Question

match each missing side length and angle with the correct value. angle measurements are rounded to the nearest hundredth

Explanation:

Step 1: Find \(LY\)

In right triangle \(LYN\) (right - angled at \(Y\)), we use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = LN = 10\) (hypotenuse) and \(b = NY = 8\) (one leg). Let \(a = LY\). Then \(LY=\sqrt{LN^{2}-NY^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6\).

Step 2: Find \(NM\)

In right triangle \(MYN\) (right - angled at \(Y\)), \(NY = 8\) (one leg) and \(YM = 15\) (the other leg). Using the Pythagorean theorem, \(NM=\sqrt{NY^{2}+YM^{2}}=\sqrt{8^{2}+15^{2}}=\sqrt{64 + 225}=\sqrt{289}=17\).

Step 3: Find \(m\angle NLY\)

In right triangle \(LYN\), \(\sin(\angle NLY)=\frac{NY}{LN}=\frac{8}{10} = 0.8\). Then \(\angle NLY=\arcsin(0.8)\approx53.13^{\circ}\).

Step 4: Find \(m\angle NMY\)

In right triangle \(MYN\), \(\sin(\angle NMY)=\frac{NY}{NM}=\frac{8}{17}\approx0.4706\). Then \(\angle NMY=\arcsin(0.4706)\approx28.07^{\circ}\).

Answer:

To solve for the missing sides and angles in the right - triangle - related figure:

For \(LY\):

We know that \(LY = LY\) (wait, no, looking at the figure, \(LY\) is part of the segment \(LM\) where \(LM = 15\) and \(YM\) can be found? Wait, actually, in right triangle \(NYL\), \(NY = 8\), \(LN=10\). By the Pythagorean theorem, \(LY=\sqrt{LN^{2}-NY^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36} = 6\).

For \(NY\) (wait, no, \(NY = 8\) is given. Wait, for \(NM\)):

In right triangle \(NYM\), \(NY = 8\), \(YM=LM - LY=15 - 6 = 9\)? Wait, no, maybe I misread. Wait, \(LY = 6\) (from above), \(YM=15 - 6=9\)? No, wait, let's re - examine. The figure has two right triangles: \(\triangle LYN\) and \(\triangle MYN\), with \(NY\perp LM\), \(NY = 8\), \(LN = 10\), \(LM=15\).

For \(\triangle LYN\) (right - angled at \(Y\)):

• \(LY=\sqrt{LN^{2}-NY^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=6\)
• \(\sin(\angle NLY)=\frac{NY}{LN}=\frac{8}{10}=0.8\), so \(\angle NLY=\arcsin(0.8)\approx53.13^{\circ}\)
• \(\cos(\angle NLY)=\frac{LY}{LN}=\frac{6}{10}=0.6\), so \(\angle LNY = 90^{\circ}-\angle NLY\approx90 - 53.13 = 36.87^{\circ}\)

For \(\triangle MYN\) (right - angled at \(Y\)):

• \(YM=LM - LY=15 - 6 = 9\)? Wait, no, that can't be. Wait, maybe \(LM = LY+YM\), and we know \(NY = 8\), \(NM\) is the hypotenuse. Wait, \(YM=15 - 6 = 9\), then \(NM=\sqrt{NY^{2}+YM^{2}}=\sqrt{8^{2}+9^{2}}=\sqrt{64 + 81}=\sqrt{145}\approx12.04\), but that's not in the options. Wait, maybe \(LM = 15\) is a mistake, or maybe I misread the figure. Wait, the options have \(17\), \(6\), \(22\), \(12\). Wait, if \(NY = 8\), \(YM = 15\) (maybe \(LM\) is not \(15\) but \(YM = 15\))? Let's try that. If \(NY = 8\), \(YM = 15\), then \(NM=\sqrt{8^{2}+15^{2}}=\sqrt{64 + 225}=\sqrt{289}=17\). Ah, that makes sense. So \(NM = 17\).

For angles:

• In \(\triangle MYN\), \(\sin(\angle NMY)=\frac{NY}{NM}=\frac{8}{17}\approx0.4706\), so \(\angle NMY=\arcsin(0.4706)\approx28.07^{\circ}\)
• \(\angle NMY\approx28.07^{\circ}\), \(\angle NLY\approx53.13^{\circ}\), \(\angle LNY\approx36.87^{\circ}\)

Matching:

• \(LY\): \(6\) (from \(\triangle LYN\) Pythagorean theorem)
• \(NM\): \(17\) (from \(\triangle MYN\) Pythagorean theorem)
• \(m\angle NLY\): \(53.13^{\circ}\) (from \(\sin\) ratio in \(\triangle LYN\))
• \(m\angle NMY\): \(28.07^{\circ}\) (from \(\sin\) ratio in \(\triangle MYN\))