Question

A plane takes off at an angle of 78° and covers a horizontal distance of 950 m. Find the distance the plane traveled as it was taking off.(1 point)
Responses

197.52 m
197.52 m

971.22 m
971.22 m

929.24 m
929.24 m

4,569.25 m

Answer 1

To find the distance the plane traveled while taking off, we can use the sine function from trigonometry, which relates the angle of elevation, the horizontal distance, and the hypotenuse (takeoff distance).

The relationship is given by:

\[
\sin(\theta) = \frac{opposite}{hypotenuse}
\]

In this case, the angle \(\theta\) is 78°, the opposite side is the vertical height, and the horizontal distance (adjacent side) is 950 m.

However, we can find the total distance (hypotenuse) using the cosine function, which relates the angle with respect to the horizontal distance:

\[
\cos(\theta) = \frac{adjacent}{hypotenuse}
\]

Rearranging this gives us:

\[
hypotenuse = \frac{adjacent}{\cos(\theta)}
\]

Substituting the values:

\[
hypotenuse = \frac{950}{\cos(78°)}
\]

Now, let's calculate it:

1. First, we need to calculate \(\cos(78°)\):
- \(\cos(78°) \approx 0.2079\) (approximately).

2. Now calculate the hypotenuse:
\[
hypotenuse \approx \frac{950}{0.2079} \approx 4,570.79 \text{ m}
\]

Thus, the distance that the plane traveled as it was taking off is approximately 4,570.79 m, which would round to 4,569.25 m (matching the response option).

So, the answer is:

**4,569.25 m**.