so \(\angle I\) and \(\angle G\) are same - side interior angles? Wait, no, in a parallelogram, consecutive angles (like \(\angle I\) and \(\angle H\), \(\angle H\) and \(\angle G\), \(\angle G\) and \(\angle F\), \(\angle F\) and \(\angle I\)) are supplementary. Wait, maybe I made a mistake. Wait, actually, in a rhombus (parallelogram), opposite angles are equal. So \(\angle I\) and \(\angle G\) – are they opposite? Wait, no, the vertices are \(F, G, H, I\), so opposite angles are \(\angle F\) and \(\angle H\), \(\angle I\) and \(\angle G\)? Wait, no, that can't be. Wait, maybe the angles given are opposite? Wait, the problem is that in the diagram, \(\angle I = 9s + 11\) and \(\angle G = 7s + 37\). Wait, maybe they are opposite angles? Wait, no, in a rhombus, opposite angles are equal. So if \(\angle I\) and \(\angle G\) are opposite, then \(9s + 11 = 7s + 37\). Wait, that makes sense. Let's try that.

So set \(9s + 11 = 7s + 37\) (since opposite angles in a rhombus are equal).

Step2: Solve for \(s\)

Subtract \(7s\) from both sides: \(9s - 7s + 11 = 37\)
Simplify: \(2s + 11 = 37\)
Subtract 11 from both sides: \(2s = 37 - 11 = 26\)
Divide by 2: \(s = \frac{26}{2} = 13\)

Step3: Find \(m\angle I\)

Now substitute \(s = 13\) into the expression for \(\angle I\): \(9s + 11\)
\(9(13) + 11 = 117 + 11 = 128^\circ\)

Wait, but let's check with the other angle. If \(s = 13\), then \(\angle G = 7(13) + 37 = 91 + 37 = 128^\circ\). Wait, but in a rhombus, consecutive angles should be supplementary. Wait, maybe I mixed up opposite and consecutive. Let's re - examine.

In a rhombus (parallelogram), consecutive angles are supplementary (sum to \(180^\circ\)), opposite angles are equal. So if \(\angle I\) and \(\angle G\) are consecutive, then \(9s + 11+7s + 37 = 180\)

Let's try that.

Step1 (corrected): Recall consecutive angles in rhombus

In a rhombus (parallelogram), consecutive angles are supplementary. So \(\angle I\) and \(\angle G\) are consecutive angles (since the rhombus is \(FGHI\), the order is \(F - G - H - I - F\), so \(\angle G\) and \(\angle I\) are consecutive, with a side \(GH\) and \(HI\) etc.). So \(m\angle I + m\angle G=180^\circ\)

So \((9s + 11)+(7s + 37)=180\)

Step2 (corrected): Solve for \(s\)

Combine like terms: \(9s+7s + 11 + 37 = 180\)
\(16s+48 = 180\)
Subtract 48 from both sides: \(16s=180 - 48 = 132\)
\(s=\frac{132}{16}=\frac{33}{4}=8.25\)? Wait, that contradicts the earlier result. There must be a mistake in identifying consecutive/opposite angles.

Wait, let's look at the diagram. The angle at \(I\) is \(9s + 11\), the angle at \(G\) is \(7s + 37\). In the rhombus \(FGHI\), the sides \(FI\) and \(GH\) are parallel, and \(FG\) and \(HI\) are parallel. So \(\angle I\) and \(\angle F\) are consecutive, \(\angle F\) and \(\angle G\) are consecutive, \(\angle G\) and \(\angle H\) are consecutive, \(\angle H\) and \(\angle I\) are consecutive. Wait, maybe \(\angle I\) and \(\angle G\) are opposite? Let's check the labels again. The rhombus is drawn with \(F\) at the top, \(G\) at the bottom right, \(H\) at the bottom left, and \(I\) at the left. So the sides: \(FG\) is the right side, \(GH\) is the bottom side, \(HI\) is the left side, \(IF\) is the top side. So \(\angle I\) is between \(HI\) and \(IF\), \(\angle G\) is between \(FG\) and \(GH\). So \(IF\parallel GH\) and \(FG\parallel HI\). So \(\angle I\) and \(\angle G\) are opposite angles (since \(IF\parallel GH\) and \(FG\parallel HI\), the angles at \(I\) and \(G\) are formed by the same pair of parallel lines and should be equal). So our first approach was correct. But when we calculated, both angles were \(128^\circ\), and if we check consecutive angles, say \(\angle I\) and \(\angle H\), they should be supplementary. But we don't have \(\angle H\)'s expression. Wait, maybe the diagram has \(\angle I\) and \(\angle G\) as opposite angles.

Let's verify with \(s = 13\):

\(\angle I=9\times13 + 11=117 + 11 = 128^\circ\)

\(\angle G=7\times13+37 = 91+37 = 128^\circ\)

If they are opposite angles, that's fine (opposite angles in a rhombus are equal). Then consecutive angles would be \(\angle I\) and \(\angle H\), which should be \(180 - 128 = 52^\circ\), and \(\angle H\) and \(\angle G\) should also be supplementary (\(52+128 = 180\)), which works.

So the correct value of \(m\angle I\) is \(128^\circ\)

Answer:

\(128\)