Question
Solve for the missing side. Round to the nearest tenth. (1 point) Responses 45.7 45.7 34.6 34.6 14.5 14.5 11.5. Angle 20, Length X 60 degrees, Width unkown
Answers
GPT-4o mini
To solve for the missing side in a right triangle where we have one angle (20 degrees), one side length (length X), and we also have another angle (60 degrees), we can use trigonometric functions.
Given:
- Angle \( A = 20^\circ \)
- Angle \( B = 60^\circ \)
- The third angle \( C = 90^\circ \)
- The side opposite to angle \( A \) is length \( X \).
Now, the side opposite angle \( B \) can be calculated using the sine function.
- \( \sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} \)
Here, if we denote:
- \( X \) as the length of the side opposite to angle \( A \) (20 degrees)
- The side opposite to angle \( B \) (60 degrees) as "unknown"
Using the triangle properties, if we know that:
- \( \sin(20^\circ) = \frac{X}{\text{hypotenuse}} \)
Let's denote the hypotenuse as \( h \), we have:
- \( h = \frac{X}{\sin(20^\circ)} \)
Now, to find the width (the opposite side to 60 degrees):
- \( \sin(60^\circ) = \frac{\text{width}}{h} \)
Substituting for \( h \):
- \( \sin(60^\circ) = \frac{\text{width}}{\frac{X}{\sin(20^\circ)}} \)
- Rearranging gives:
\[ \text{width} = h \cdot \sin(60^\circ) = \left(\frac{X}{\sin(20^\circ)}\right) \cdot \sin(60^\circ) \]
Finally, we can evaluate this by plugging in the values for \( \sin(20^\circ) \approx 0.3420 \) and \( \sin(60^\circ) \approx 0.8660 \):
\[ \text{width} = \frac{X \cdot \sin(60^\circ)}{\sin(20^\circ)} \approx \frac{X \cdot 0.8660}{0.3420} \approx 2.53X \]
Therefore, the width in terms of \( X \) is \( 2.53X \).
If we assume a value for \( X \) we would plug in \( X \) into \( 2.53X \) to find the width.
Since I don't have the value for \( X \) initially, please provide the value to finalize the calculation. If you are looking for lengths with given options, you may have \( X \) or final numbers to provide.
Given:
- Angle \( A = 20^\circ \)
- Angle \( B = 60^\circ \)
- The third angle \( C = 90^\circ \)
- The side opposite to angle \( A \) is length \( X \).
Now, the side opposite angle \( B \) can be calculated using the sine function.
- \( \sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} \)
Here, if we denote:
- \( X \) as the length of the side opposite to angle \( A \) (20 degrees)
- The side opposite to angle \( B \) (60 degrees) as "unknown"
Using the triangle properties, if we know that:
- \( \sin(20^\circ) = \frac{X}{\text{hypotenuse}} \)
Let's denote the hypotenuse as \( h \), we have:
- \( h = \frac{X}{\sin(20^\circ)} \)
Now, to find the width (the opposite side to 60 degrees):
- \( \sin(60^\circ) = \frac{\text{width}}{h} \)
Substituting for \( h \):
- \( \sin(60^\circ) = \frac{\text{width}}{\frac{X}{\sin(20^\circ)}} \)
- Rearranging gives:
\[ \text{width} = h \cdot \sin(60^\circ) = \left(\frac{X}{\sin(20^\circ)}\right) \cdot \sin(60^\circ) \]
Finally, we can evaluate this by plugging in the values for \( \sin(20^\circ) \approx 0.3420 \) and \( \sin(60^\circ) \approx 0.8660 \):
\[ \text{width} = \frac{X \cdot \sin(60^\circ)}{\sin(20^\circ)} \approx \frac{X \cdot 0.8660}{0.3420} \approx 2.53X \]
Therefore, the width in terms of \( X \) is \( 2.53X \).
If we assume a value for \( X \) we would plug in \( X \) into \( 2.53X \) to find the width.
Since I don't have the value for \( X \) initially, please provide the value to finalize the calculation. If you are looking for lengths with given options, you may have \( X \) or final numbers to provide.