Question

what are the correct trigonometric ratios that could be used to determine the length of ln? check all that apply. □ $\sin(20^\circ) = \frac{ln}{8}$ □ $\cos(70^\circ) = \frac{8}{ln}$ □ $\tan(70^\circ) = \frac{ln}{mn}$ □ $\sin(20^\circ) = \frac{8}{ln}$ □ $\cos(70^\circ) = \frac{ln}{8}$ (with a right triangle diagram labeled l, n, m, right angle at n, angle at l is 70°, angle at m is 20°, hypotenuse lm is 8)

Explanation:

Step1: Analyze the right triangle

In right triangle \( \triangle LNM \), \( \angle N = 90^\circ \), hypotenuse \( LM = 8 \), \( \angle L = 70^\circ \), \( \angle M = 20^\circ \). We need to find the trigonometric ratios for \( LN \).

Step2: Recall trigonometric ratios

- Sine of an angle: \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \)
- Cosine of an angle: \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \)
- Tangent of an angle: \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \)

Step3: Analyze each option

- Option 1: \( \sin(20^\circ)=\frac{LN}{8} \)
- For \( \angle M = 20^\circ \), the opposite side to \( 20^\circ \) is \( LN \), hypotenuse is \( LM = 8 \). So \( \sin(20^\circ)=\frac{LN}{8} \), this is correct.
- Option 2: \( \cos(70^\circ)=\frac{8}{LN} \)
- For \( \angle L = 70^\circ \), adjacent side to \( 70^\circ \) is \( LN \), hypotenuse is \( LM = 8 \). So \( \cos(70^\circ)=\frac{LN}{8} \), not \( \frac{8}{LN} \), this is incorrect.
- Option 3: \( \tan(70^\circ)=\frac{LN}{MN} \)
- For \( \angle L = 70^\circ \), opposite side to \( 70^\circ \) is \( MN \), adjacent side is \( LN \). So \( \tan(70^\circ)=\frac{MN}{LN} \), not \( \frac{LN}{MN} \), this is incorrect.
- Option 4: \( \sin(20^\circ)=\frac{8}{LN} \)
- As in option 1, \( \sin(20^\circ)=\frac{LN}{8} \), not \( \frac{8}{LN} \), this is incorrect.
- Option 5: \( \cos(70^\circ)=\frac{LN}{8} \)
- For \( \angle L = 70^\circ \), adjacent side is \( LN \), hypotenuse is \( 8 \). So \( \cos(70^\circ)=\frac{LN}{8} \), this is correct. Wait, earlier analysis for option 2 was wrong. Wait, let's re - check.
- Wait, \( \cos(70^\circ)=\frac{\text{adjacent to }70^\circ}{\text{hypotenuse}} \). Adjacent to \( 70^\circ \) (angle at L) is LN, hypotenuse is LM = 8. So \( \cos(70^\circ)=\frac{LN}{8} \), which can be rewritten as \( \cos(70^\circ)=\frac{LN}{8} \), or \( 8=\frac{LN}{\cos(70^\circ)} \), but the option 2 was \( \cos(70^\circ)=\frac{8}{LN} \), which is \( LN\cos(70^\circ)=8 \), \( LN = \frac{8}{\cos(70^\circ)} \), which is different from \( \cos(70^\circ)=\frac{LN}{8} \). Wait, I made a mistake earlier. Let's re - analyze option 5: \( \cos(70^\circ)=\frac{LN}{8} \)
- Yes, for \( \angle L = 70^\circ \), adjacent side (LN) over hypotenuse (8) is \( \cos(70^\circ) \), so \( \cos(70^\circ)=\frac{LN}{8} \), this is correct. And option 1: \( \sin(20^\circ)=\frac{LN}{8} \), since \( \sin(20^\circ)=\cos(70^\circ) \) (because \( \sin(\theta)=\cos(90^\circ - \theta) \), \( 90 - 20 = 70 \)), so both option 1 and option 5 are correct? Wait, let's re - check the triangle.
- Wait, \( \angle L = 70^\circ \), \( \angle M = 20^\circ \), \( \angle N = 90^\circ \). So:
- For \( \angle M = 20^\circ \):
- Opposite: \( LN \)
- Adjacent: \( MN \)
- Hypotenuse: \( LM = 8 \)
- So \( \sin(20^\circ)=\frac{LN}{8} \) (correct, option 1)
- \( \cos(20^\circ)=\frac{MN}{8} \)
- \( \tan(20^\circ)=\frac{LN}{MN} \)
- For \( \angle L = 70^\circ \):
- Opposite: \( MN \)
- Adjacent: \( LN \)
- Hypotenuse: \( LM = 8 \)
- So \( \sin(70^\circ)=\frac{MN}{8} \)
- \( \cos(70^\circ)=\frac{LN}{8} \) (correct, option 5)
- \( \tan(70^\circ)=\frac{MN}{LN} \)

So the correct options are \( \sin(20^\circ)=\frac{LN}{8} \) and \( \cos(70^\circ)=\frac{LN}{8} \) (the fifth option, since \( \cos(70^\circ)=\frac{LN}{8} \) is equivalent to the fifth option? Wait the fifth option is \( \cos(70^\circ)=\frac{LN}{8} \), yes. And the first option is \( \sin(20^\circ)=\frac{LN}{8} \). Wait the options are:

1. \( \sin(20^\circ)=\frac{LN}{8} \) - correct
2. \( \cos(70^\circ)=\frac{8}{LN} \) - incorrect
3. \( \tan(70^\circ)=\frac{LN}{MN} \) - incorrect
4. \( \sin(20^\circ)=\frac{8}{LN} \) - incorrect
5. \( \cos(70^\circ)=\frac{LN}{8} \) - correct (since \( \cos(70^\circ)=\sin(20^\circ) \))

So the correct options are the first one (\( \sin(20^\circ)=\frac{LN}{8} \)) and the fifth one (\( \cos(70^\circ)=\frac{LN}{8} \))? Wait the original options:

First option: \( \sin(20^\circ)=\frac{LN}{8} \) - correct

Fifth option: \( \cos(70^\circ)=\frac{LN}{8} \) - correct

Because \( \sin(20^\circ)=\cos(70^\circ) \), so both are correct.

Answer:

\( \sin(20^\circ)=\frac{LN}{8} \) (first option) and \( \cos(70^\circ)=\frac{LN}{8} \) (fifth option)

So the correct options are:

• \( \sin(20^\circ)=\frac{LN}{8} \)
• \( \cos(70^\circ)=\frac{LN}{8} \) (the fifth option: \( \cos(70^\circ)=\frac{LN}{8} \))