Question

geometry
using sine, cosine, and tangent
find the missing side. round to the nearest tenth.
1)
2)
3)
4)
5)
6)

Explanation:

Let's solve each problem step by step using sine, cosine, or tangent.

Problem 1:

We have a right triangle with hypotenuse \( 10 \) and angle \( 36^\circ \). The side \( x \) is opposite the \( 36^\circ \) angle.
Using the sine function: \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
So, \( \sin(36^\circ) = \frac{x}{10} \)
Multiply both sides by \( 10 \): \( x = 10 \sin(36^\circ) \)
Calculate \( \sin(36^\circ) \approx 0.5878 \)
\( x \approx 10 \times 0.5878 = 5.878 \approx 5.9 \)

Problem 2:

Right triangle with angle \( 59^\circ \), opposite side \( 17 \), and hypotenuse \( x \).
Using sine: \( \sin(59^\circ) = \frac{17}{x} \)
Solve for \( x \): \( x = \frac{17}{\sin(59^\circ)} \)
\( \sin(59^\circ) \approx 0.8572 \)
\( x \approx \frac{17}{0.8572} \approx 19.8 \)

Problem 3:

Right triangle with hypotenuse \( 19 \), angle \( 68^\circ \), and adjacent side \( x \).
Using cosine: \( \cos(68^\circ) = \frac{x}{19} \)
Solve for \( x \): \( x = 19 \cos(68^\circ) \)
\( \cos(68^\circ) \approx 0.3746 \)
\( x \approx 19 \times 0.3746 \approx 7.1174 \)? Wait, no—wait, the triangle: if it's a right triangle with right angle, angle \( 68^\circ \), hypotenuse \( 19 \), then adjacent to \( 68^\circ \) is \( x \), opposite is... Wait, maybe I misread. Wait, the triangle has right angle, angle \( 68^\circ \), hypotenuse \( 19 \), so adjacent to \( 68^\circ \) is \( x \)? Wait, no—wait, the diagram: maybe the hypotenuse is \( 19 \), angle \( 68^\circ \), and \( x \) is adjacent? Wait, no, let's check again. Wait, the user's work has \( 19 \cos(68^\circ) \approx 7.1 \)? No, maybe I made a mistake. Wait, the user's work says \( x \approx 17.6 \). Wait, maybe the angle is \( 68^\circ \), hypotenuse \( 19 \), and \( x \) is opposite? Wait, no—wait, let's recalculate. If angle is \( 68^\circ \), hypotenuse \( 19 \), then opposite side is \( 19 \sin(68^\circ) \approx 19 \times 0.9272 \approx 17.6 \). Ah, yes! So I mixed up adjacent and opposite. So correct: \( \sin(68^\circ) = \frac{x}{19} \), so \( x = 19 \sin(68^\circ) \approx 19 \times 0.9272 \approx 17.6 \) (rounded to nearest tenth: \( 17.6 \))

Problem 4:

Right triangle with hypotenuse \( 10 \), angle \( 66^\circ \), and adjacent side \( x \).
Using cosine: \( \cos(66^\circ) = \frac{x}{10} \)
Solve for \( x \): \( x = 10 \cos(66^\circ) \)
\( \cos(66^\circ) \approx 0.4067 \)
\( x \approx 10 \times 0.4067 = 4.067 \approx 4.1 \)

Problem 5:

Right triangle with adjacent side \( 17 \), angle \( 37^\circ \), and hypotenuse \( x \).
Using cosine: \( \cos(37^\circ) = \frac{17}{x} \)
Solve for \( x \): \( x = \frac{17}{\cos(37^\circ)} \)
\( \cos(37^\circ) \approx 0.7986 \)
\( x \approx \frac{17}{0.7986} \approx 21.3 \)

Problem 6:

Right triangle with adjacent side \( x \), opposite side \( 12 \), and angle \( 75^\circ \).
Using tangent: \( \tan(75^\circ) = \frac{12}{x} \)
Solve for \( x \): \( x = \frac{12}{\tan(75^\circ)} \)
\( \tan(75^\circ) \approx 3.732 \)
\( x \approx \frac{12}{3.732} \approx 3.2 \)

Final Answers:

1. \( \boldsymbol{5.9} \)
2. \( \boldsymbol{19.8} \)
3. \( \boldsymbol{17.6} \)
4. \( \boldsymbol{4.1} \)
5. \( \boldsymbol{21.3} \)
6. \( \boldsymbol{3.2} \)

Answer:

Let's solve each problem step by step using sine, cosine, or tangent.

