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directions: solve for x. round to the nearest tenth.
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To solve for \( x \) in each right triangle, we use trigonometric ratios (sine, cosine, tangent) based on the given angle and sides. Let's solve each problem one by one:
Problem 1: Right triangle with angle \( 40^\circ \), hypotenuse \( 37 \), and \( x \) as the opposite side.
Trig Ratio: \( \sin(40^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{37} \)
\( x = 37 \cdot \sin(40^\circ) \approx 37 \cdot 0.6428 \approx 23.8 \)
Problem 2: Right triangle with angle \( 67^\circ \), hypotenuse \( 29 \), and \( x \) as the adjacent side.
Trig Ratio: \( \cos(67^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{29} \)
\( x = 29 \cdot \cos(67^\circ) \approx 29 \cdot 0.3907 \approx 11.3 \)
Problem 3: Right triangle with angle \( 70^\circ \), opposite side \( 12 \), and \( x \) as the hypotenuse.
Trig Ratio: \( \sin(70^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{x} \)
\( x = \frac{12}{\sin(70^\circ)} \approx \frac{12}{0.9397} \approx 12.8 \)
Problem 4: Right triangle with angle \( 37^\circ \), hypotenuse \( 16 + x \)? Wait, no—wait, the diagram shows a right triangle with angle \( 37^\circ \), one leg \( 16 \), and hypotenuse \( x \)? Wait, no, let’s re-examine. Wait, the diagram has a right angle, angle \( 37^\circ \), and one leg \( 16 \), hypotenuse \( x \)? Wait, no—maybe the angle is \( 37^\circ \), adjacent side \( 16 \), hypotenuse \( x \)? Wait, no, the right angle is at the bottom, so:
Trig Ratio: \( \cos(37^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} \)? Wait, no—wait, the leg adjacent to \( 37^\circ \) is \( 16 \), and the hypotenuse is \( x \)? Wait, no, maybe the angle is \( 37^\circ \), opposite side \( 16 \), hypotenuse \( x \)? Wait, the diagram is a right triangle with angle \( 37^\circ \), one leg \( 16 \), and hypotenuse \( x \). Let’s assume:
\( \sin(37^\circ) = \frac{16}{x} \) → \( x = \frac{16}{\sin(37^\circ)} \approx \frac{16}{0.6018} \approx 26.6 \)? Wait, no—wait, maybe the angle is \( 37^\circ \), adjacent side \( 16 \), hypotenuse \( x \): \( \cos(37^\circ) = \frac{16}{x} \) → \( x = \frac{16}{\cos(37^\circ)} \approx \frac{16}{0.7986} \approx 20.0 \)? Wait, the original diagram is a bit unclear, but let’s check the next problem.
Problem 5: Right triangle with angle \( 58^\circ \), adjacent side \( 22 \), and \( x \) as the opposite side.
Trig Ratio: \( \tan(58^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{22} \)
\( x = 22 \cdot \tan(58^\circ) \approx 22 \cdot 1.6003 \approx 35.2 \)
Problem 6: Right triangle with angle \( 51^\circ \), adjacent side \( 15 \), and \( x \) as the opposite side.
Trig Ratio: \( \tan(51^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{15} \)
\( x = 15 \cdot \tan(51^\circ) \approx 15 \cdot 1.2349 \approx 18.5 \)
Problem 7: Right triangle with angle \( 57^\circ \), opposite side \( 45 \), and \( x \) as the hypotenuse.
Trig Ratio: \( \sin(57^\circ) = \frac{45}{x} \) → \( x = \frac{45}{\sin(57^\circ)} \approx \frac{45}{0.8387} \approx 53.7 \)
Problem 8: Right triangle with angle \( 24^\circ \), hypotenuse \( 9 \), and \( x \) as the opposite side.
Trig Ratio: \( \sin(24^\circ) = \frac{x}{9} \) → \( x = 9 \cdot \sin(24^\circ) \approx 9 \cdot 0.4067 \approx 3.66 \approx 3.7 \)
Problem 9: Right triangle with hypotenuse \( 42 \), angle \( 24^\circ \), and \( x \) as the opposite side.
Trig Ratio: \( \sin(24^\circ) = \frac{x}{42} \) → \( x = 42 \cdot \sin(24^\circ) \approx 42 \cdot 0.4067 \approx 17.1 \)
Problem 10: Right triangle with adjace…
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To solve for \( x \) in each right triangle, we use trigonometric ratios (sine, cosine, tangent) based on the given angle and sides. Let's solve each problem one by one:
Problem 1: Right triangle with angle \( 40^\circ \), hypotenuse \( 37 \), and \( x \) as the opposite side.
Trig Ratio: \( \sin(40^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{37} \)
\( x = 37 \cdot \sin(40^\circ) \approx 37 \cdot 0.6428 \approx 23.8 \)
Problem 2: Right triangle with angle \( 67^\circ \), hypotenuse \( 29 \), and \( x \) as the adjacent side.
