To solve this problem, we need to find \(\sin L\), \(\cos L\), \(\tan L\), \(\sin M\), \(\cos M\), and \(\tan M\) given the values \(\ell=12\), \(m=12\sqrt{3}\), and \(n=24\). We'll express each trigonometric ratio both as a fraction and as a decimal to the nearest hundredth.
### Calculating \(\sin L\), \(\cos L\), and \(\tan L\)
1. \(\sin L\):
- \(\sin L = \frac{\ell}{n} = \frac{12}{24} = \frac{1}{2}\)
- As a decimal: \(\frac{1}{2} = 0.50\)
2. \(\cos L\):
- \(\cos L = \frac{m}{n} = \frac{12\sqrt{3}}{24} = \frac{\sqrt{3}}{2}\)
- As a decimal: \(\frac{\sqrt{3}}{2} \approx 0.87\)
3. \(\tan L\):
- \(\tan L = \frac{\sin L}{\cos L} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\)
- As a decimal: \(\frac{\sqrt{3}}{3} \approx 0.58\)
### Calculating \(\sin M\), \(\cos M\), and \(\tan M\)
1. \(\sin M\):
- \(\sin M = \frac{m}{n} = \frac{12\sqrt{3}}{24} = \frac{\sqrt{3}}{2}\)
- As a decimal: \(\frac{\sqrt{3}}{2} \approx 0.87\)
2. \(\cos M\):
- \(\cos M = \frac{\ell}{n} = \frac{12}{24} = \frac{1}{2}\)
- As a decimal: \(\frac{1}{2} = 0.50\)
3. \(\tan M\):
- \(\tan M = \frac{\sin M}{\cos M} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}\)
- As a decimal: \(\sqrt{3} \approx 1.73\)
### Summary of Results:
- \(\sin L = \frac{1}{2} \approx 0.50 \)
- \(\cos L = \frac{\sqrt{3}}{2} \approx 0.87 \)
- \(\tan L = \frac{\sqrt{3}}{3} \approx 0.58 \)
- \(\sin M = \frac{\sqrt{3}}{2} \approx 0.87 \)
- \(\cos M = \frac{1}{2} \approx 0.50 \)
- \(\tan M = \sqrt{3} \approx 1.73 \)
Based on the above calculations, the correct answer is:
c [tex]\(\sin L=\frac{1}{2} \approx 0.50 ; \cos L=\frac{\sqrt{3}}{2} \approx 0.87 ; \tan L=\frac{1}{\sqrt{3}}\)[/tex] or [tex]\(\frac{\sqrt{3}}{3} \approx 0.58 ; \sin M=\frac{\sqrt{3}}{2} \approx 0.87 ; \cos M=\frac{1}{2} \approx 0.50 ; tan M=\sqrt{3} \approx 1.73\)[/tex]