Problem 1:

We have a right triangle with hypotenuse \( 10 \) and angle \( 36^\circ \). The side \( x \) is opposite the \( 36^\circ \) angle.
Using the sine function: \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)
So, \( \sin(36^\circ) = \frac{x}{10} \)
Multiply both sides by \( 10 \): \( x = 10 \sin(36^\circ) \)
Calculate \( \sin(36^\circ) \approx 0.5878 \)
\( x \approx 10 \times 0.5878 = 5.878 \approx 5.9 \)

Problem 2:

Right triangle with angle \( 59^\circ \), opposite side \( 17 \), and hypotenuse \( x \).
Using sine: \( \sin(59^\circ) = \frac{17}{x} \)
Solve for \( x \): \( x = \frac{17}{\sin(59^\circ)} \)
\( \sin(59^\circ) \approx 0.8572 \)
\( x \approx \frac{17}{0.8572} \approx 19.8 \)

Problem 3:

Right triangle with hypotenuse \( 19 \), angle \( 68^\circ \), and adjacent side \( x \).
Using cosine: \( \cos(68^\circ) = \frac{x}{19} \)
Solve for \( x \): \( x = 19 \cos(68^\circ) \)
\( \cos(68^\circ) \approx 0.3746 \)
\( x \approx 19 \times 0.3746 \approx 7.1174 \)? Wait, no—wait, the triangle: if it's a right triangle with right angle, angle \( 68^\circ \), hypotenuse \( 19 \), then adjacent to \( 68^\circ \) is \( x \), opposite is... Wait, maybe I misread. Wait, the triangle has right angle, angle \( 68^\circ \), hypotenuse \( 19 \), so adjacent to \( 68^\circ \) is \( x \)? Wait, no—wait, the diagram: maybe the hypotenuse is \( 19 \), angle \( 68^\circ \), and \( x \) is adjacent? Wait, no, let's check again. Wait, the user's work has \( 19 \cos(68^\circ) \approx 7.1 \)? No, maybe I made a mistake. Wait, the user's work says \( x \approx 17.6 \). Wait, maybe the angle is \( 68^\circ \), hypotenuse \( 19 \), and \( x \) is opposite? Wait, no—wait, let's recalculate. If angle is \( 68^\circ \), hypotenuse \( 19 \), then opposite side is \( 19 \sin(68^\circ) \approx 19 \times 0.9272 \approx 17.6 \). Ah, yes! So I mixed up adjacent and opposite. So correct: \( \sin(68^\circ) = \frac{x}{19} \), so \( x = 19 \sin(68^\circ) \approx 19 \times 0.9272 \approx 17.6 \) (rounded to nearest tenth: \( 17.6 \))

Problem 4:

Right triangle with hypotenuse \( 10 \), angle \( 66^\circ \), and adjacent side \( x \).
Using cosine: \( \cos(66^\circ) = \frac{x}{10} \)
Solve for \( x \): \( x = 10 \cos(66^\circ) \)
\( \cos(66^\circ) \approx 0.4067 \)
\( x \approx 10 \times 0.4067 = 4.067 \approx 4.1 \)

Problem 5:

Right triangle with adjacent side \( 17 \), angle \( 37^\circ \), and hypotenuse \( x \).
Using cosine: \( \cos(37^\circ) = \frac{17}{x} \)
Solve for \( x \): \( x = \frac{17}{\cos(37^\circ)} \)
\( \cos(37^\circ) \approx 0.7986 \)
\( x \approx \frac{17}{0.7986} \approx 21.3 \)

Problem 6:

Right triangle with adjacent side \( x \), opposite side \( 12 \), and angle \( 75^\circ \).
Using tangent: \( \tan(75^\circ) = \frac{12}{x} \)
Solve for \( x \): \( x = \frac{12}{\tan(75^\circ)} \)
\( \tan(75^\circ) \approx 3.732 \)
\( x \approx \frac{12}{3.732} \approx 3.2 \)

Final Answers:

1. \( \boldsymbol{5.9} \)
2. \( \boldsymbol{19.8} \)
3. \( \boldsymbol{17.6} \)
4. \( \boldsymbol{4.1} \)
5. \( \boldsymbol{21.3} \)
6. \( \boldsymbol{3.2} \)