Question

date
quiz 8-2: trigonometry
directions: find the indicated trigonometric ratio as a fraction in simplest form.

1. sin l =
2. tan n =
3. cos l =
4. sin n =

directions: solve for x. round all answers to the nearest tenth.

1. x =
2. x =

Explanation:

Step1: Analyze the first triangle (for trigonometric ratios)

First, we need to find the hypotenuse of the right triangle \( \triangle JLN \). Given the legs are \( 6 \) and \( 8 \)? Wait, no, looking at the diagram, the legs are \( 6 \) (LN) and \( 8 \)? Wait, no, the base is \( 8 \)? Wait, the diagram shows \( JN = 10 \), \( LN = 6 \), so by Pythagoras, \( JL = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \). So the triangle has legs \( 6 \) and \( 8 \), hypotenuse \( 10 \).

Step2: Solve \( \sin L \)

\( \sin L = \frac{\text{opposite}}{\text{hypotenuse}} \). For \( \angle L \), the opposite side is \( JN = 8 \)? Wait, no, \( \angle L \) is at vertex \( L \), so the sides: adjacent to \( L \) is \( LN = 6 \), opposite is \( JN = 8 \), hypotenuse \( JL = 10 \)? Wait, no, \( JN \) is the base, length \( 10 \), \( LN \) is vertical leg \( 6 \), so \( JL \) is the other leg, \( 8 \). So \( \angle L \): opposite side is \( JN = 10 \)? No, wait, right triangle: right angle at \( L \)? Wait, the diagram has a right angle at \( L \)? Wait, the triangle is labeled with \( L \) as the right angle? Wait, the diagram shows \( \angle L \) with a right angle symbol? Wait, the original diagram: \( J \), \( L \), \( N \), with \( L \) having a right angle. So \( \triangle JLN \) is right-angled at \( L \). So \( JL \) and \( LN \) are legs, \( JN \) is hypotenuse. So \( JL = 8 \), \( LN = 6 \), \( JN = 10 \) (since \( 6^2 + 8^2 = 36 + 64 = 100 = 10^2 \)).

So \( \sin L \): in right triangle, \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \). For \( \angle L \), the angle at \( L \), so the sides: opposite side to \( L \) is \( JN \) (length \( 10 \)? No, \( JN \) is hypotenuse. Wait, no, right angle at \( L \), so the two legs are \( JL \) and \( LN \), hypotenuse \( JN \). So \( JL = 8 \), \( LN = 6 \), \( JN = 10 \). So \( \angle L \): the angle at \( L \), so the sides: adjacent to \( L \) is \( LN = 6 \), opposite is \( JL = 8 \), hypotenuse \( JN = 10 \). So \( \sin L = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{JL}{JN} = \frac{8}{10} = \frac{4}{5} \).

Step3: Solve \( \tan N \)

\( \tan N = \frac{\text{opposite}}{\text{adjacent}} \). For \( \angle N \), right angle at \( L \), so opposite side to \( N \) is \( JL = 8 \), adjacent is \( LN = 6 \). So \( \tan N = \frac{8}{6} = \frac{4}{3} \).

Step4: Solve \( \cos L \)

\( \cos L = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{LN}{JN} = \frac{6}{10} = \frac{3}{5} \).

Step5: Solve \( \sin N \)

\( \sin N = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{JL}{JN} = \frac{8}{10} = \frac{4}{5} \)? Wait, no, \( \angle N \): opposite side is \( JL = 8 \), hypotenuse \( JN = 10 \), so \( \sin N = \frac{8}{10} = \frac{4}{5} \)? Wait, no, \( \angle N \): adjacent is \( LN = 6 \), opposite is \( JL = 8 \), so \( \sin N = \frac{8}{10} = \frac{4}{5} \), \( \cos N = \frac{6}{10} = \frac{3}{5} \).

Step6: Solve problem 5 (triangle with \( 30^\circ \), hypotenuse \( 14 \), find \( x \) (adjacent?))

The triangle is right-angled, angle \( 30^\circ \), hypotenuse \( 14 \), \( x \) is the adjacent side to \( 30^\circ \). So \( \cos 30^\circ = \frac{x}{14} \), so \( x = 14 \cos 30^\circ = 14 \times \frac{\sqrt{3}}{2} = 7\sqrt{3} \approx 12.1 \). Wait, or is \( x \) the opposite? Wait, the angle is \( 30^\circ \), the side labeled \( 14 \) is the hypotenuse, and \( x \) is the side opposite? No, the diagram: right angle, angle \( 30^\circ \), side \( 14 \) is the hypotenuse, \( x \) is the side adjacent to \( 30^\circ \). So \( \cos 30^\circ = \frac{x}{14} \), so \( x = 14 \times \cos 30^\circ \approx 14 \times 0.866 \approx 12.1 \).

Step7: Solve problem 6 (triangle with \( 54^\circ \), leg \( 22 \), find \( x \) (opposite?))

Right-angled triangle, angle \( 54^\circ \), leg \( 22 \) is adjacent to \( 54^\circ \), \( x \) is opposite. So \( \tan 54^\circ = \frac{x}{22} \), so \( x = 22 \tan 54^\circ \approx 22 \times 1.376 \approx 30.3 \). Or if \( 22 \) is opposite, then \( \tan 54^\circ = \frac{22}{x} \), \( x = \frac{22}{\tan 54^\circ} \approx \frac{22}{1.376} \approx 16.0 \). Wait, the diagram: right angle, angle \( 54^\circ \), side \( 22 \) is one leg, \( x \) is the other leg. Let's see: angle \( 54^\circ \), so if \( 22 \) is adjacent to \( 54^\circ \), then \( x = 22 \tan 54^\circ \approx 22 \times 1.376 \approx 30.3 \). If \( 22 \) is opposite, \( x = 22 \cot 54^\circ \approx 16.0 \). Wait, the diagram: the side labeled \( 22 \) is adjacent to \( 54^\circ \), so \( x \) is opposite. So \( \tan 54^\circ = \frac{x}{22} \), \( x = 22 \times 1.3764 \approx 30.3 \).

Answer:

1. \( \sin L = \frac{4}{5} \)
2. \( \tan N = \frac{4}{3} \)
3. \( \cos L = \frac{3}{5} \)
4. \( \sin N = \frac{4}{5} \)
5. \( x \approx 12.1 \)
6. \( x \approx 30.3 \)