Question

1. what is the measure of angle x?

options: 30°, 45°, 60°, 90°

Explanation:

Step1: Analyze triangle IJL

In right triangle \(IJL\), \(IJ = 11\) cm, \(JL\) is a side, and angle at \(J\) is \(45^\circ\). In a right triangle, if one angle is \(45^\circ\), it's an isosceles right triangle, so \(IJ = IL = 11\) cm. Also, \(JL=\sqrt{IJ^{2}+IL^{2}}=\sqrt{11^{2} + 11^{2}}=11\sqrt{2}\) cm, but maybe we can relate to triangle \(KJL\).

Step2: Analyze triangle KJL

In triangle \(KJL\), \(KJ\) and \(JL\): Wait, \(IJ = 11\), \(KJ = 22\)? Wait, no, \(IJ = 11\) cm, and in triangle \(JIL\), right - angled at \(I\), angle at \(J\) is \(45^\circ\), so \(IL = IJ = 11\) cm. Now, in triangle \(KJL\), \(JL\) and \(KJ\): Wait, \(KJ\) is adjacent, \(JL\) is... Wait, maybe look at triangle \(HJL\) first. In triangle \(HJL\), right - angled at \(I\), angle at \(H\) is \(30^\circ\), so \(JL=\frac{1}{2}HJ\) (since in a \(30 - 60 - 90\) triangle, the side opposite \(30^\circ\) is half the hypotenuse). But \(IJ = 11\), and in triangle \(JIL\), it's a \(45 - 45 - 90\) triangle, so \(JL = 11\sqrt{2}\)? Wait, no, maybe \(KJ = 22\) cm? Wait, the length of \(KJ\) is \(22\) cm? Wait, no, the diagram shows \(KJ\) as a side, and \(IJ = 11\) cm. Wait, maybe \(JL = 11\) cm? Wait, no, let's re - examine.

Wait, in triangle \(JIL\), right - angled at \(I\), \(\angle IJL=45^\circ\), so \(\angle ILJ = 45^\circ\), so \(IJ = IL = 11\) cm. Now, in triangle \(HJL\), right - angled at \(I\), \(\angle H = 30^\circ\), so \(\angle HLJ=60^\circ\) (since the sum of angles in a triangle is \(180^\circ\), \(90^\circ+30^\circ+\angle HLJ = 180^\circ\), so \(\angle HLJ = 60^\circ\)).

Now, in triangle \(KJL\), we can see that \(JL\) and \(KJ\): Wait, \(KJ = 22\) cm? Wait, no, the length of \(KJ\) is \(22\) cm? Wait, the diagram has \(KJ = 22\) cm and \(IJ = 11\) cm. Wait, maybe \(JL = 11\) cm? No, let's think about the angles.

Wait, another approach: In triangle \(JIL\), \(\angle IJL = 45^\circ\), right - angled at \(I\), so \(JL = IJ\sqrt{2}=11\sqrt{2}\)? No, maybe the key is that in triangle \(KJL\), we can find angle \(x\). Wait, triangle \(KJL\): if \(JL\) is equal to \(IJ = 11\) cm, and \(KJ = 22\) cm? Wait, no, maybe \(JL\) is \(11\) cm, and \(KJ = 22\) cm? Wait, no, let's look at the angles.

