Question
A viewer is looking up at a 70°
angle to watch a plane fly towards an airport terminal. His distance from the plane is 2000m. Someone who is standing at the terminal can watch the plane by looking up at a 50°
angle. To solve this problem, will you use Law of Sines or Law of Cosines? How far is the viewer from the terminal? You will select two answers.
angle to watch a plane fly towards an airport terminal. His distance from the plane is 2000m. Someone who is standing at the terminal can watch the plane by looking up at a 50°
angle. To solve this problem, will you use Law of Sines or Law of Cosines? How far is the viewer from the terminal? You will select two answers.
Answers
GPT-4o mini
To solve this problem, we first need to visualize the scenario. We have two points of interest: the viewer and the terminal. The viewer is looking up at a plane at a 70° angle from the horizontal. The viewer is 2000 m away from the plane. Meanwhile, someone at the terminal is looking at the plane at a 50° angle.
From this information, we can form a triangle:
- One angle is 70° (the angle of elevation from the viewer to the plane).
- The other angle is 50° (the angle of elevation from the terminal to the plane).
- The side opposite the 70° angle is the distance from the viewer to the plane, which is 2000 m.
Now, we can calculate the angle at the terminal:
- The sum of angles in a triangle is 180°. So, the angle at the terminal, denoted as angle C, is:
\[
C = 180° - 70° - 50° = 60°
\]
Now, we can apply the **Law of Sines** to find the distance from the viewer to the terminal (let's denote it as \(a\)). In terms of the sides and angles, we have:
- \(a\) (the distance from the viewer to the terminal)
- \(b\) (the distance from the terminal to the plane which we can find using the Law of Sines)
- angle A = 50° (angle at the terminal)
- angle C = 60° (angle opposite the distance \(a\))
- opposite side to angle A is the distance from the viewer to the plane (2000 m).
Using the Law of Sines:
\[
\frac{a}{\sin(50°)} = \frac{2000}{\sin(60°)}
\]
Now we can solve for \(a\):
\[
a = \frac{2000 \cdot \sin(50°)}{\sin(60°)}
\]
Calculating the values:
- \(\sin(50°) \approx 0.7660\)
- \(\sin(60°) \approx 0.8660\)
Substituting into the equation:
\[
a \approx \frac{2000 \cdot 0.7660}{0.8660} \approx \frac{1532}{0.8660} \approx 1766.76 \text{ m}
\]
Thus, the distance from the viewer to the terminal is approximately 1766.76 m.
### Summary
- **Use:** Law of Sines
- **Distance from viewer to terminal:** Approximately 1766.76 m.
From this information, we can form a triangle:
- One angle is 70° (the angle of elevation from the viewer to the plane).
- The other angle is 50° (the angle of elevation from the terminal to the plane).
- The side opposite the 70° angle is the distance from the viewer to the plane, which is 2000 m.
Now, we can calculate the angle at the terminal:
- The sum of angles in a triangle is 180°. So, the angle at the terminal, denoted as angle C, is:
\[
C = 180° - 70° - 50° = 60°
\]
Now, we can apply the **Law of Sines** to find the distance from the viewer to the terminal (let's denote it as \(a\)). In terms of the sides and angles, we have:
- \(a\) (the distance from the viewer to the terminal)
- \(b\) (the distance from the terminal to the plane which we can find using the Law of Sines)
- angle A = 50° (angle at the terminal)
- angle C = 60° (angle opposite the distance \(a\))
- opposite side to angle A is the distance from the viewer to the plane (2000 m).
Using the Law of Sines:
\[
\frac{a}{\sin(50°)} = \frac{2000}{\sin(60°)}
\]
Now we can solve for \(a\):
\[
a = \frac{2000 \cdot \sin(50°)}{\sin(60°)}
\]
Calculating the values:
- \(\sin(50°) \approx 0.7660\)
- \(\sin(60°) \approx 0.8660\)
Substituting into the equation:
\[
a \approx \frac{2000 \cdot 0.7660}{0.8660} \approx \frac{1532}{0.8660} \approx 1766.76 \text{ m}
\]
Thus, the distance from the viewer to the terminal is approximately 1766.76 m.
### Summary
- **Use:** Law of Sines
- **Distance from viewer to terminal:** Approximately 1766.76 m.