Question

what are the correct trigonometric ratios that could be used to determine the length of ln? choose two correct answers. \\(\cos(70^\circ) = \frac{8}{ln}\\) \\(\sin(20^\circ) = \frac{ln}{8}\\) \\(\tan(70^\circ) = \frac{ln}{nq}\\) \\(\cos(70^\circ) = \frac{ln}{8}\\) \\(\sin(20^\circ) = \frac{8}{ln}\\)

Explanation:

To solve this, we analyze the right triangle \( \triangle LNO \) (assuming \( O \) is the right - angled vertex? Wait, no, the right angle is at \( N \), so \( \angle N = 90^{\circ} \), \( \angle L = 70^{\circ} \), then \( \angle O=20^{\circ} \) (since the sum of angles in a triangle is \( 180^{\circ} \), \( 180 - 90 - 70=20^{\circ} \)) and the hypotenuse is \( LO = 8 \) (assuming the side labeled \( 8 \) is the hypotenuse), \( LN \) is one of the legs.

Recall Trigonometric Ratios:

In a right - triangle, for an acute angle \( \theta \):

• \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \)
• \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \)
• \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \)

Analyze each option:

1. Option 1: \( \cos(70^{\circ})=\frac{8}{LN} \)

• For \( \angle L = 70^{\circ} \), the adjacent side to \( \angle L \) is \( LN \), and the hypotenuse is \( LO = 8 \). But \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \), so \( \cos(70^{\circ})=\frac{LN}{8} \), not \( \frac{8}{LN} \). So this option is incorrect.

1. Option 2: \( \sin(20^{\circ})=\frac{LN}{8} \)

• For \( \angle O = 20^{\circ} \), the opposite side to \( \angle O \) is \( LN \), and the hypotenuse is \( LO = 8 \). Since \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \), \( \sin(20^{\circ})=\frac{LN}{8} \). This option is correct.

1. Option 3: \( \tan(70^{\circ})=\frac{LN}{MN} \)

• We don't know what \( MN \) is. Also, for \( \angle L = 70^{\circ} \), \( \tan(70^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{ON}{LN} \) (assuming \( ON \) is the opposite side to \( \angle L \)). So this option is incorrect.

1. Option 4: \( \cos(70^{\circ})=\frac{LN}{8} \)

• For \( \angle L = 70^{\circ} \), the adjacent side is \( LN \) and the hypotenuse is \( LO = 8 \). Since \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \), \( \cos(70^{\circ})=\frac{LN}{8} \). This option is correct.

1. Option 5: \( \sin(20^{\circ})=\frac{8}{LN} \)

• For \( \angle O = 20^{\circ} \), \( \sin(20^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{LN}{8} \), not \( \frac{8}{LN} \). So this option is incorrect.

The two correct answers are:

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

Answer:

To solve this, we analyze the right triangle \( \triangle LNO \) (assuming \( O \) is the right - angled vertex? Wait, no, the right angle is at \( N \), so \( \angle N = 90^{\circ} \), \( \angle L = 70^{\circ} \), then \( \angle O=20^{\circ} \) (since the sum of angles in a triangle is \( 180^{\circ} \), \( 180 - 90 - 70=20^{\circ} \)) and the hypotenuse is \( LO = 8 \) (assuming the side labeled \( 8 \) is the hypotenuse), \( LN \) is one of the legs.

Recall Trigonometric Ratios:

In a right - triangle, for an acute angle \( \theta \):

• \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \)
• \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \)
• \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \)

Analyze each option:

1. Option 1: \( \cos(70^{\circ})=\frac{8}{LN} \)

• For \( \angle L = 70^{\circ} \), the adjacent side to \( \angle L \) is \( LN \), and the hypotenuse is \( LO = 8 \). But \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \), so \( \cos(70^{\circ})=\frac{LN}{8} \), not \( \frac{8}{LN} \). So this option is incorrect.

1. Option 2: \( \sin(20^{\circ})=\frac{LN}{8} \)

• For \( \angle O = 20^{\circ} \), the opposite side to \( \angle O \) is \( LN \), and the hypotenuse is \( LO = 8 \). Since \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \), \( \sin(20^{\circ})=\frac{LN}{8} \). This option is correct.

1. Option 3: \( \tan(70^{\circ})=\frac{LN}{MN} \)

• We don't know what \( MN \) is. Also, for \( \angle L = 70^{\circ} \), \( \tan(70^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{ON}{LN} \) (assuming \( ON \) is the opposite side to \( \angle L \)). So this option is incorrect.

1. Option 4: \( \cos(70^{\circ})=\frac{LN}{8} \)

• For \( \angle L = 70^{\circ} \), the adjacent side is \( LN \) and the hypotenuse is \( LO = 8 \). Since \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \), \( \cos(70^{\circ})=\frac{LN}{8} \). This option is correct.

1. Option 5: \( \sin(20^{\circ})=\frac{8}{LN} \)

• For \( \angle O = 20^{\circ} \), \( \sin(20^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{LN}{8} \), not \( \frac{8}{LN} \). So this option is incorrect.

The two correct answers are:

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