Trig Ratio: \( \cos(67^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{29} \)
\( x = 29 \cdot \cos(67^\circ) \approx 29 \cdot 0.3907 \approx 11.3 \)
Problem 3: Right triangle with angle \( 70^\circ \), opposite side \( 12 \), and \( x \) as the hypotenuse.
Trig Ratio: \( \sin(70^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{x} \)
\( x = \frac{12}{\sin(70^\circ)} \approx \frac{12}{0.9397} \approx 12.8 \)
Problem 4: Right triangle with angle \( 37^\circ \), hypotenuse \( 16 + x \)? Wait, no—wait, the diagram shows a right triangle with angle \( 37^\circ \), one leg \( 16 \), and hypotenuse \( x \)? Wait, no, let’s re-examine. Wait, the diagram has a right angle, angle \( 37^\circ \), and one leg \( 16 \), hypotenuse \( x \)? Wait, no—maybe the angle is \( 37^\circ \), adjacent side \( 16 \), hypotenuse \( x \)? Wait, no, the right angle is at the bottom, so:
Trig Ratio: \( \cos(37^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} \)? Wait, no—wait, the leg adjacent to \( 37^\circ \) is \( 16 \), and the hypotenuse is \( x \)? Wait, no, maybe the angle is \( 37^\circ \), opposite side \( 16 \), hypotenuse \( x \)? Wait, the diagram is a right triangle with angle \( 37^\circ \), one leg \( 16 \), and hypotenuse \( x \). Let’s assume:
\( \sin(37^\circ) = \frac{16}{x} \) → \( x = \frac{16}{\sin(37^\circ)} \approx \frac{16}{0.6018} \approx 26.6 \)? Wait, no—wait, maybe the angle is \( 37^\circ \), adjacent side \( 16 \), hypotenuse \( x \): \( \cos(37^\circ) = \frac{16}{x} \) → \( x = \frac{16}{\cos(37^\circ)} \approx \frac{16}{0.7986} \approx 20.0 \)? Wait, the original diagram is a bit unclear, but let’s check the next problem.
Problem 5: Right triangle with angle \( 58^\circ \), adjacent side \( 22 \), and \( x \) as the opposite side.
Trig Ratio: \( \tan(58^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{22} \)
\( x = 22 \cdot \tan(58^\circ) \approx 22 \cdot 1.6003 \approx 35.2 \)
Problem 6: Right triangle with angle \( 51^\circ \), adjacent side \( 15 \), and \( x \) as the opposite side.
Trig Ratio: \( \tan(51^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{15} \)
\( x = 15 \cdot \tan(51^\circ) \approx 15 \cdot 1.2349 \approx 18.5 \)
Problem 7: Right triangle with angle \( 57^\circ \), opposite side \( 45 \), and \( x \) as the hypotenuse.
Trig Ratio: \( \sin(57^\circ) = \frac{45}{x} \) → \( x = \frac{45}{\sin(57^\circ)} \approx \frac{45}{0.8387} \approx 53.7 \)
Problem 8: Right triangle with angle \( 24^\circ \), hypotenuse \( 9 \), and \( x \) as the opposite side.
Trig Ratio: \( \sin(24^\circ) = \frac{x}{9} \) → \( x = 9 \cdot \sin(24^\circ) \approx 9 \cdot 0.4067 \approx 3.66 \approx 3.7 \)
Problem 9: Right triangle with hypotenuse \( 42 \), angle \( 24^\circ \), and \( x \) as the opposite side.
Trig Ratio: \( \sin(24^\circ) = \frac{x}{42} \) → \( x = 42 \cdot \sin(24^\circ) \approx 42 \cdot 0.4067 \approx 17.1 \)
Problem 10: Right triangle with adjacent side \( 13 \), angle \( 37^\circ \), and \( x \) as the hypotenuse.
Trig Ratio: \( \cos(37^\circ) = \frac{13}{x} \) → \( x = \frac{13}{\cos(37^\circ)} \approx \frac{13}{0.7986} \approx 16.3 \)
Final Answers (Rounded to the Nearest Tenth):
- \( \boldsymbol{23.8} \)
- \( \boldsymbol{11.3} \)
- \( \boldsymbol{12.8} \)
- \( \boldsymbol{20.0} \) (assuming \( \cos(37^\circ) = \frac{16}{x} \))
- \( \boldsymbol{35.2} \)
- \( \boldsymbol{18.5} \)
- \( \boldsymbol{53.7} \)
- \( \boldsymbol{3.7} \)
- \( \boldsymbol{17.1} \)
- \( \boldsymbol{16.3} \)
(Note: Some diagrams may have ambiguous labeling; adjust trig ratios based on actual side-angle relationships. If a diagram is unclear, re-verify the angle and side positions.)