Wait, in triangle \(HJL\), \(\angle H = 30^\circ\), right - angled at \(I\), so \(\angle HLJ=60^\circ\). In triangle \(JIL\), \(\angle ILJ = 45^\circ\), so \(\angle KLJ=\angle HLJ-\angle ILJ = 60^\circ - 45^\circ=15^\circ\)? No, that doesn't seem right. Wait, maybe triangle \(KJL\) is a \(30 - 60 - 90\) triangle? Wait, no, let's check the sides. If \(IJ = 11\) cm, and \(KJ = 22\) cm, and \(JL = 11\) cm (from triangle \(JIL\), since it's a \(45 - 45 - 90\) triangle, \(JL=\sqrt{11^{2}+11^{2}} = 11\sqrt{2}\), no). Wait, maybe the length of \(JL\) is equal to \(IJ = 11\) cm, and \(KJ = 22\) cm, so in triangle \(KJL\), \(\sin x=\frac{JL}{KJ}\). If \(JL = 11\) and \(KJ = 22\), then \(\sin x=\frac{11}{22}=\frac{1}{2}\), so \(x = 30^\circ\)? Wait, no, \(\sin30^\circ=\frac{1}{2}\), but if \(JL\) is opposite angle \(x\), and \(KJ\) is the hypotenuse? Wait, no, triangle \(KJL\) is right - angled at \(J\)? Wait, the right angle is at \(J\) (the red square at \(J\) for triangle \(KJL\)), so triangle \(KJL\) is right - angled at \(J\). So in right - triangle \(KJL\), right - angled at \(J\), \(\sin x=\frac{JL}{KJ}\). If \(JL = 11\) and \(KJ = 22\), then \(\sin x=\frac{11}{22}=\frac{1}{2}\), so \(x = 30^\circ\)? Wait, no, \(\sin30^\circ=\frac{1}{2}\), but wait, in triangle \(JIL\), \(JL\) is the hypotenuse, \(JL=\sqrt{IJ^{2}+IL^{2}}=\sqrt{11^{2}+11^{2}} = 11\sqrt{2}\approx15.56\), and \(KJ = 22\). Then \(\sin x=\frac{JL}{KJ}=\frac{11\sqrt{2}}{22}=\frac{\sqrt{2}}{2}\), so \(x = 45^\circ\)? No, that's not. Wait, maybe I made a mistake.

Wait, let's start over. In triangle \(JIL\), right - angled at \(I\), \(\angle IJL = 45^\circ\), so \(IJ = IL = 11\) cm, so \(JL = 11\sqrt{2}\) cm. In triangle \(HJL\), right - angled at \(I\), \(\angle H = 30^\circ\), so \(JL=\frac{1}{2}HJ\) (since in a \(30 - 60 - 90\) triangle, the side opposite \(30^\circ\) is half the hypotenuse). So \(HJ = 2JL=22\sqrt{2}\) cm. Now, in triangle \(KJL\), \(KJ = 22\) cm, \(JL = 11\sqrt{2}\) cm? No, that doesn't match. Wait, maybe \(KJ = 22\) cm and \(JL = 11\) cm. Wait, the diagram shows \(KJ = 22\) cm (the length from \(K\) to \(J\)) and \(IJ = 11\) cm (from \(I\) to \(J\)). Since \(IJ\) and \(IL\) are equal (from the \(45 - 45 - 90\) triangle), \(IL = 11\) cm. And in triangle \(HJL\), \(JL\) is the side opposite \(30^\circ\), so \(JL=\frac{1}{2}HJ\). But \(HJ = HI + IJ\), and \(HI\) can be found from the \(30 - 60 - 90\) triangle: \(HI=\sqrt{HJ^{2}-JL^{2}}=\sqrt{(2JL)^{2}-JL^{2}}=\sqrt{3}JL\). But maybe we don't need that.

Wait, the key is that in triangle \(KJL\), right - angled at \(J\), if \(JL = 11\) cm and \(KJ = 22\) cm, then \(\sin x=\frac{JL}{KJ}=\frac{11}{22}=\frac{1}{2}\), so \(x = 30^\circ\)? Wait, no, \(\sin30^\circ=\frac{1}{2}\), but if \(JL\) is opposite angle \(x\) and \(KJ\) is the hypotenuse, then \(x = 30^\circ\). Wait, but let's check the angles. Alternatively, since in triangle \(HJL\), angle at \(H\) is \(30^\circ\), and in triangle \(JIL\), angle at \(J\) is \(45^\circ\), maybe triangle \(KJL\) has angle \(x = 30^\circ\)? Wait, the options are \(30^\circ\), \(45^\circ\), \(60^\circ\), \(90^\circ\). Wait, if we consider that \(JL\) is half of \(KJ\) (if \(KJ = 22\) and \(JL = 11\)), then in a right - triangle, the angle opposite the side that is half the hypotenuse is \(30^\circ\). So angle \(x = 30^\circ\).

Answer:

\(30^\circ\) (corresponding to the option with \(30^\circ\))