Question

1. a plane takes off from an airport and climbs at a steady rate to a height of 3.4 kilometres over a ground distance of 8 kilometres.

at what angle to the ground does the plane take off?

remember to draw a diagram of this situation before trying to solve the problem.

Explanation:

Model the physical scenario

We model the plane's path using a right-angled triangle.
The ground distance is the adjacent side:

$$\text{Adjacent} = 8\text{ km}$$

The height is the opposite side:

$$\text{Opposite} = 3.4\text{ km}$$

Let \(\theta\) represent the takeoff angle to the ground.

Apply trigonometric ratios

Using Right Triangle Trigonometry, we relate the sides to the angle:

$$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$$

Substitute the given values:

$$\tan(\theta) = \frac{3.4}{8} = 0.425$$

Solve for the angle

Using Inverse Trigonometric Functions, we find \(\theta\):

$$\theta = \arctan(0.425)$$

Calculate the value in degrees:

$$\theta \approx 23.025^\circ$$

Rounding to one decimal place gives:

$$\theta \approx 23.0^\circ$$

Answer:

The plane takes off at an angle of approximately \(23.0^\circ\) to the ground.